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\begin_body

\begin_layout Title
Teorija Analize 1 —
 IŠRM 2023/24
\end_layout

\begin_layout Author

\noun on
Anton Luka Šijanec
\end_layout

\begin_layout Date
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
today
\end_layout

\end_inset


\end_layout

\begin_layout Abstract
Povzeto po zapiskih s predavanj profesorja Oliverja Dragičevića.
\end_layout

\begin_layout Standard
\begin_inset CommandInset toc
LatexCommand tableofcontents

\end_inset


\end_layout

\begin_layout Section
Števila
\end_layout

\begin_layout Definition*
Množica je matematični objekt,
 ki predstavlja skupino elementov.
 Če element 
\begin_inset Formula $a$
\end_inset

 pripada množici 
\begin_inset Formula $A$
\end_inset

,
 pišemo 
\begin_inset Formula $a\in A$
\end_inset

,
 sicer pa 
\begin_inset Formula $a\not\in A$
\end_inset

.
 Množica 
\begin_inset Formula $B$
\end_inset

 je podmnožica množice 
\begin_inset Formula $A$
\end_inset

,
 pišemo 
\begin_inset Formula $B\subset A$
\end_inset

,
 če 
\begin_inset Formula $\forall b\in B:b\in A$
\end_inset

.
 Presek 
\begin_inset Formula $B$
\end_inset

 in 
\begin_inset Formula $C$
\end_inset

 označimo 
\begin_inset Formula $B\cap C\coloneqq\left\{ x;x\in B\wedge x\in C\right\} $
\end_inset

.
 Unijo 
\begin_inset Formula $B$
\end_inset

 in 
\begin_inset Formula $C$
\end_inset

 označimo 
\begin_inset Formula $B\cup C\coloneqq\left\{ x;x\in B\vee x\in C\right\} $
\end_inset

.
 Razliko/komplement 
\begin_inset Quotes gld
\end_inset


\begin_inset Formula $B$
\end_inset

 manj/brez 
\begin_inset Formula $C$
\end_inset


\begin_inset Quotes grd
\end_inset

 označimo 
\begin_inset Formula $B\setminus C\coloneqq\left\{ x;x\in B\wedge x\not\in C\right\} $
\end_inset

.
\end_layout

\begin_layout Subsection
Realna števila
\end_layout

\begin_layout Standard
Množico realnih števil označimo 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
 V njej obstajata binarni operaciji seštevanje 
\begin_inset Formula $a+b$
\end_inset

 in množenje 
\begin_inset Formula $a\cdot b$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Lastnosti seštevanja
\end_layout

\begin_layout Axiom
Komutativnost:
 
\begin_inset Formula $\forall a,b\in\mathbb{R}:a+b=b+a$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Asociativnost:
 
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a+\left(b+c\right)=\left(a+b\right)+c$
\end_inset

,
 torej je 
\begin_inset Formula $a+\cdots+z$
\end_inset

 dobro definiran izraz.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Obstoj enote:
 
\begin_inset Formula $\exists0\in\mathbb{R}\forall a\in\mathbb{R}:a+0=a$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Obstoj inverzov:
 
\begin_inset Formula $\forall a\in\mathbb{R}\exists b\in\mathbb{R}\ni a+b=0$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Claim*
Inverz je enoličen.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $a,b,c\in\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $a+b=0$
\end_inset

 in 
\begin_inset Formula $a+c=0$
\end_inset

.
 Tedaj 
\begin_inset Formula $b=b+0=b+a+c=0+c=c$
\end_inset

.
\end_layout

\begin_layout Corollary*
Inverz je funkcija in aditivni inverz 
\begin_inset Formula $a$
\end_inset

 označimo z 
\begin_inset Formula $-a$
\end_inset

.
 Pri zapisu 
\begin_inset Formula $a+\left(-b\right)$
\end_inset

 običajno 
\begin_inset Formula $+$
\end_inset

 izpustimo in pišemo 
\begin_inset Formula $a-b$
\end_inset

,
 čemur pravimo odštevanje 
\begin_inset Formula $b$
\end_inset

 od 
\begin_inset Formula $a$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $\forall a\in\mathbb{R}:a=-\left(-a\right)$
\end_inset

.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $b=-a$
\end_inset

 in 
\begin_inset Formula $c=-b=-\left(-a\right)$
\end_inset

.
 Tedaj velja 
\begin_inset Formula $c-a=c+b=-\left(-a\right)-a=0$
\end_inset

 in 
\begin_inset Formula $a=0+a=c-a+a=c+\left(-a\right)+a=c+0=c=-\left(-a\right)$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $-\left(b+c\right)=-b-c$
\end_inset


\end_layout

\begin_layout Proof
Velja 
\begin_inset Formula $b+c+\left(-b-c\right)=b+c+\left(\left(-b\right)+\left(-c\right)\right)=b+\left(-b\right)+c+\left(-c\right)=0$
\end_inset

,
 torej je 
\begin_inset Formula $b+c$
\end_inset

 inverz od 
\begin_inset Formula $\left(-b-c\right)$
\end_inset

,
 torej je 
\begin_inset Formula $-\left(b+c\right)=-b-c$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsubsection
Lastnosti množenja
\end_layout

\begin_layout Axiom
Komutativnost:
 
\begin_inset Formula $\forall a,b\in\mathbb{R}:ab=ba$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Asociativnost:
 
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:a\left(bc\right)=\left(ab\right)c$
\end_inset

,
 torej je 
\begin_inset Formula $a\cdots z$
\end_inset

 dobro definiran izraz.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Obstoj enote:
 
\begin_inset Formula $\exists1\in\mathbb{R}\forall a\in\mathbb{R}:a1=a$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Obstoj inverzov:
 
\begin_inset Formula $\forall a\in\mathbb{R}\setminus\left\{ 0\right\} \exists b\in\mathbb{R}\setminus\left\{ 0\right\} \ni:ab=1$
\end_inset


\end_layout

\begin_deeper
\begin_layout Claim*
Inverz je enoličen.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $a,b,c\in\mathbb{R}\setminus\left\{ 0\right\} $
\end_inset

 in 
\begin_inset Formula $ab=1$
\end_inset

 in 
\begin_inset Formula $ac=1$
\end_inset

.
 Tedaj 
\begin_inset Formula $b=b1=bac=1c=c$
\end_inset

.
\end_layout

\begin_layout Corollary*
Inverz je funkcija in multiplikativni inverz 
\begin_inset Formula $a$
\end_inset

 označimo z 
\begin_inset Formula $a^{-1}$
\end_inset

.
 Pri zapisu 
\begin_inset Formula $a\cdot b^{-1}$
\end_inset

 lahko 
\begin_inset Formula $\cdot$
\end_inset

 izpustimo in pišemo 
\begin_inset Formula $a/b$
\end_inset

,
 čemur pravimo deljenje 
\begin_inset Formula $a$
\end_inset

 z 
\begin_inset Formula $b$
\end_inset

 za neničeln 
\begin_inset Formula $b$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsubsection
Skupne lastnosti v 
\begin_inset Formula $\mathbb{R}$
\end_inset


\end_layout

\begin_layout Axiom
\begin_inset Formula $1\not=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Axiom
Distributivnost:
 
\begin_inset Formula $\forall a,b,c\in\mathbb{R}:\left(a+b\right)c=ac+bc$
\end_inset


\end_layout

\begin_layout Paragraph
Urejenost 
\begin_inset Formula $\mathbb{R}$
\end_inset


\end_layout

\begin_layout Standard
Realna števila delimo na pozitivna 
\begin_inset Formula $\mathbb{R}_{+}\coloneqq\left\{ x\in\mathbb{R};x>0\right\} $
\end_inset

,
 negativna 
\begin_inset Formula $\mathbb{R}_{-}\coloneqq\left\{ x\in\mathbb{R};x<0\right\} $
\end_inset

 in ničlo 
\begin_inset Formula $0$
\end_inset

.
 Če je 
\begin_inset Formula $x\in\mathbb{\mathbb{R}}_{+}\cup\left\{ 0\right\} $
\end_inset

,
 pišemo 
\begin_inset Formula $x\geq0$
\end_inset

,
 če je 
\begin_inset Formula $x\in\mathbb{R}_{-}\cup\left\{ 0\right\} $
\end_inset

,
 pišemo 
\begin_inset Formula $x\leq0$
\end_inset

.
\end_layout

\begin_layout Axiom
Če je 
\begin_inset Formula $a\not=0$
\end_inset

,
 je natanko eno izmed 
\begin_inset Formula $\left\{ a,-a\right\} $
\end_inset

 pozitivno,
 imenujemo ga absolutna vrednost 
\begin_inset Formula $a$
\end_inset

 (pišemo 
\begin_inset Formula $\left|a\right|$
\end_inset

),
 in natanko eno negativno,
 pišemo 
\begin_inset Formula $-\left|a\right|$
\end_inset

.
\end_layout

\begin_layout Definition*
\begin_inset Formula $\left|0\right|=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Za 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

 se 
\begin_inset Formula $\left|a-b\right|$
\end_inset

 imenuje razdalja.
\end_layout

\begin_layout Axiom
\begin_inset Formula $\forall a,b\in\mathbb{R}:a,b>0\Rightarrow\left(a+b>0\right)\wedge\left(ab>0\right)$
\end_inset

.
\end_layout

\begin_layout Definition*
Za 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

:
 
\begin_inset Formula $a$
\end_inset

 je večje od 
\begin_inset Formula $b$
\end_inset

,
 oznaka 
\begin_inset Formula $a>b\Leftrightarrow a-b>0$
\end_inset

.
 
\begin_inset Formula $a$
\end_inset

 je manjše od 
\begin_inset Formula $b$
\end_inset

,
 oznaka 
\begin_inset Formula $a<b\Leftrightarrow a-b<0$
\end_inset

.
 Podobno 
\begin_inset Formula $\leq$
\end_inset

 in 
\begin_inset Formula $\geq$
\end_inset

.
\end_layout

\begin_layout Claim*
Trikotniška neenakost.
 
\begin_inset Formula $\forall a,b\in\mathbb{R}$
\end_inset

:
 
\begin_inset Formula $\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|\leq\left|a\right|+\left|b\right|$
\end_inset

.
\end_layout

\begin_layout Proof
Dokažimo desni neenačaj.
 Vemo 
\begin_inset Formula $ab\leq\left|ab\right|$
\end_inset

 in 
\begin_inset Formula $\left|a\right|=\sqrt{a^{2}}$
\end_inset

.
 Naj bo 
\begin_inset Formula $a^{2}+2ab+b^{2}\leq\left|a\right|^{2}+2\left|a\right|\left|b\right|+\left|b\right|^{2}$
\end_inset

,
 torej 
\begin_inset Formula $\left(a+b\right)^{2}\leq\left(\left|a\right|+\left|b\right|\right)^{2}$
\end_inset

,
 korenimo:
 
\begin_inset Formula $\left|a+b\right|\leq\left|a\right|+\left|b\right|$
\end_inset

.
\end_layout

\begin_layout Subsubsection
Intervali
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $a<b$
\end_inset

.
 Označimo odprti interval 
\begin_inset Formula $\left(a,b\right)\coloneqq\left\{ x\in\mathbb{R};a<x<b\right\} $
\end_inset

,
 zaprti 
\begin_inset Formula $\left[a,b\right]\coloneqq\left\{ x\in\mathbb{R};a\leq x\leq b\right\} $
\end_inset

,
 polodprti 
\begin_inset Formula $(a,b]\coloneqq\left\{ x\in\mathbb{R};a<x\leq b\right\} $
\end_inset

 in podobno 
\begin_inset Formula $[a,b)$
\end_inset

.
 
\begin_inset Formula $\left(a,\infty\right)\coloneqq\left\{ x\in\mathbb{R};x>a\right\} $
\end_inset

 in podobno 
\begin_inset Formula $[a,\infty)$
\end_inset

.
\end_layout

\begin_layout Subsection
Temeljne številske podmnožice
\end_layout

\begin_layout Subsubsection
Naravna števila 
\begin_inset Formula $\mathbb{N}$
\end_inset


\end_layout

\begin_layout Definition*
\begin_inset Formula $\mathbb{N}\coloneqq\left\{ 1,1+1,1+1+1,1+1+1+1,\dots\right\} $
\end_inset


\end_layout

\begin_layout Paragraph
Matematična indukcija
\end_layout

\begin_layout Standard
Če je 
\begin_inset Formula $A\subseteq\mathbb{N}$
\end_inset

 in velja 
\begin_inset Formula $1\in A$
\end_inset

 (baza) in 
\begin_inset Formula $a\in A\Rightarrow a+1\in A$
\end_inset

 (korak),
 tedaj 
\begin_inset Formula $A=\mathbb{N}$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset

.
\end_layout

\begin_layout Proof
\begin_inset Formula $A\coloneqq\left\{ n\in\mathbb{N};\text{velja trditev za }n\right\} $
\end_inset

.
 Dokažimo 
\begin_inset Formula $A=\mathbb{N}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Baza:
 
\begin_inset Formula $1=\frac{1\cdot2}{2}=1$
\end_inset

.
\end_layout

\begin_layout Itemize
Korak:
 Predpostavimo 
\begin_inset Formula $1+2+3+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset

.
 Prištejmo 
\begin_inset Formula $n+1$
\end_inset

:
 
\begin_inset Formula 
\[
1+2+3+\cdots+n+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\left(n+1\right)=\frac{n\left(n+1\right)}{2}+\frac{2\left(n+1\right)}{2}=\frac{\left(n+2\right)\left(n+1\right)}{2}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Subsubsection
Cela števila 
\begin_inset Formula $\mathbb{Z}$
\end_inset


\end_layout

\begin_layout Standard
Množica 
\begin_inset Formula $\mathbb{N}$
\end_inset

 je zaprta za seštevanje in množenje,
 torej 
\begin_inset Formula $\forall a,b\in\mathbb{N}:a+b\in\mathbb{N}\wedge ab\in\mathbb{N}$
\end_inset

,
 ni pa zaprta za odštevanje,
 ker recimo 
\begin_inset Formula $5-3\not\in\mathbb{N}$
\end_inset

.
 Zapremo jo za odštevanje in dobimo množico 
\begin_inset Formula $\mathbb{Z}$
\end_inset

.
\end_layout

\begin_layout Definition*
\begin_inset Formula $\mathbb{Z}\coloneqq\left\{ a-b;b,a\in\mathbb{N}\right\} $
\end_inset


\end_layout

\begin_layout Subsubsection
Racionalna števila 
\begin_inset Formula $\mathbb{Q}$
\end_inset


\end_layout

\begin_layout Standard
Najmanjša podmnožica 
\begin_inset Formula $\mathbb{R}$
\end_inset

,
 ki vsebuje 
\begin_inset Formula $\mathbb{Z}$
\end_inset

 in je zaprta za deljenje,
 je 
\begin_inset Formula $\mathbb{Q}$
\end_inset

.
\end_layout

\begin_layout Definition*
\begin_inset Formula $\mathbb{Q}\coloneqq\left\{ a/b;a\in\mathbb{Z},b\in\mathbb{Z}\setminus\left\{ 0\right\} \right\} $
\end_inset

.
\end_layout

\begin_layout Standard
Velja 
\begin_inset Formula $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Claim*
Za 
\begin_inset Formula $a\in\mathbb{Q}$
\end_inset

,
 
\begin_inset Formula $b\not\in\mathbb{Q}$
\end_inset

 velja 
\begin_inset Formula $a+b\not\in\mathbb{Q}$
\end_inset

 in 
\begin_inset Formula $a\not=0\Rightarrow ab\not\in\mathbb{Q}$
\end_inset

.
\end_layout

\begin_layout Proof
PDDRAA 
\begin_inset Formula $a+b\in\mathbb{Q}$
\end_inset

.
 Tedaj 
\begin_inset Formula $a+b-a\in\mathbb{Q}$
\end_inset

,
 tedaj 
\begin_inset Formula $b\in\mathbb{Q}$
\end_inset

,
 kar je 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

.
 PDDRAA 
\begin_inset Formula $ab\in\mathbb{Q}$
\end_inset

.
 Tedaj 
\begin_inset Formula $\frac{ab}{a}\in\mathbb{Q}$
\end_inset

,
 tedaj 
\begin_inset Formula $b\in\mathbb{Q}$
\end_inset

,
 kar je 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

.
\end_layout

\begin_layout Subsection
\begin_inset CommandInset label
LatexCommand label
name "subsec:Omejenost-množic"

\end_inset

Omejenost množic
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset

.
 
\begin_inset Formula $A$
\end_inset

 je navzgor omejena 
\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\leq m$
\end_inset

.
 Takemu 
\begin_inset Formula $m$
\end_inset

 pravimo zgornja meja.
 Najmanjši zgornji meji 
\begin_inset Formula $A$
\end_inset

 pravimo supremum ali natančna zgornja meja množice 
\begin_inset Formula $A$
\end_inset

,
 označimo 
\begin_inset Formula $\sup A$
\end_inset

.
 Če je zgornja meja 
\begin_inset Formula $A$
\end_inset

 (
\begin_inset Formula $m$
\end_inset

) element 
\begin_inset Formula $A$
\end_inset

,
 je maksimum množice 
\begin_inset Formula $A$
\end_inset

,
 označimo 
\begin_inset Formula $m=\max A$
\end_inset

.
 Če množica ni navzgor omejena,
 pišemo 
\begin_inset Formula $\sup A=\infty$
\end_inset

.
\end_layout

\begin_layout Standard
Če 
\begin_inset Formula $s=\sup A\in\mathbb{R}$
\end_inset

,
 mora veljati 
\begin_inset Formula $\forall a\in A:a\leq s$
\end_inset

 in 
\begin_inset Formula $\forall\varepsilon>0\exists b\in A\ni:b>s-\varepsilon$
\end_inset

,
 torej za vsak neničeln 
\begin_inset Formula $\varepsilon$
\end_inset

 
\begin_inset Formula $s-\varepsilon$
\end_inset

 ni več natančna zgornja meja za 
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset

.
 
\begin_inset Formula $A$
\end_inset

 je navzdol omejena 
\begin_inset Formula $\Leftrightarrow\exists m\in\mathbb{R}\forall a\in A:a\geq m$
\end_inset

.
 Takemu 
\begin_inset Formula $m$
\end_inset

 pravimo spodnja meja.
 Največji spodnji meji 
\begin_inset Formula $A$
\end_inset

 pravimo infimum ali natančna spodnja meja množice 
\begin_inset Formula $A$
\end_inset

,
 označimo 
\begin_inset Formula $\inf A$
\end_inset

.
 Če je spodnja meja 
\begin_inset Formula $A$
\end_inset

 (
\begin_inset Formula $m$
\end_inset

) element 
\begin_inset Formula $A$
\end_inset

,
 je minimum množice 
\begin_inset Formula $A$
\end_inset

,
 označimo 
\begin_inset Formula $m=\min A$
\end_inset

.
 Če množica ni navzdol omejena,
 pišemo 
\begin_inset Formula $\inf A=-\infty$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $A\subset\mathbb{R}$
\end_inset

 je omejena,
 če je hkrati navzgor in navzdol omejena.
\end_layout

\begin_layout Axiom
\begin_inset CommandInset label
LatexCommand label
name "axm:Dedekind.-Vsaka-navzgor"

\end_inset

Dedekind.
 Vsaka navzgor omejena množica v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 ima natančno zgornjo mejo v 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Remark*
Za 
\begin_inset Formula $\mathbb{Q}$
\end_inset

 aksiom 
\begin_inset CommandInset ref
LatexCommand ref
reference "axm:Dedekind.-Vsaka-navzgor"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 ne velja.
 Če 
\begin_inset Formula $B\subset\mathbb{Q}$
\end_inset

,
 se lahko zgodi,
 da 
\begin_inset Formula $\sup B\not\in\mathbb{Q}$
\end_inset

.
 Primer:
 
\begin_inset Formula $B\coloneqq\left\{ q\in\mathbb{Q};q^{2}\leq2\right\} $
\end_inset

.
 
\begin_inset Formula $\sup B=\sqrt{2}\not\in\mathbb{Q}$
\end_inset

.
\end_layout

\begin_layout Example*

\end_layout

\begin_layout Subsection
Decimalni zapis
\end_layout

\begin_layout Definition*
\begin_inset Formula $\forall x\in\mathbb{R}^{+}\exists!m\in\mathbb{N}\cup\left\{ 0\right\} ,d_{1},d_{2},\dots\in\left\{ 0..9\right\} $
\end_inset

,
 ki število natančno določajo.
 Pišemo 
\begin_inset Formula $x=m,d_{1}d_{2}\dots$
\end_inset

.
 Natančno določitev mislimo v smislu:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $m\leq x<m+1$
\end_inset

 —
 s tem se izognemo dvojnemu zapisu 
\begin_inset Formula $1=0,\overline{9}$
\end_inset

 in 
\begin_inset Formula $1=1,\overline{0}$
\end_inset

.
\end_layout

\begin_layout Itemize
\begin_inset Formula $[m,m+1)$
\end_inset

 razdelimo na 10 enako dolgih polodprtih intervalov 
\begin_inset Formula $I_{0},\dots,I_{9}$
\end_inset

.
 
\begin_inset Formula $x$
\end_inset

 leži na natanko enem izmed njih,
 indeks njega je 
\begin_inset Formula $d_{1}$
\end_inset

.
 Nadaljujemo tako,
 da 
\begin_inset Formula $I_{d_{1}}$
\end_inset

 razdelimo zopet na 10 delov itd.
\end_layout

\end_deeper
\begin_layout Definition*
Števila 
\begin_inset Formula $x\in\mathbb{R^{-}}$
\end_inset

 pišemo tako,
 da zapišemo decimalni zapis števila 
\begin_inset Formula $-x$
\end_inset

 in predenj zapišemo 
\begin_inset Formula $-$
\end_inset

.
\end_layout

\begin_layout Definition*
Če se decimalke v zaporedju 
\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 ponavljajo,
 uporabimo periodični zapis,
 denimo 
\begin_inset Formula $5,01\overline{763}\in\mathbb{Q}$
\end_inset

.
\end_layout

\begin_layout Subsection
Kompleksna števila
\end_layout

\begin_layout Definition*
Vpeljimo število 
\begin_inset Formula $i$
\end_inset

 z lastnostjo 
\begin_inset Formula $i^{2}=-1$
\end_inset

,
 da je 
\begin_inset Formula $i$
\end_inset

 rešitev enačbe 
\begin_inset Formula $x^{2}+1=0$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $i\not\in\mathbb{R}$
\end_inset


\end_layout

\begin_layout Proof
Sicer bi veljajo 
\begin_inset Formula $i^{2}\geq0$
\end_inset

,
 kar po definiciji ne velja.
\end_layout

\begin_layout Definition*
Kompleksna števila so 
\begin_inset Formula $\mathbb{C}\coloneqq\left\{ a+bi;a,b\in\mathbb{R}\right\} $
\end_inset

.
 
\begin_inset Formula $bi$
\end_inset

 je še nedefinirano,
 zato za kompleksna števila definirano seštevanje in množenje za 
\begin_inset Formula $z=a+bi$
\end_inset

 in 
\begin_inset Formula $w=c+di$
\end_inset

:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $z+w\coloneqq\left(a+c\right)+\left(b+d\right)i$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $zw\coloneqq\left(a+bi\right)\left(c+di\right)=ac+adi+bic+bidi=ac+adi+bic-bd=\left(ac-bd\right)+\left(ad+bc\right)i$
\end_inset


\end_layout

\end_deeper
\begin_layout Definition*
Definiramo še konjugirano vrednost 
\begin_inset Formula $z\in\mathbb{C}$
\end_inset

:
 
\begin_inset Formula $\overline{z}\coloneqq a-bi$
\end_inset

 in označimo 
\begin_inset Formula $\left|z\right|^{2}=z\overline{z}$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $z\overline{z}=a^{2}+b^{2}\geq0$
\end_inset

 za 
\begin_inset Formula $z=a+bi$
\end_inset

.
\end_layout

\begin_layout Proof
\begin_inset Formula $\left(a+bi\right)\left(a-bi\right)=a^{2}+abi-bia-bibi=a^{2}+b^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
Velja,
 da je 
\begin_inset Formula $\mathbb{R}\subset\mathbb{C}$
\end_inset

 v smislu identifikacije 
\begin_inset Formula $\mathbb{R}$
\end_inset

 z množico 
\begin_inset Formula $\mathbb{C}$
\end_inset

:
 
\begin_inset Formula $\mathbb{R}=\left\{ a+0i;a\in\mathbb{R}\right\} $
\end_inset

,
 torej smo 
\begin_inset Formula $\mathbb{R}$
\end_inset

 razširili v 
\begin_inset Formula $\mathbb{C}$
\end_inset

,
 kjer ima vsak polinom vedno rešitev.
\end_layout

\begin_layout Subsubsection
Deljenje v 
\begin_inset Formula $\mathbb{C}$
\end_inset


\end_layout

\begin_layout Standard
Za 
\begin_inset Formula $w,z\in\mathbb{C},w\not=0$
\end_inset

 iščemo 
\begin_inset Formula $x\in\mathbb{C}\ni:wx=z$
\end_inset

.
 Ločimo dva primera:
\end_layout

\begin_layout Itemize
\begin_inset Formula $w\in\mathbb{R}\setminus\left\{ 0\right\} $
\end_inset

:
 definiramo 
\begin_inset Formula $x=\frac{z}{w}\coloneqq\frac{a}{w}+\frac{b}{w}i$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $w\in\mathbb{C}\setminus\left\{ 0\right\} $
\end_inset

 (splošno):
 
\begin_inset Formula $wx=z\overset{/\cdot\overline{w}}{\Longrightarrow}w\overline{w}x=z\overline{w}\Rightarrow\left|w\right|^{2}x=z\overline{w}\Rightarrow x=\frac{z\overline{w}}{\left|w\right|^{2}}$
\end_inset

,
 z 
\begin_inset Formula $\left|w\right|^{2}$
\end_inset

 pa znamo deliti,
 ker je realen.
\end_layout

\begin_layout Subsubsection
Lastnosti v 
\begin_inset Formula $\mathbb{C}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $+$
\end_inset

 in 
\begin_inset Formula $\cdot$
\end_inset

 sta komutativni,
 asociativni,
 distributivni,
 
\begin_inset Formula $0$
\end_inset

 je aditivna enota,
 
\begin_inset Formula $1$
\end_inset

 je multiplikativna.
\end_layout

\begin_layout Definition*
Za 
\begin_inset Formula $z=a+bi$
\end_inset

 vpeljemo 
\begin_inset Formula $\Re z=a$
\end_inset

 in 
\begin_inset Formula $\Im z=b$
\end_inset

.
\end_layout

\begin_layout Remark*
Opazimo 
\begin_inset Formula $\Re z=\frac{z+\overline{z}}{2}$
\end_inset

,
 
\begin_inset Formula $\Im z=\frac{z-\overline{z}}{2}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Remark*
\begin_inset Formula $\mathbb{C}$
\end_inset

 si lahko predstavljamo kot urejene pare;
 
\begin_inset Formula $a+bi$
\end_inset

 ustreza paru 
\begin_inset Formula $\left(a,b\right)$
\end_inset

.
 Tako 
\begin_inset Formula $\mathbb{C}$
\end_inset

 enačimo/identificiramo z 
\begin_inset Formula $\mathbb{R}^{2}\coloneqq\left\{ \left(a,b\right);a,b\in\mathbb{R}\right\} $
\end_inset

,
 s čimer dobimo geometrično predstavitev 
\begin_inset Formula $\mathbb{C}$
\end_inset

 kot vektorje v 
\begin_inset Formula $\mathbb{R}^{2}$
\end_inset

.
\end_layout

\begin_layout Definition*
Za 
\begin_inset Formula $z=a+bi$
\end_inset

,
 predstavljen z vektorjem s komponentami 
\begin_inset Formula $\left(a,b\right)$
\end_inset

,
 velja 
\begin_inset Formula $a=\left|z\right|\cos\varphi$
\end_inset

 in 
\begin_inset Formula $v=\left|z\right|\sin\varphi$
\end_inset

.
 Kotu 
\begin_inset Formula $\varphi$
\end_inset

 pravimo argument kompleksnega števila 
\begin_inset Formula $z$
\end_inset

,
 oznaka 
\begin_inset Formula $\arg z$
\end_inset

.
\end_layout

\begin_layout Corollary*
\begin_inset Formula $z=$
\end_inset


\begin_inset Formula $\left|z\right|\left(\cos\varphi+i\sin\varphi\right)$
\end_inset

.
 Velja
\begin_inset Foot
status open

\begin_layout Plain Layout
TODO DOPISATI ZAKAJ (v bistvu še jaz ne vem).
 ne razumem.
\end_layout

\end_inset

 
\begin_inset Formula $\left(\cos\varphi+i\sin\varphi\right)\left(\cos\psi+i\sin\psi\right)=\cos\left(\varphi+\psi\right)+i\sin\left(\varphi+\psi\right)$
\end_inset

,
 zato lahko pišemo 
\begin_inset Formula $e^{i\varphi}=\cos\varphi+i\sin\varphi$
\end_inset

.
 Množenje kompleksnh števil 
\begin_inset Formula $z=\left|z\right|e^{i\varphi}$
\end_inset

 in 
\begin_inset Formula $w=\left|w\right|e^{i\psi}$
\end_inset

 vrne število 
\begin_inset Formula $zw$
\end_inset

,
 za katero velja:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\left|zw\right|=\left|z\right|\left|w\right|$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $\arg zw=\arg z+\arg w$
\end_inset

 (do periode 
\begin_inset Formula $2\pi$
\end_inset

 natančno)
\end_layout

\end_deeper
\begin_layout Section
Zaporedja
\end_layout

\begin_layout Definition*
Funkcija 
\begin_inset Formula $a:\mathbb{N}\to\mathbb{R}$
\end_inset

 se imenuje realno zaporedje,
 oznaka 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 
\begin_inset Formula $a_{n}$
\end_inset

 je funkcijska vrednost pri 
\begin_inset Formula $n$
\end_inset

.
\end_layout

\begin_layout Example*
\begin_inset Formula $a_{n}=n$
\end_inset

:
 
\begin_inset Formula $1,2,3,\dots$
\end_inset

;
 
\begin_inset Formula $a_{n}=\left(-1\right)^{n}n^{2}$
\end_inset

:
 
\begin_inset Formula $-1,4,-9,16,-25,\dots$
\end_inset

;
 
\begin_inset Formula $a_{n}=\cos\left(\frac{\pi}{2}n\right)=0,-1,0,1,0,-1,\dots$
\end_inset


\end_layout

\begin_layout Standard
Zaporedje lahko podamo rekurzivno.
 Podamo prvi člen ali nekaj prvih členov in pravilo,
 kako iz prejšnjih členov dobiti naslednje.
\end_layout

\begin_layout Example*
\begin_inset Formula $a_{1}=0,a_{n+1}=a_{n}+n$
\end_inset

 da zaporedje 
\begin_inset Formula $a_{n}=\frac{n\left(n+1\right)}{2}$
\end_inset

.
 
\begin_inset Formula $a_{0}=0,a_{n+1}=\sqrt{b+a_{n}}$
\end_inset

 da zaporedje 
\begin_inset Formula $0,\sqrt{b},\sqrt{b+\sqrt{b}},\sqrt{b+\sqrt{b+\sqrt{b}}},\dots$
\end_inset

.
 Fibbonacijevo zaporedje:
 
\begin_inset Formula $a_{1}=a_{2}=1,a_{n+1}=a_{n}+a_{n-1}$
\end_inset

 da zaporedje 
\begin_inset Formula $1,1,2,3,5,8,\dots$
\end_inset


\end_layout

\begin_layout Subsection
Posebni tipi zaporedij
\end_layout

\begin_layout Definition*
Zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 je aritmetično,
 če 
\begin_inset Formula $\exists k\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}-a_{n}=k$
\end_inset

.
 Tedaj 
\begin_inset Formula $a_{n+1}=a_{n}+k=a_{1}+nd$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 je geometrično,
 če 
\begin_inset Formula $\exists\lambda\in\mathbb{R}\forall n\in\mathbb{N}:a_{n+1}=a_{n}\lambda$
\end_inset

.
 Tedaj 
\begin_inset Formula $a_{n}=\lambda^{n-1}a_{1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 je navzdol oz.
 navzgor omejeno,
 če je množica vseh členov tega taporedja navzgor oz.
 navzdol omejena (glej 
\begin_inset CommandInset ref
LatexCommand ref
reference "subsec:Omejenost-množic"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

).
 Podobno z množico členov definiramo supremum,
 infimum,
 maksimum in infimum zaporedja.
\end_layout

\begin_layout Definition*
Zaporedje je naraščajoče,
 če 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\geq a_{n}$
\end_inset

,
 padajoče,
 če 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}\leq a_{n}$
\end_inset

,
 strogo naraščajoče,
 če 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}>a_{n}$
\end_inset

,
 strogo padajoče podobno,
 monotono,
 če je naraščajoče ali padajoče in strogo monotono,
 če je strogo naraščajoče ali strogo padajoče.
\end_layout

\begin_layout Subsection
Limita zaporedja
\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $U\subseteq\mathbb{R}$
\end_inset

 je odprta,
 če 
\begin_inset Formula $\forall u\in U\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $U\subseteq\mathbb{R}$
\end_inset

 je zaprta,
 če je 
\begin_inset Formula $U^{\mathcal{C}}\coloneqq\mathbb{R}\setminus U$
\end_inset

 odprta.
\end_layout

\begin_layout Claim*
Odprt interval je odprta množica.
\end_layout

\begin_layout Proof
Za poljubna 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

,
 
\begin_inset Formula $b>a$
\end_inset

,
 naj bo 
\begin_inset Formula $u\in\left(a,b\right)$
\end_inset

 poljuben.
 Ustrezen 
\begin_inset Formula $r$
\end_inset

 je 
\begin_inset Formula $\min\left\{ \left|r-a\right|,\left|r-b\right|\right\} $
\end_inset

,
 da je 
\begin_inset Formula $\left(u-r,u+r\right)\subseteq U$
\end_inset

.
\end_layout

\begin_layout Claim*
Zaprt interval je zaprt.
\end_layout

\begin_layout Proof
Naj bosta 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

 poljubna in 
\begin_inset Formula $b>a$
\end_inset

.
 Dokazujemo,
 da je 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 zaprt,
 torej da je 
\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}=\left(-\infty,a\right)\cup\left(b,\infty\right)$
\end_inset

 odprta množica.
 Za poljuben 
\begin_inset Formula $u\in\left[a,b\right]^{\mathcal{C}}$
\end_inset

 velja,
 da je bodisi 
\begin_inset Formula $\in\left(-\infty,a\right)$
\end_inset

 bodisi 
\begin_inset Formula $\left(b,\infty\right)$
\end_inset

,
 kajti 
\begin_inset Formula $\left(-\infty,a\right)\cap\left(b,\infty\right)=\emptyset$
\end_inset

.
 Po prejšnji trditvi v obeh primerih velja 
\begin_inset Formula $\exists r>0\ni:\left(u-r,u+r\right)\subseteq U$
\end_inset

,
 torej je 
\begin_inset Formula $\left[a,b\right]^{\mathcal{C}}$
\end_inset

 res odprta,
 torej je 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 res zaprta.
\end_layout

\begin_layout Definition*
Množica 
\begin_inset Formula $B$
\end_inset

 je okolica točke 
\begin_inset Formula $t\in\mathbb{R}$
\end_inset

,
 če vsebuje kakšno odprto množico 
\begin_inset Formula $U$
\end_inset

,
 ki vsebuje 
\begin_inset Formula $t$
\end_inset

,
 torej 
\begin_inset Formula $t\in U^{\text{odp.}}\subseteq B\subseteq\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Definition*
\begin_inset Formula $L\in\mathbb{R}$
\end_inset

 je limita zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{R}}$
\end_inset

 
\begin_inset Formula $\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon$
\end_inset

.
 ZDB 
\begin_inset Formula $\forall V$
\end_inset

 okolica 
\begin_inset Formula $L\in\mathbb{R}\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in V$
\end_inset

,
 pravimo,
 da 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 konvergira k 
\begin_inset Formula $L$
\end_inset

 in pišemo 
\begin_inset Formula $L\coloneqq\lim_{n\to\infty}a_{n}$
\end_inset

 ali drugače 
\begin_inset Formula $a_{n}\underset{n\to\infty}{\longrightarrow}L$
\end_inset

.
 Če zaporedje ima limito,
 pravimo,
 da je konvergentno,
 sicer je divergentno.
\end_layout

\begin_layout Claim*
Konvergentno zaporedje v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 ima natanko eno limito.
\end_layout

\begin_layout Proof
Naj bosta 
\begin_inset Formula $J$
\end_inset

 in 
\begin_inset Formula $L$
\end_inset

 limiti zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\to\infty}$
\end_inset

.
 Torej
\begin_inset Foot
status open

\begin_layout Plain Layout
Ko trdimo,
 da obstaja 
\begin_inset Formula $n_{0}$
\end_inset

,
 še ne vemo,
 ali sta za 
\begin_inset Formula $L$
\end_inset

 in 
\begin_inset Formula $J$
\end_inset

 ta 
\begin_inset Formula $n_{0}$
\end_inset

 ista.
 Ampak trditev še vedno velja,
 ker lahko vzamemo večjega izmed njiju,
 ako bi bila drugačna.
\end_layout

\end_inset

 po definiciji 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{n}-L\right|<\varepsilon\wedge\left|a_{n}-J\right|<\varepsilon$
\end_inset

.
 Velja torej 
\begin_inset Formula $\forall\varepsilon>0:\left|J-L\right|<\varepsilon$
\end_inset

.
 PDDRAA 
\begin_inset Formula $J\not=L$
\end_inset

.
 Tedaj 
\begin_inset Formula $\left|J-L\right|\not=0$
\end_inset

,
 naj bo 
\begin_inset Formula $\left|J-L\right|=k$
\end_inset

.
 Tedaj 
\begin_inset Formula $\exists\varepsilon>0:\left|J-L\right|\not<\varepsilon$
\end_inset

,
 ustrezen 
\begin_inset Formula $\varepsilon$
\end_inset

 je na primer 
\begin_inset Formula $\frac{\left|J-L\right|}{2}$
\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset CommandInset label
LatexCommand label
name "Konvergentno-zaporedje-v-R-je-omejeno"

\end_inset

Konvergentno zaporedje v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 je omejeno.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
\end_inset

.
 Znotraj intervala 
\begin_inset Formula $\left(L-1,L+1\right)$
\end_inset

 so vsi členi zaporedja razen končno mnogo (
\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
\end_inset

).
 
\begin_inset Formula $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$
\end_inset

 je unija dveh omejenih množic;
 
\begin_inset Formula $\left(L-1,L+1\right)$
\end_inset

 in 
\begin_inset Formula $\left\{ a_{1},\dots,a_{n_{0}}\right\} $
\end_inset

,
 zato je tudi sama omejena.
\end_layout

\begin_layout Theorem*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hypertarget{pmkdlim}{Naj bosta}
\end_layout

\end_inset

 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 in 
\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 konvergentni zaporedji v 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
 Tedaj so tudi 
\begin_inset Formula $\left(a_{n}*b_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 konvergentna in velja 
\begin_inset Formula $\lim_{n\to\infty}a_{n}*b_{n}=\lim_{n\to\infty}a_{n}*\lim_{n\to\infty}b_{n}$
\end_inset

 za 
\begin_inset Formula $*\in\left\{ +,-,\cdot\right\} $
\end_inset

.
 Če je 
\begin_inset Formula $\lim_{n\to\infty}b_{n}\not=0$
\end_inset

,
 isto velja tudi za deljenje.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $a_{n}\to A$
\end_inset

 in 
\begin_inset Formula $b_{n}\to B$
\end_inset

 oziroma 
\begin_inset Formula $\forall\varepsilon>0\exists n_{1},n_{2}\ni:\left(n>n_{1}\Rightarrow\left|a_{n}-A\right|<\varepsilon\right)\wedge\left(n>n_{2}\Rightarrow\left|b_{n}-B\right|<\varepsilon\right)$
\end_inset

,
 torej 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}=\max\left\{ n_{1},n_{2}\right\} \ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|<\varepsilon\wedge\left|b_{n}-B\right|<\varepsilon$
\end_inset

.
 Dokažimo za vse operacije:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $+$
\end_inset

 Po predpostavki velja 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\ni:n>n_{0}\Rightarrow\left|a_{n}-A\right|+\left|a_{n}-B\right|<2\varepsilon$
\end_inset

.
 Oglejmo si sedaj
\begin_inset Formula 
\[
\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|=\left|\left(a_{n}-A\right)+\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
\]

\end_inset

in uporabimo še prejšnjo trditev,
 torej 
\begin_inset Formula $\forall2\varepsilon\exists n_{0}\ni:\left|\left(a_{n}+b_{n}\right)-\left(A+B\right)\right|\leq2\varepsilon$
\end_inset

,
 s čimer dokažemo 
\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $-$
\end_inset

 Oglejmo si
\begin_inset Formula 
\[
\left|\left(a_{n}-b_{n}\right)-\left(A-B\right)\right|=\left|a_{n}-b_{n}-A+B\right|=\left|\left(a_{n}-A\right)+\left(-\left(b_{n}-B\right)\right)\right|\leq\left|a_{n}-A\right|+\left|b_{n}-B\right|
\]

\end_inset

in nato kot zgoraj.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\cdot$
\end_inset

 Oglejmo si 
\begin_inset Formula 
\[
\left|a_{n}b_{n}-AB\right|=\left|a_{n}b_{n}-Ab_{n}+Ab_{n}-AB\right|=\left|\left(a_{n}-A\right)b_{n}+A\left(b_{n}-B\right)\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|.
\]

\end_inset

Od prej vemo,
 da sta zaporedji omejeni,
 ker sta konvergentni,
 zato 
\begin_inset Formula $\exists M>0\forall n\in\mathbb{N}:\left|b_{n}\right|\leq M$
\end_inset

.
 Naj bo 
\begin_inset Formula $\varepsilon>0$
\end_inset

 poljuben 
\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
\end_inset

 taka,
 da 
\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon}{2M}$
\end_inset

 in 
\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|A\right|}$
\end_inset

.
 Tedaj za 
\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
\end_inset

 velja
\begin_inset Formula $\left|a_{n}b_{n}-AB\right|\leq\left|a_{n}-A\right|\left|b_{n}\right|+\left|A\right|\left|b_{n}-B\right|<\frac{\varepsilon}{2M}M+\left|A\right|\frac{\varepsilon}{2\left|A\right|}=\varepsilon$
\end_inset

,
 skratka 
\begin_inset Formula $\left|a_{n}b_{n}-AB\right|<\varepsilon$
\end_inset

,
 s čimer dokažemo 
\begin_inset Formula $\left(a_{n}+b_{n}\right)\to A+B$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $/$
\end_inset

 Ker je 
\begin_inset Formula $B\not=0$
\end_inset

,
 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|b_{n}\right|\geq\frac{\left|B\right|}{2}>0$
\end_inset

.
 ZDB vsi členi zaporedja razen končno mnogo so v poljubno majhni okolici 
\begin_inset Formula $\left|B\right|$
\end_inset

.
 Če torej vzamemo točko na polovici med 0 in 
\begin_inset Formula $\left|B\right|$
\end_inset

,
 to je 
\begin_inset Formula $\frac{\left|B\right|}{2}$
\end_inset

,
 bo neskončno mnogo absolutnih vrednosti členov večjih od 
\begin_inset Formula $\frac{\left|B\right|}{2}$
\end_inset

.
 Pri razumevanju pomaga številska premica.
 Nadalje uporabimo predpostavko z 
\begin_inset Formula $\varepsilon=\frac{\left|B\right|}{2}$
\end_inset

,
 torej je za 
\begin_inset Formula $n>n_{0}:$
\end_inset

 
\begin_inset Formula $\left|B-b_{n}\right|<\frac{\left|B\right|}{2}$
\end_inset

 in velja 
\begin_inset Formula 
\[
\left|b_{n}\right|=\left|B-\left(B-b_{n}\right)\right|=\left|B+\left(-\left(B-b_{n}\right)\right)\right|\overset{\text{trik. neen.}}{\geq}\left|\left|B\right|-\left|B-b_{n}\right|\right|=\left|B\right|-\left|B-b_{n}\right|\overset{\text{predp.}}{>}\left|B\right|-\frac{\left|B\right|}{2}=\frac{\left|B\right|}{2},
\]

\end_inset

skratka 
\begin_inset Formula $\left|b_{n}\right|>\frac{\left|B\right|}{2}$
\end_inset

.
 Če spet izpustimo končno začetnih členov,
 velja
\begin_inset Formula 
\[
\frac{a_{n}}{b_{n}}-\frac{A}{B}=\frac{a_{n}B-Ab_{n}}{b_{n}B}\overset{\text{prištejemo in odštejemo člen}}{=}\frac{\left(a_{n}-A\right)B+A\left(B-b_{n}\right)}{Bb_{n}}=\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A/B}{b_{n}}\left(B-b_{n}\right)
\]

\end_inset

sedaj uporabimo na obeh straneh absolutno vrednost:
\begin_inset Formula 
\[
\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|=\left|\frac{1}{b_{n}}\left(a_{n}-A\right)+\frac{A}{Bb_{n}}\left(B-b_{n}\right)\right|\leq\frac{1}{\left|b_{n}\right|}\left|a_{n}-A\right|+\frac{\left|A\right|}{\left|B\right|\left|b_{n}\right|}\left|B-b_{n}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|
\]

\end_inset

skratka 
\begin_inset Formula $\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{2}{\left|B\right|}\left|a_{n}-A\right|+\frac{2\left|A\right|}{\left|B\right|^{2}}\left|B-B_{n}\right|$
\end_inset

.
 Opazimo,
 da 
\begin_inset Formula $\frac{2}{\left|B\right|}$
\end_inset

 in 
\begin_inset Formula $\frac{2\left|A\right|}{\left|B\right|^{2}}$
\end_inset

 nista odvisna od 
\begin_inset Formula $n$
\end_inset

.
 Sedaj vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

 in 
\begin_inset Formula $n_{1},n_{2}\in\mathbb{N}$
\end_inset

 takšna,
 da velja:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $n\geq n_{1}\Rightarrow\left|a_{n}-A\right|<\frac{\varepsilon\left|B\right|}{4}$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $n\geq n_{2}\Rightarrow\left|b_{n}-B\right|<\frac{\varepsilon\left|B\right|^{2}}{4\left|A\right|}$
\end_inset


\end_layout

\begin_layout Standard
Tedaj iz zgornje ocene sledi za 
\begin_inset Formula $n\geq\max\left\{ n_{0},n_{1},n_{2}\right\} $
\end_inset

:
\begin_inset Formula 
\[
\left|\frac{a_{n}}{b_{n}}-\frac{A}{B}\right|<\frac{\cancel{2}}{\cancel{\left|B\right|}}\cdot\frac{\varepsilon\cancel{\left|B\right|}}{\cancelto{2}{4}}+\frac{\cancel{2}\cancel{\left|A\right|}}{\cancel{\left|B\right|^{2}}}\cdot\frac{\varepsilon\cancel{\left|B\right|^{2}}}{\cancelto{2}{4}\cancel{\left|A\right|}}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,
\]

\end_inset

s čimer dokažemo 
\begin_inset Formula $a_{n}/b_{n}\to A/B$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout Example*
Naj bo 
\begin_inset Formula $a>0$
\end_inset

.
 Izračunajmo
\begin_inset Formula 
\[
\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{\cdots}}}}\eqqcolon\alpha.
\]

\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula $\alpha$
\end_inset

 je torej 
\begin_inset Formula $\lim_{n\to\infty}x_{n}$
\end_inset

,
 kjer je 
\begin_inset Formula $x_{0}=0,x_{1}=\sqrt{a},x_{2}=\sqrt{a+\sqrt{a}},x_{3}=\sqrt{a+\sqrt{a+\sqrt{a}}},\dots,x_{n+1}=\sqrt{a+x_{n}}$
\end_inset

.
 Iz zadnjega sledi 
\begin_inset Formula $x_{n+1}^{2}=a+x_{n}$
\end_inset

.
 Če torej limita 
\begin_inset Formula $\alpha\coloneqq\lim x_{n}$
\end_inset

 obstaja,
 mora veljati 
\begin_inset Formula $\alpha^{2}=a+\alpha$
\end_inset

 oziroma 
\begin_inset Formula $\alpha_{1,2}=\frac{1\pm\sqrt{1+4a}}{2}$
\end_inset

.
 Opcija z minusom ni mogoča,
 ker je zaporedje 
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 očitno pozitivno.
 Če torej limita obstaja (
\series bold
česar še nismo dokazali
\series default
),
 je enaka 
\begin_inset Formula $\frac{1+\sqrt{1+4a}}{2}$
\end_inset

,
 za primer 
\begin_inset Formula $a=2$
\end_inset

 je torej 
\begin_inset Formula $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}=2$
\end_inset

.
\end_layout

\begin_layout Remark*
Lahko se zgodi,
 da limita rekurzivno podanega zaporedja ne obstaja,
 čeprav jo znamo izračunati,
 če bi obstajala.
 Na primer 
\begin_inset Formula $y_{1}\coloneqq1$
\end_inset

,
 
\begin_inset Formula $y_{n+1}=1-2y_{n}$
\end_inset

 nam da zaporedje 
\begin_inset Formula $1,-1,3,-5,11,\dots$
\end_inset

,
 kar očitno nima limite.
 Če bi limita obstajala,
 bi zanjo veljalo 
\begin_inset Formula $\beta=1-2\beta$
\end_inset

 oz.
 
\begin_inset Formula $3\beta=1$
\end_inset

,
 
\begin_inset Formula $\beta=\frac{1}{3}$
\end_inset

.
 Navedimo torej nekaj zadostnih in potrebnih pogojev za konvergenco zaporedij.
\end_layout

\begin_layout Theorem*
Naj bo 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 monotono realno zaporedje.
 Če narašča,
 ima limito 
\begin_inset Formula $\lim_{n\to\infty}a_{n}=\sup\left\{ a_{n},n\in\mathbb{N}\right\} $
\end_inset

.
 Če pada,
 ima limito 
\begin_inset Formula $\lim_{n\to\infty}a_{n}=\inf\left\{ a_{n},n\in\mathbb{N}\right\} $
\end_inset

.
 (
\begin_inset Formula $\sup$
\end_inset

 in 
\begin_inset Formula $\inf$
\end_inset

 imata lahko tudi vrednost 
\begin_inset Formula $\infty$
\end_inset

 in 
\begin_inset Formula $-\infty$
\end_inset

 —
 zaporedje s tako limito ni konvergentno v 
\begin_inset Formula $\mathbb{R}$
\end_inset

).
\end_layout

\begin_layout Proof
Denimo,
 da 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 narašča.
 Pišimo 
\begin_inset Formula $s\coloneqq\sup_{n\in\mathbb{N}}a_{n}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Tedaj 
\begin_inset Formula $s-\varepsilon$
\end_inset

 ni zgornja meja za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

,
 zato 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s-\varepsilon<a_{n_{0}}$
\end_inset

.
 Ker pa je 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

naraščajoče,
 sledi 
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\geq a_{n_{0}}>s-\varepsilon$
\end_inset

.
 Hkrati je 
\begin_inset Formula $a_{n}\leq s$
\end_inset

,
 saj je 
\begin_inset Formula $s$
\end_inset

 zgornja meja.
 Torej 
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in(s-\varepsilon,s]\subset\left(s-\varepsilon,s+\varepsilon\right)$
\end_inset

,
 s čimer dokažemo konvergenco.
\end_layout

\begin_layout Proof
Denimo sedaj,
 da 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 pada.
 Dokaz je povsem analogen.
 Pišimo 
\begin_inset Formula $m\coloneqq\inf_{n\in\mathbb{N}}a_{n}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Tedaj 
\begin_inset Formula $m+\varepsilon$
\end_inset

 ni spodnja meja za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

,
 zato 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:m+\varepsilon>a_{n_{0}}$
\end_inset

.
 Ker pa je 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 padajoče,
 sledi 
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\leq a_{n_{0}}<m+\varepsilon$
\end_inset

.
 Hkrati je 
\begin_inset Formula $a_{n}\geq m$
\end_inset

,
 saj je 
\begin_inset Formula $m$
\end_inset

 spodnja meja.
 Torej 
\begin_inset Formula $\forall n\geq n_{0}:a_{n}\in[m,m+\varepsilon)\subset\left(m-\varepsilon,m+\varepsilon\right)$
\end_inset

.
\end_layout

\begin_layout Corollary*
Za monotono zaporedje velja,
 da je v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 konvergentno natanko tedaj,
 ko je omejeno.
\end_layout

\begin_layout Example*
Naj bo,
 kot prej,
 
\begin_inset Formula $a>0$
\end_inset

 in 
\begin_inset Formula $x_{0}=0,x_{n+1}=\sqrt{a+x_{n}}$
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $\left(x_{n}\right)_{n}$
\end_inset

 konvergentno.
 Dovolj je pokazati,
 da je naraščajoče in navzgor omejeno.
\end_layout

\begin_deeper
\begin_layout Itemize
Naraščanje z indukcijo:
 Baza:
 
\begin_inset Formula $0=x_{0}>x_{1}=\sqrt{a}$
\end_inset

.
 Dokažimo 
\begin_inset Formula $x_{n+1}-x_{n}>0$
\end_inset

.
\begin_inset Formula 
\[
\left(x_{n+1}-x_{n}\right)\left(x_{n+1}+x_{n}\right)=x_{n+1}^{2}-x_{n}^{2}=\left(a+x_{n}\right)-\left(a+x_{n-1}\right)=x_{n}-x_{n-1}
\]

\end_inset

Ker je zaporedje pozitivno,
 je 
\begin_inset Formula $x_{n+1}+x_{n}>0$
\end_inset

.
 Desna stran je po I.
 P.
 pozitivna,
 torej tudi 
\begin_inset Formula $x_{n+1}-x_{n}>0$
\end_inset

.
\end_layout

\begin_layout Itemize
Omejenost:
 Če je zaporedje res omejeno,
 je po zgornjem tudi konvergentno in je 
\begin_inset Formula $\sup_{n\in\mathbb{N}}x_{n}=\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}\leq\frac{1+\sqrt{1+4a+4a^{2}}}{2}=\frac{1+\sqrt{\left(2a+1\right)^{2}}}{}=1+a$
\end_inset

.
 Uganili smo neko zgornjo mejo.
 Domnevamo,
 da 
\begin_inset Formula $\forall n\in\mathbb{N}:x_{n}\leq1+a$
\end_inset

.
 Dokažimo to z indukcijo:
 Baza:
 
\begin_inset Formula $0=x_{0}<1+a$
\end_inset

.
 Po I.
 P.
 
\begin_inset Formula $x_{n}>1+a$
\end_inset

.
 Korak:
\begin_inset Formula 
\[
x_{n+1}=\sqrt{x_{n}+a}\leq\sqrt{1+a+a}=\sqrt{1+2a}<\sqrt{1+2a+2a^{2}}=1+a
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Example*
S tem smo dokazali,
 da 
\begin_inset Formula $\lim_{n\to\infty}x_{n}=\frac{1+\sqrt{1+4a}}{2}$
\end_inset

.
\end_layout

\begin_layout Example*
To lahko dokažemo tudi na alternativen način.
 Vidimo,
 da je edini kandidat za limito,
 če obstaja 
\begin_inset Formula $L=\frac{1+\sqrt{1+4a}}{2}$
\end_inset

 in da torej velja 
\begin_inset Formula $L^{2}=a+L$
\end_inset

.
 Preverimo,
 da je 
\begin_inset Formula $L$
\end_inset

 res limita:
 
\begin_inset Formula 
\[
x_{n+1}-L=\sqrt{a+x_{n}}-L=\frac{\left(\sqrt{a+x_{n}}-L\right)\left(\sqrt{a+x_{n}}+L\right)}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-L^{2}}{\sqrt{a+x_{n}}+L}=\frac{\left(a+x_{n}\right)-\left(a+L\right)}{\sqrt{a+x_{n}}+L}=\frac{x_{n}-L}{\sqrt{a+x_{n}}+L}.
\]

\end_inset

Vpeljimo sedaj 
\begin_inset Formula $y_{n}\coloneqq x_{n}-L$
\end_inset

.
 Sledi 
\begin_inset Formula $\left|y_{n+1}\right|\leq\frac{\left|y_{n}\right|}{\sqrt{a+x_{n}}+L}\leq\frac{\left|y_{n}\right|}{L}$
\end_inset

.
 Ker je 
\begin_inset Formula $\left|y_{0}\right|=L$
\end_inset

,
 dobimo
\begin_inset Foot
status open

\begin_layout Plain Layout
Za razumevanje si oglej nekaj členov rekurzivnega zaporedje 
\begin_inset Formula $y_{0}=L,y_{n}=\frac{\left|y_{n+1}\right|}{L}$
\end_inset

.
 Začnemo z 1 in nato vsakič delimo z 
\begin_inset Formula $L$
\end_inset

.
\end_layout

\end_inset

 oceno 
\begin_inset Formula $\left|y_{n}\right|\leq\frac{1}{L^{n-1}}$
\end_inset

 oziroma 
\begin_inset Formula $\left|x_{n}-L\right|\leq\frac{1}{L^{n-1}}$
\end_inset

.
 Ker iz definicije 
\begin_inset Formula $L$
\end_inset

 sledi 
\begin_inset Formula $L>1$
\end_inset

,
 je 
\begin_inset Formula $L^{n}\to\infty$
\end_inset

 za 
\begin_inset Formula $n\to\infty$
\end_inset

,
 torej smo dokazali,
 da 
\begin_inset Formula $\left|x_{n}-L\right|$
\end_inset

 eksponentno pada proti 0 za 
\begin_inset Formula $n\to\infty$
\end_inset

.
 Eksponentno padanje 
\begin_inset Formula $\left|x_{n}-L\right|$
\end_inset

 proti 0 je dovolj,
 da rečemo,
 da zaporedje konvergira k 
\begin_inset Formula $L$
\end_inset


\begin_inset Foot
status open

\begin_layout Plain Layout
a res,
 vprašaj koga.
 ne razumem.
 zakaj.
 TODO.
\end_layout

\end_inset

.
\end_layout

\begin_layout Claim*
\begin_inset Formula $\lim_{n\to\infty}\sin n$
\end_inset

 in 
\begin_inset Formula $\lim_{n\to\infty}\cos n$
\end_inset

 ne obstajata.
\end_layout

\begin_layout Proof
Pišimo 
\begin_inset Formula $a_{n}=\sin n$
\end_inset

 in 
\begin_inset Formula $b_{n}=\cos n$
\end_inset

.
 Iz adicijskih izrekov dobimo 
\begin_inset Formula $a_{n+1}=\sin\left(n+1\right)=\sin n\cos1+\cos n\sin1=a_{n}\cos1+b_{n}\sin1$
\end_inset

.
 Torej 
\begin_inset Formula $b_{n}=\frac{a_{n+1}-a_{n}\cos1}{\sin1}$
\end_inset

.
 Torej če 
\begin_inset Formula $\exists a\coloneqq\lim_{n\to\infty}a_{n},a\in\mathbb{R}$
\end_inset

,
 potem tudi 
\begin_inset Formula $\exists b\coloneqq\lim_{n\to\infty}b_{n},b\in\mathbb{R}$
\end_inset

.
 Podobno iz adicijske formule za 
\begin_inset Formula $\cos\left(n+1\right)$
\end_inset

 sledi 
\begin_inset Formula $a_{n}=\frac{b_{n}\cos1-b_{n+1}}{\sin1}$
\end_inset

,
 torej če 
\begin_inset Formula $\exists b$
\end_inset

,
 potem tudi 
\begin_inset Formula $\exists a$
\end_inset

.
 Iz obojega sledi,
 da 
\begin_inset Formula $\exists a\Leftrightarrow\exists b$
\end_inset

.
 Posledično,
 če 
\begin_inset Formula $a$
\end_inset

 in 
\begin_inset Formula $b$
\end_inset

 obstajata,
 iz zgornjih obrazcev za 
\begin_inset Formula $a_{n}$
\end_inset

 in 
\begin_inset Formula $b_{n}$
\end_inset

 sledi,
 da za
\begin_inset Formula 
\[
\lambda=\frac{1-\cos1}{\sin1}\in\left(0,1\right)
\]

\end_inset

velja 
\begin_inset Formula $b=\lambda a$
\end_inset

 in 
\begin_inset Formula $a=-\lambda b$
\end_inset

 in zato 
\begin_inset Formula $b=\lambda\left(-\lambda b\right)$
\end_inset

 oziroma 
\begin_inset Formula $1=-\lambda^{2}$
\end_inset

,
 torej 
\begin_inset Formula $-1=\lambda^{2}$
\end_inset

,
 torej 
\begin_inset Formula $\lambda=i$
\end_inset

,
 kar je v protislovju z 
\begin_inset Formula $\lambda\in\left(0,1\right)$
\end_inset

.
 Podobno za 
\begin_inset Formula $a=-\lambda\left(\lambda a\right)$
\end_inset

 oziroma 
\begin_inset Formula $1=-\lambda^{2}$
\end_inset

,
 kar je zopet 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

.
 Edina druga opcija je,
 da je 
\begin_inset Formula $a=b=0$
\end_inset

.
 Hkrati pa vemo,
 da 
\begin_inset Formula $a_{n}^{2}+b_{n}^{2}=1$
\end_inset

,
 zato 
\begin_inset Formula $a+b=1$
\end_inset

,
 kar ni mogoče za ničelna 
\begin_inset Formula $a$
\end_inset

 in 
\begin_inset Formula $b$
\end_inset

.
 Torej 
\begin_inset Formula $a$
\end_inset

 in 
\begin_inset Formula $b$
\end_inset

 ne obstajata.
\end_layout

\begin_layout Subsection
Eulerjevo število
\end_layout

\begin_layout Theorem*
Bernoullijeva neenakost.
 
\begin_inset Formula $\forall\alpha\leq1,n\in\mathbb{N}$
\end_inset

 velja 
\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
\end_inset

.
\end_layout

\begin_layout Proof
Z indukcijo na 
\begin_inset Formula $n$
\end_inset

 ob fiksnem 
\begin_inset Formula $\alpha$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Baza:
 
\begin_inset Formula $n=1$
\end_inset

:
 
\begin_inset Formula $\left(1-\alpha\right)^{1}=1-1\alpha$
\end_inset

.
 Velja celo enakost.
\end_layout

\begin_layout Itemize
I.
 P.:
 Velja 
\begin_inset Formula $\left(1-\alpha\right)^{n}\geq1-n\alpha$
\end_inset


\end_layout

\begin_layout Itemize
Korak:
 
\begin_inset Formula $\left(1-\alpha\right)^{n+1}=\left(1-\alpha\right)\left(1-\alpha\right)^{n}\geq\left(1-\alpha\right)$
\end_inset


\begin_inset Formula $\left(1-n\alpha\right)=1-n\alpha-\alpha-n\alpha^{2}=1-\left(n+1\right)\alpha-n\alpha^{2}\geq1-\left(n+1\right)\alpha$
\end_inset

,
 torej ocena velja tudi za 
\begin_inset Formula $n+1$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Definition*
Vpeljimo oznaki:
\end_layout

\begin_deeper
\begin_layout Itemize
Za 
\begin_inset Formula $n\in\mathbb{N}$
\end_inset

 označimo 
\begin_inset Formula $n!=1\cdot2\cdot3\cdot\cdots\cdot n$
\end_inset

 (pravimo 
\begin_inset Formula $n-$
\end_inset

faktoriala oziroma 
\begin_inset Formula $n-$
\end_inset

fakulteta).
 Ker velja 
\begin_inset Formula $n!=n\cdot\left(n-1\right)!$
\end_inset

 za 
\begin_inset Formula $n\geq2$
\end_inset

,
 je smiselno definirati še 
\begin_inset Formula $0!=1$
\end_inset

.
\end_layout

\begin_layout Itemize
Za 
\begin_inset Formula $n,k\in\mathbb{N}$
\end_inset

 označimo še binomski simbol:
 
\begin_inset Formula $\binom{n}{k}\coloneqq\frac{n!}{k!\left(n-k\right)!}$
\end_inset

 (pravimo 
\begin_inset Formula $n$
\end_inset

 nad 
\begin_inset Formula $k$
\end_inset

).
\end_layout

\begin_layout Itemize
Če je 
\begin_inset Formula $\left(a_{k}\right)_{k}$
\end_inset

 neko zaporedje (lahko tudi končno),
 lahko pišemo 
\begin_inset Formula $\sum_{k=1}^{n}a_{k}\coloneqq a_{1}+a_{2}+\cdots+a_{n}$
\end_inset

 (pravimo summa) in 
\begin_inset Formula $\prod_{k=1}^{n}a_{k}\coloneqq a_{1}\cdot\cdots\cdot a_{n}$
\end_inset

 (pravimo produkt).
\end_layout

\end_deeper
\begin_layout Example*
\begin_inset Formula $\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$
\end_inset

 in 
\begin_inset Formula $\prod_{k=1}^{n}k=1\cdot2\cdot3\cdot\cdots\cdot n=n!$
\end_inset

.
\end_layout

\begin_layout Theorem*
Binomska formula.
 
\begin_inset Formula $\forall a,b\in\mathbb{R},n\in\mathbb{N}:\left(a+b\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$
\end_inset

.
\end_layout

\begin_layout Proof
Indukcija po 
\begin_inset Formula $n$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Itemize
Baza 
\begin_inset Formula $n=1$
\end_inset

:
 
\begin_inset Formula $\sum_{k=0}^{1}\binom{n}{k}a^{k}b^{n-k}=\binom{1}{0}a^{0}b^{1-0}+\binom{1}{1}a^{1}b^{1-1}=a+b=\left(a+b\right)^{1}$
\end_inset


\end_layout

\begin_layout Itemize
I.
 P.
 
\begin_inset Formula $\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\left(a+b\right)^{n}$
\end_inset


\end_layout

\begin_layout Itemize
Korak:
 
\begin_inset Formula 
\[
\left(a+b\right)^{n+1}=\left(a+b\right)\left(a+b\right)^{n}=\left(a+b\right)\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=
\]

\end_inset

sedaj naj bo 
\begin_inset Formula $m=k+1$
\end_inset

 v levem členu:
\begin_inset Formula 
\[
=\sum_{m=1}^{n+1}\binom{n}{m-1}a^{m}b^{n-\left(m-1\right)}+\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k+1}=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=
\]

\end_inset

Sedaj obravnavajmo le izraz v oglatih oklepajih:
\begin_inset Formula 
\[
\binom{n}{k-1}+\binom{n}{k}=\frac{n!}{\left(k-1\right)!\left(n-k+1\right)!}+\frac{n!}{k!\left(n-k\right)!}=\frac{kn!}{k!\left(n-k+1\right)!}+\frac{n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=\frac{kn!+n!\left(n-k+1\right)}{k!\left(n-k+1\right)!}=
\]

\end_inset


\begin_inset Formula 
\[
=\frac{n!\left(\cancel{k+}n\cancel{-k}+1\right)}{k!\left(n+1-k\right)!}=\frac{n!\left(n+1\right)}{k!\left(n+1-k\right)!}=\frac{\left(n+1\right)!}{k!\left(n+1-k\right)}=\binom{n+1}{k}
\]

\end_inset

in skratka dobimo 
\begin_inset Formula $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$
\end_inset

.
 Vstavimo to zopet v naš zgornji račun:
\begin_inset Formula 
\[
\cdots=a^{n+1}+\sum_{k=1}^{n}\left[\binom{n}{k-1}+\binom{n}{k}\right]a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=
\]

\end_inset


\begin_inset Formula 
\[
=a^{n+1}+\sum_{k=1}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}+b^{n+1}=a^{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}a^{k}b^{n-k+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{k}b^{n-k+1}
\]

\end_inset


\end_layout

\end_deeper
\begin_layout Theorem*
Bernoulli.
 Zaporedje 
\begin_inset Formula $a_{n}\coloneqq\left(1+\frac{1}{n}\right)^{n}$
\end_inset

 je konvergentno.
\end_layout

\begin_layout Proof
Dokazali bomo,
 da je naraščajoče in omejeno.
\end_layout

\begin_deeper
\begin_layout Itemize
Naraščanje:
 Dokazujemo,
 da za 
\begin_inset Formula $n\geq2$
\end_inset

 velja 
\begin_inset Formula $a_{n}\geq a_{n-1}$
\end_inset

 oziroma 
\begin_inset Formula 
\[
\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n-1}{n-1}+\frac{1}{n-1}\right)^{n-1}=\left(\frac{n}{n-1}\right)^{n-1}=\left(\frac{n-1}{n}\right)^{1-n}=\left(1-\frac{1}{n}\right)^{1-n}
\]

\end_inset


\begin_inset Formula 
\[
\left(1+\frac{1}{n}\right)^{n}\overset{?}{\geq}\left(1-\frac{1}{n}\right)^{1-n}
\]

\end_inset


\begin_inset Formula 
\[
\left(1+\frac{1}{n^{2}}\right)^{n}=\left(\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right)\right)^{n}=\left(1+\frac{1}{n}\right)^{n}\left(1-\frac{1}{n}\right)^{n}\overset{?}{\geq}1-\frac{1}{n}
\]

\end_inset


\begin_inset Formula 
\[
\left(1+\frac{1}{n^{2}}\right)^{n}\overset{?}{\geq}1-\frac{1}{n},
\]

\end_inset

kar je poseben primer Bernoullijeve neenakosti za 
\begin_inset Formula $\alpha=\frac{1}{n^{2}}$
\end_inset

.
\end_layout

\begin_layout Itemize
Omejenost:
 Po binomski formuli je 
\begin_inset Formula 
\[
a_{n}=\left(1+\frac{1}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{n}\right)^{k}=\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=1+1+\sum_{k=2}^{n}\frac{n!}{k!\left(n-k\right)!n^{k}}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n\left(n-1\right)\left(n-2\right)\cdots\left(n-k+1\right)}{n^{k}}=
\]

\end_inset


\begin_inset Formula 
\[
=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\cdots\cdot\frac{n-k+1}{n}=2+\sum_{k=2}^{n}\frac{1}{k!}\cdot\cancel{\left(1-0\right)}\cdot\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdot\cdots\cdot\left(1-\frac{k-1}{n}\right)=
\]

\end_inset


\begin_inset Formula 
\[
=2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<2+\sum_{k=2}^{n}\frac{1}{k!}<2+\sum_{k=2}^{n}\frac{1}{2^{k-1}}=\cdots
\]

\end_inset

Opomnimo,
 da je 
\begin_inset Formula $1-\frac{j}{n}<1$
\end_inset

,
 zato 
\begin_inset Formula $\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)<1$
\end_inset

 (prvi neenačaj) ter 
\begin_inset Formula $k!=1\cdot2\cdot3\cdot\cdots\cdot k\geq1\cdot2\cdot2\cdot\cdots\cdot2=2^{k-1}$
\end_inset

 (drugi).
 Sedaj si z indukcijo dokažimo 
\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
\end_inset

:
\end_layout

\begin_deeper
\begin_layout Itemize
Baza:
 
\begin_inset Formula $n=2$
\end_inset

:
 
\begin_inset Formula $\frac{1}{2^{2-1}}=1-\frac{1}{2^{2-1}}=1-\frac{1}{2}=\frac{1}{2}$
\end_inset

.
 Velja!
\end_layout

\begin_layout Itemize
I.
 P.:
 
\begin_inset Formula $\sum_{k=2}^{n}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}$
\end_inset


\end_layout

\begin_layout Itemize
Korak:
 
\begin_inset Formula $\sum_{k=2}^{n+1}\frac{1}{2^{k-1}}=1-\frac{1}{2^{n-1}}+\frac{1}{2^{n+1-1}}=1-2\cdot2^{-n}+2^{-n}=1+2^{-n}\left(1-2\right)=1+2^{-n}$
\end_inset


\end_layout

\begin_layout Standard
In nadaljujmo z računanjem:
\begin_inset Formula 
\[
\cdots=2+1-\frac{1}{2^{n-1}}=3-\frac{1}{2^{n-1}},
\]

\end_inset

s čimer dobimo zgornjo mejo 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}<3$
\end_inset

.
 Ker je očitno 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}>0$
\end_inset

,
 je torej zaporedje omejeno in ker je tudi monotono po prejšnjem izreku konvergira.
\end_layout

\end_deeper
\end_deeper
\begin_layout Definition*
Označimo število 
\begin_inset Formula $e\coloneqq\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$
\end_inset

 in ga imenujemo Eulerjevo število.
 Velja 
\begin_inset Formula $e\approx2,71828$
\end_inset

.
\end_layout

\begin_layout Remark*
V dokazu vidimo moč izreka 
\begin_inset Quotes gld
\end_inset

omejenost in monotonost 
\begin_inset Formula $\Rightarrow$
\end_inset

 konvergenca
\begin_inset Quotes grd
\end_inset

,
 saj nam omogoča dokazati konvergentnost zaporedja brez kandidata za limito.
 Jasno je,
 da ne bi mogli vnaprej uganiti,
 da je limita ravno 
\shape italic
transcendentno število
\shape default
 
\begin_inset Formula $e$
\end_inset

.
\end_layout

\begin_layout Definition*
Podzaporedje zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 je poljubno zaporedje oblike 
\begin_inset Formula $\left(a_{\varphi\left(n\right)}\right)_{n\in\mathbb{N}}$
\end_inset

,
 kjer je 
\begin_inset Formula $\varphi:\mathbb{N}\to\mathbb{N}$
\end_inset

 strogo naraščajoča funkcija.
\end_layout

\begin_layout Theorem*
Če je 
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}$
\end_inset

,
 tedaj je 
\begin_inset Formula $L$
\end_inset

 tudi limita vsakega podzaporedja.
\end_layout

\begin_layout Proof
Po predpostavki velja 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}:\left|a_{n}-L\right|<\varepsilon$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Po predpostavki obstaja 
\begin_inset Formula $n_{0}\in\mathbb{N}$
\end_inset

,
 da bodo vsi členi zaporedja po 
\begin_inset Formula $n_{0}-$
\end_inset

tem v 
\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
\end_inset

.
 Iz definicijskega območja 
\begin_inset Formula $\varphi$
\end_inset

 vzemimo poljuben element 
\begin_inset Formula $n_{1}$
\end_inset

,
 da velja 
\begin_inset Formula $n_{1}\geq n_{0}$
\end_inset

.
 Gotovo obstaja,
 ker je definicijsko območje števno neskončne moči in s pogojem 
\begin_inset Formula $n_{1}\geq n_{0}$
\end_inset

 onemogočimo izbiro le končno mnogo elementov.
 
\begin_inset Note Note
status open

\begin_layout Plain Layout
Če slednji ne obstaja,
 je v 
\begin_inset Formula $D_{\varphi}$
\end_inset

 končno mnogo elementov,
 tedaj vzamemo 
\begin_inset Formula $n_{1}\coloneqq\max D_{\varphi}+1$
\end_inset

 in je pogoj za limito izpolnjen na prazno.
 Sicer pa v
\end_layout

\end_inset

Velja 
\begin_inset Formula $\forall n\in\mathbb{N}:n>n_{1}\Rightarrow\left|a_{\varphi n}-L\right|<\varepsilon$
\end_inset

,
 ker je 
\begin_inset Formula $\varphi$
\end_inset

 strogo naraščajoča in izbiramo le elemente podzaporedja,
 ki so v izvornem zaporedju za 
\begin_inset Formula $n_{0}-$
\end_inset

tim členom in zato v 
\begin_inset Formula $\left(L-\varepsilon,L+\varepsilon\right)$
\end_inset

.
\end_layout

\begin_layout Example*
\begin_inset Formula $\lim_{n\to\infty}\frac{1}{2n+3}=\lim_{n\to\infty}\frac{1}{n}=0$
\end_inset

 za zaporedje 
\begin_inset Formula $a_{n}=\frac{1}{n}$
\end_inset

 in podzaporedje 
\begin_inset Formula $a_{\varphi n}$
\end_inset

,
 kjer je 
\begin_inset Formula $\varphi\left(n\right)=2n+3$
\end_inset

.
\end_layout

\begin_layout Theorem*
Karakterizacija limite s podzaporedji.
 Naj bo 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 realno zaporedje in 
\begin_inset Formula $L\in\mathbb{R}$
\end_inset

.
 Tedaj 
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Leftrightarrow$
\end_inset

 za vsako podzaporedje 
\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
\end_inset

 zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n}$
\end_inset

 obstaja njegovo podzaporedje 
\begin_inset Formula $\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}$
\end_inset

,
 ki konvergira k 
\begin_inset Formula $L$
\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Dokazano poprej.
 Limita se pri prehodu na podzaporedje ohranja.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 PDDRAA 
\begin_inset Formula $a_{n}\not\to L$
\end_inset

.
 Tedaj 
\begin_inset Formula $\exists\varepsilon>0$
\end_inset

 in podzaporedje 
\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}\ni:\forall k\in\mathbb{N}:\left|a_{n_{k}}-K\right|>\varepsilon$
\end_inset

 (*)
\begin_inset Note Note
status open

\begin_layout Plain Layout
tu je na 
\begin_inset Quotes gld
\end_inset

Zaporedja 2
\begin_inset Quotes grd
\end_inset

 napaka,
 neenačaj obrne v drugo smer
\end_layout

\end_inset

.
 Po predpostavki sedaj 
\begin_inset Formula $\exists\left(a_{n_{k_{l}}}\right)_{l\in\mathbb{N}}\ni:\lim_{l\to\infty}a_{n_{k_{l}}}=L$
\end_inset

.
 To pa je v protislovju z (*),
 torej je začetna predpostavka 
\begin_inset Formula $a_{n}\not\to L$
\end_inset

 napačna,
 torej 
\begin_inset Formula $a_{n}\to L$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Subsection
Stekališča
\end_layout

\begin_layout Definition*
Točka 
\begin_inset Formula $s\in\mathbb{R}$
\end_inset

 je stekališče zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subset\mathbb{R}$
\end_inset

,
 če v vsaki okolici te točke leži neskončno členov zaporedja.
\end_layout

\begin_layout Remark*
Pri limiti zahtevamo več;
 da izven vsake okolice limite leži le končno mnogo členov.
\end_layout

\begin_layout Example*
Primeri stekališč.
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $L=\lim_{n\to\infty}a_{n}\Rightarrow L$
\end_inset

 je stekališče za 
\begin_inset Formula $a_{n}$
\end_inset


\end_layout

\begin_layout Enumerate
\begin_inset Formula $0,1,0,1,\dots$
\end_inset

 stekališči sta 
\begin_inset Formula $\left\{ 0,1\right\} $
\end_inset

 in zaporedje nima limite (ni konvergentno)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $1,1,2,1,2,3,1,2,3,4,\dots$
\end_inset

 ima neskončno stekališč,
 
\begin_inset Formula $\mathbb{N}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $b_{n}=n-$
\end_inset

to racionalno število
\begin_inset Foot
status open

\begin_layout Plain Layout
Racionalnih števil je števno mnogo,
 zato jih lahko linearno uredimo in oštevilčimo.
\end_layout

\end_inset

 ima neskončno stekališč,
 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Remark*
Limita je stakališče,
 stekališče pa ni nujno limita.
 Poleg tega,
 če se spomnimo,
 velja,
 da vsota konvergentnih zaporedij konvergira k vsoti njunih limit,
 ni pa nujno res,
 da so stekališča vsote dveh zaporedij paroma vsote stekališč teh dveh zaporedij.
 Primer:
 
\begin_inset Formula $a_{n}=\left(-1\right)^{n}$
\end_inset

 in 
\begin_inset Formula $b_{n}=-\left(-1\right)^{n}$
\end_inset

.
 Njuni stekališči sta 
\begin_inset Formula $\left\{ -1,1\right\} $
\end_inset

,
 toda 
\begin_inset Formula $a_{n}+b_{n}=0$
\end_inset

 ima le stekališče 
\begin_inset Formula $\left\{ 0\right\} $
\end_inset

,
 ne pa tudi 
\begin_inset Formula $\left\{ 1,-1,2,-2\right\} $
\end_inset

.
\end_layout

\begin_layout Theorem*
\begin_inset Formula $S$
\end_inset

 je stekališče 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow S$
\end_inset

 je limita nekega podzaporedja 
\begin_inset Formula $a_{n}$
\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Očitno.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Definirajmo 
\begin_inset Formula $\forall m\in\mathbb{N}:U_{m}\coloneqq\left(S-\frac{1}{m},S+\frac{1}{m}\right)$
\end_inset

.
 Ker je 
\begin_inset Formula $S$
\end_inset

 stekališče,
 
\begin_inset Formula $\forall m\in\mathbb{N}\exists a_{k_{m}}\in U_{m}$
\end_inset

.
 Podzaporedje 
\begin_inset Formula $\left(a_{k_{m}}\right)_{m\in\mathbb{N}}$
\end_inset

 konvergira k 
\begin_inset Formula $S$
\end_inset

,
 kajti 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a_{k_{n}}-S\right|<\frac{1}{n}<\varepsilon$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Corollary*
Če je 
\begin_inset Formula $L$
\end_inset

 limita nekega zaporedja,
 je 
\begin_inset Formula $L$
\end_inset

 edino njegovo stekališče.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $a_{n}\to L$
\end_inset

.
 Naj bo 
\begin_inset Formula $S$
\end_inset

 stekališče za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 Po izreku zgoraj je 
\begin_inset Formula $S$
\end_inset

 limita nekega podzaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 Toda limita vsakega podzaporedja je enaka limiti zaporedja,
 iz katerega to podzaporedje izhaja,
 če ta limita obstaja.
 Potemtakem je 
\begin_inset Formula $S=L$
\end_inset

.
\end_layout

\begin_layout Theorem*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hypertarget{bw}{Bolzano-Weierstraß}
\end_layout

\end_inset

.
 Eksistenčni izrek.
 Vsako omejeno zaporedje v realnih številih ima kakšno stekališče v 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Proof
Označimo 
\begin_inset Formula $m_{0}\coloneqq\inf_{n\in\mathbb{N}}a_{n},M_{0}\coloneqq\sup_{n\in\mathbb{N}}a_{n},I_{0}\coloneqq\left[m_{0},M_{0}\right]$
\end_inset

.
 Očitno je 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in I_{0}$
\end_inset

.
 Interval 
\begin_inset Formula $I_{0}$
\end_inset

 razdelimo na dve polovici:
 
\begin_inset Formula $I_{0}=\left[m_{0},\frac{m_{0}+M_{0}}{2}\right]\cup\left[\frac{m_{0}+M_{0}}{2},M_{0}\right]$
\end_inset

.
 Izberemo polovico (vsaj ena obstaja),
 v kateri leži neskončno mnogo členov,
 in jo označimo z 
\begin_inset Formula $I_{1}$
\end_inset

.
 Spet jo razdelimo na pol in z 
\begin_inset Formula $I_{2}$
\end_inset

 označimo tisto polovico,
 v kateri leži neskončno mnogo členov.
 Postopek ponavljamo in dobimo zaporedje zaprtih intervalov 
\begin_inset Formula $\left(I_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 in velja 
\begin_inset Formula $I_{0}\supset I_{1}\supset I_{2}\supset\cdots$
\end_inset

 ter 
\begin_inset Formula $\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset

.
\end_layout

\begin_layout Proof
Označimo sedaj 
\begin_inset Formula $I_{n}\eqqcolon\left[l_{n},d_{n}\right]$
\end_inset

.
 Iz konstrukcije je očitno,
 da 
\begin_inset Formula $\left(l_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 narašča in 
\begin_inset Formula $\left(d_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 pada ter da sta obe zaporedji omejeni.
 Posledično 
\begin_inset Formula $\exists l\coloneqq\lim_{n\to\infty}l_{n},d\coloneqq\lim_{n\to\infty}d_{n}$
\end_inset

.
 Iz 
\begin_inset Formula $l_{n}\leq l\leq d\leq d_{n}$
\end_inset

 sledi ocena 
\begin_inset Formula $d-l\leq l_{n}-d_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
\end_inset

,
 kar konvergira k 0.
 Posledično 
\begin_inset Formula $d=l$
\end_inset

.
\end_layout

\begin_layout Proof
Treba je pokazati še,
 da je 
\begin_inset Formula $d=l$
\end_inset

 stekališče za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Ker je 
\begin_inset Formula $l=\lim_{n\to\infty}l_{n}\Rightarrow\exists n_{1}\in\mathbb{N}\ni:l_{n_{1}}>l-\varepsilon$
\end_inset

 in ker je 
\begin_inset Formula $d=\lim_{n\to\infty}d_{n}\Rightarrow\exists n_{2}\in\mathbb{N}\ni:d_{n_{2}}<d-\varepsilon$
\end_inset

.
 Torej 
\begin_inset Formula $\left[l_{n_{1}},d_{n_{2}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset

.
 Torej za 
\begin_inset Formula $n_{0}\coloneqq\max\left\{ n_{1},n_{2}\right\} $
\end_inset

 velja 
\begin_inset Formula $I_{n_{0}}=\left[l_{n_{0}},d_{n_{n}}\right]\subset\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset

.
 Ker 
\begin_inset Formula $I_{n_{0}}$
\end_inset

 po konstrukciji vsebuje neskončno mnogo elementov,
 jih torej tudi 
\begin_inset Formula $\left(l-\varepsilon,d+\varepsilon\right)$
\end_inset

 oziroma poljubno majhna okolica 
\begin_inset Formula $d=l$
\end_inset

,
 torej je 
\begin_inset Formula $d=l$
\end_inset

 stekališče za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
\end_layout

\begin_layout Corollary*
Če je 
\begin_inset Formula $s\in\mathbb{R}$
\end_inset

 edino stekališče omejenega zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

,
 tedaj je 
\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
\end_inset

.
\end_layout

\begin_layout Proof
Naj bo 
\begin_inset Formula $s$
\end_inset

 stekališče 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 PDDRAA 
\begin_inset Formula $a_{n}\not\to s$
\end_inset

.
 Tedaj 
\begin_inset Formula $\exists\varepsilon>0\ni:$
\end_inset

 izven 
\begin_inset Formula $\left(s-\varepsilon,s+\varepsilon\right)$
\end_inset

 se nahaja neskončno mnogo členov zaporedja.
 Ti členi sami zase tvorijo omejeno zaporedje,
 ki ima po 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{bw}{B.-W.}
\end_layout

\end_inset

 izreku stekališče.
 Slednje gotovo ne more biti enako 
\begin_inset Formula $s$
\end_inset

,
 torej imamo vsaj dve stekališči,
 kar je v je v 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

 s predpostavko.
\end_layout

\begin_layout Definition*
Pravimo,
 da ima realno zaporedje:
\end_layout

\begin_deeper
\begin_layout Itemize
stekališče v 
\begin_inset Formula $\infty$
\end_inset

,
 če 
\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
\end_inset

 vsebuje neskončno mnogo členov zapopredja
\end_layout

\begin_layout Itemize
limito v 
\begin_inset Formula $\infty$
\end_inset

,
 če 
\begin_inset Formula $\forall M>0:\left(M,\infty\right)$
\end_inset

 vsebuje vse člene zaporedja od nekega indeksa dalje
\end_layout

\begin_layout Standard
in podobno za 
\begin_inset Formula $-\infty$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Remark*
Povezava s pojmom realnega stekališča/limite:
 okolice 
\begin_inset Quotes gld
\end_inset

točke
\begin_inset Quotes grd
\end_inset

 
\begin_inset Formula $\infty$
\end_inset

 so intervali oblike 
\begin_inset Formula $\left(M,\infty\right)$
\end_inset

.
 To je smiselno,
 saj biti 
\begin_inset Quotes gld
\end_inset

blizu 
\begin_inset Formula $\infty$
\end_inset


\begin_inset Quotes grd
\end_inset

 pomeni bizi zelo velik,
 kar je ravno biti v 
\begin_inset Formula $\left(M,\infty\right)$
\end_inset

za poljubno velik 
\begin_inset Formula $M$
\end_inset

.
 
\begin_inset Quotes gld
\end_inset

Okolica točke 
\begin_inset Formula $\infty$
\end_inset


\begin_inset Quotes grd
\end_inset

 so torej vsi intervali oblike 
\begin_inset Formula $\left(M,\infty\right)$
\end_inset

.
\end_layout

\begin_layout Subsection
Limes superior in limes inferior
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

realno zaporedje.
 Tvorimo novo zaporedje 
\begin_inset Formula $s_{n}\coloneqq\sup\left\{ a_{k};k\geq n\right\} $
\end_inset

.
 Očitno je padajoče (
\begin_inset Formula $s_{1}\geq s_{2}\geq s_{3}\geq\cdots$
\end_inset

),
 ker je supremum množice vsaj supremum njene stroge podmnožice.
 Zaporedje 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 ima limito,
 ki ji rečemo limes superior oziroma zgornja limita zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 in označimo 
\begin_inset Formula $\limsup_{n\to\infty}a_{n}=\overline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}s_{n}$
\end_inset

 in velja,
 da leži v 
\begin_inset Formula $\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
\end_inset

.
 Podobno definiramo tudi limes inferior oz.
 spodnjo limito zaporedja:
 
\begin_inset Formula $\liminf_{n\to\infty}a_{n}=\underline{\lim_{n\to\infty}}a_{n}\coloneqq\lim_{n\to\infty}\left(\inf_{k\geq n}a_{k}\right)=\sup_{n\in\mathbb{N}}\left(\inf_{k\geq n}a_{k}\right)$
\end_inset

.
\end_layout

\begin_layout Remark*
Za razliko od običajne limite,
 ki lahko ne obstaja,
 
\begin_inset Formula $\limsup$
\end_inset

 in 
\begin_inset Formula $\liminf$
\end_inset

 vedno obstajata.
\end_layout

\begin_layout Claim*
\begin_inset Formula $\limsup_{n\to\infty}a_{n}$
\end_inset

 je največje stekališče zaporedja 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 in 
\begin_inset Formula $\liminf_{n\to\infty}$
\end_inset

 najmanjše.
\end_layout

\begin_layout Proof
Označimo 
\begin_inset Formula $s\coloneqq\limsup_{n\to\infty}a_{n}$
\end_inset

.
 Za 
\begin_inset Formula $\liminf$
\end_inset

 je dokaz analogen in ga ne bomo pisali.
 Dokazujemo,
 da je 
\begin_inset Formula $s$
\end_inset

 stekališče in 
\begin_inset Formula $\forall t>s:t$
\end_inset

 ni stekališče.
 Ločimo primere:
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s\in\mathbb{R}$
\end_inset

 Naj bo 
\begin_inset Formula $\varepsilon>0$
\end_inset

 poljuben.
 Ker
\begin_inset Foot
status open

\begin_layout Plain Layout
Infimum padajočega konvergentnega zaporedja je očitno njegova limita.
\end_layout

\end_inset

 je 
\begin_inset Formula $s=\inf s_{n}$
\end_inset

,
 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in[s,s+\varepsilon)$
\end_inset

.
 Ker 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 pada proti 
\begin_inset Formula $s$
\end_inset

,
 sledi 
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow s_{n}\in[s,s+\varepsilon)$
\end_inset

.
 Po definiciji 
\begin_inset Formula $s_{n}$
\end_inset

 velja 
\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right)\geq n\ni:s_{n}-\varepsilon<a_{N\left(n\right)}$
\end_inset

.
 Torej imamo 
\begin_inset Formula $s-\varepsilon\leq s_{n}-\varepsilon<a_{N\left(n\right)}\leq s_{n}<s+\varepsilon$
\end_inset

 (zadnji neenačaj za 
\begin_inset Formula $n\geq n_{0}$
\end_inset

),
 skratka 
\begin_inset Formula $a_{N\left(n\right)}-s<\varepsilon$
\end_inset

 oziroma 
\begin_inset Formula $\forall n\geq n_{0}:\left|a_{N\left(n\right)}-s\right|<\varepsilon$
\end_inset

.
 Ker je 
\begin_inset Formula $N\left(n\right)\geq n$
\end_inset

,
 je 
\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
\end_inset

 neskončna množica,
 torej je neskončno mnogo členov v poljubni okolici 
\begin_inset Formula $s$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Standard
Treba je dokazati še,
 da 
\begin_inset Formula $\forall t>s:t$
\end_inset

 ni stekališče.
 Naj bo 
\begin_inset Formula $t>s$
\end_inset

.
 Označimo 
\begin_inset Formula $\delta\coloneqq t-s>0$
\end_inset

.
 Po definiciji
\begin_inset Foot
status open

\begin_layout Plain Layout
\begin_inset Formula $s$
\end_inset

 je limita zaporedja 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

,
 zato v poljubno majhni okolici obstaja tak 
\begin_inset Formula $s_{n_{1}}$
\end_inset

.
 
\begin_inset Formula $s_{n_{1}}$
\end_inset

 torej tu najdemo v 
\begin_inset Formula $[s,s+\frac{\delta}{2})$
\end_inset

.
\end_layout

\end_inset

 
\begin_inset Formula $s$
\end_inset

 
\begin_inset Formula $\exists n_{1}\in\mathbb{N}\ni:s\leq s_{n_{1}}<s+\frac{\delta}{2}<s+t$
\end_inset

.
 Ker 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 pada proti 
\begin_inset Formula $s$
\end_inset

,
 sledi 
\begin_inset Formula $\forall n\geq n_{1}:s\leq s_{n}<s+\frac{\delta}{2}$
\end_inset

.
 Po definiciji
\begin_inset Foot
status open

\begin_layout Plain Layout
\begin_inset Formula $s_{n}$
\end_inset

 je supremum členov od vključno 
\begin_inset Formula $n$
\end_inset

 dalje
\end_layout

\end_inset

 
\begin_inset Formula $s_{n}$
\end_inset

 sledi 
\begin_inset Formula $\forall n\geq n_{1}:a_{n}\leq s+\frac{\delta}{2}$
\end_inset

.
 Za takšne 
\begin_inset Formula $n$
\end_inset

 je 
\begin_inset Formula $\left|t-a_{n}\right|=t-a_{n}\geq t-\left(s+\frac{\delta}{2}\right)=\frac{\delta}{2}$
\end_inset

.
 Torej v 
\begin_inset Formula $\frac{\delta}{2}-$
\end_inset

okolici točke 
\begin_inset Formula $t$
\end_inset

 leži kvečjemu končno mnogo členov zaporedja oziroma členi 
\begin_inset Formula $\left(a_{1},a_{2},\dots,a_{n_{1}-1}\right)$
\end_inset

.
 Torej 
\begin_inset Formula $t$
\end_inset

 ni stekališče za 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s=\infty$
\end_inset

 Naj bo 
\begin_inset Formula $M>0$
\end_inset

 poljuben.
 Ker je 
\begin_inset Formula $s=\inf s_{n}$
\end_inset

,
 velja 
\begin_inset Formula $\forall n\in\mathbb{N}:s_{n}=\infty$
\end_inset

.
 Po definiciji 
\begin_inset Formula $s_{n}=\infty$
\end_inset

 velja 
\begin_inset Formula $\forall n\in\mathbb{N}\exists N\left(n\right):a_{N\left(n\right)}>M$
\end_inset

.
 Ker je 
\begin_inset Formula $N\left(n\right)\geq n$
\end_inset

,
 je 
\begin_inset Formula $\left\{ N\left(n\right);n\in\mathbb{N}\right\} $
\end_inset

 neskončna množica,
 torej je neskončno mnogo členov v 
\begin_inset Formula $\left(M,\infty\right)$
\end_inset

 za poljuben 
\begin_inset Formula $M$
\end_inset

,
 torej je 
\begin_inset Formula $s=\infty$
\end_inset

 res stekališče.
\end_layout

\begin_deeper
\begin_layout Standard
Večjih stekališč od 
\begin_inset Formula $\infty$
\end_inset

 očitno ni.
\end_layout

\end_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $s=-\infty$
\end_inset

 Naj bo 
\begin_inset Formula $m<0$
\end_inset

 poljuben.
 Ker je 
\begin_inset Formula $s=\inf s_{n}$
\end_inset

,
 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:s_{n_{0}}\in\left(-\infty,m\right)$
\end_inset

 Ker 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 pada proti 
\begin_inset Formula $s=-\infty$
\end_inset

,
 sledi 
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}:s_{n}\in\left(-\infty,m\right)$
\end_inset

.
 Po definiciji 
\begin_inset Formula $s_{n}$
\end_inset

 velja 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n}\in\left(-\infty,m\right)$
\end_inset

.
 Ker je za poljuben 
\begin_inset Formula $m$
\end_inset

 neskončno mnogo členov v 
\begin_inset Formula $\left(-\infty,m\right)$
\end_inset

,
 je 
\begin_inset Formula $s=-\infty$
\end_inset

 res stekališče.
\end_layout

\end_deeper
\begin_layout Subsection
Cauchyjev pogoj
\end_layout

\begin_layout Definition*
Zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 ustreza Cauchyjevemu pogoju (oz.
 je Cauchyjevo),
 če 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\varepsilon$
\end_inset

.
 ZDB Dovolj pozni členi so si poljubno blizu.
\end_layout

\begin_layout Claim*
Zaporedje v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 je konvergentno 
\begin_inset Formula $\Leftrightarrow$
\end_inset

 je Cauchyjevo.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Če 
\begin_inset Formula $a_{n}\to L$
\end_inset

,
 tedaj 
\begin_inset Formula $\left|a_{m}-a_{n}\right|=\left|\left(a_{m}-L\right)+\left(L-a_{n}\right)\right|\leq\left|a_{m}-\varepsilon\right|+\left|a_{n}-\varepsilon\right|$
\end_inset

.
 Cauchyjev pogoj sledi iz definicije limite za 
\begin_inset Formula $\frac{\varepsilon}{2}$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Če je zaporedje Cauchyjevo,
 je omejeno:
 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|\leq1$
\end_inset

.
 V posebnem,
 
\begin_inset Formula $m=n_{0}$
\end_inset

,
 
\begin_inset Formula $\left|a_{n_{0}}-a_{n}\right|\leq1$
\end_inset

 oziroma 
\begin_inset Formula $\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow a_{n}\in\left[a_{n_{0}}-1,a_{n_{0}}+1\right]$
\end_inset

.
 Preostali členi tvorijo končno veliko množico,
 ki ima 
\begin_inset Formula $\min$
\end_inset

 in 
\begin_inset Formula $\max$
\end_inset

,
 torej je 
\begin_inset Formula $\left\{ a_{k};k\in\mathbb{N}\right\} =\left\{ a_{1},a_{2},\dots,a_{n_{0}-1}\right\} \cup\left\{ a_{k};k\in\mathbb{N},k\geq n_{0}\right\} $
\end_inset

 tudi omejena.
 Po 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{bw}{izreku od prej}
\end_layout

\end_inset

 sledi,
 da ima zaporedje stekališče 
\begin_inset Formula $s$
\end_inset

.
 Dokažimo,
 da je 
\begin_inset Formula $s=\lim_{n\to\infty}a_{n}$
\end_inset

.
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Ker je 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 Cauchyjevo,
 
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\ni:\forall m,n\in\mathbb{N}:m,n\geq n_{0}\Rightarrow\left|a_{m}-a_{n}\right|<\frac{\varepsilon}{2}$
\end_inset

.
 Po definiciji 
\begin_inset Formula $s$
\end_inset

 
\begin_inset Formula $\exists n_{1}\geq n_{0}\ni:\left|a_{n_{1}}-s\right|<\frac{\varepsilon}{2}$
\end_inset

.
 Sledi 
\begin_inset Formula $\forall n\geq n_{0}:\left|a_{n}-s\right|=\left|a_{n}-s+s-a_{n_{1}}\right|\leq\left|a_{n}-s\right|+\left|s-a_{n_{1}}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Remark*
Moč izreka je v tem,
 da lahko konvergenco preverjamo tudi tedaj,
 ko nimamo kandidatov za limito.
\end_layout

\begin_layout Section
Številske vrste
\end_layout

\begin_layout Standard
Kako sešteti neskončno mnogo števil?
 Nadgradimo pristop končnih vsot na neskončne vsote!
\end_layout

\begin_layout Definition*
Imejmo zaporedje 
\begin_inset Formula $\left(a_{k}\right)_{k\in\mathbb{N}},a_{k}\in\mathbb{R}$
\end_inset

.
 Izraz 
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset

 se imenuje vrsta s členi 
\begin_inset Formula $a_{j}$
\end_inset

.
 Pomen izraza opredelimo na naslednjo način:
\end_layout

\begin_layout Definition*
Tvorimo novo zaporedje,
 pravimo mu zaporedje delnih vsot vrste:
 
\begin_inset Formula $s_{1}=a_{1}$
\end_inset

,
 
\begin_inset Formula $s_{2}=a_{1}+a_{2}$
\end_inset

,
 
\begin_inset Formula $s_{3}=a_{1}+a_{2}+a_{3}$
\end_inset

,
 ...,
 
\begin_inset Formula $s_{n}=a_{1}+a_{2}+\cdots+a_{n}=\sum_{j=1}^{n}a_{j}$
\end_inset

 —
 številu 
\begin_inset Formula $s_{n}$
\end_inset

 pravimo 
\begin_inset Formula $n-$
\end_inset

ta delna vsota.
\end_layout

\begin_layout Definition*
Vrsta je konvergentna,
 če je v 
\begin_inset Formula $\mathbb{R}$
\end_inset

 konvergentno zaporedje 
\begin_inset Formula $\left(s_{n}\right)_{n\in\mathbb{N}}$
\end_inset

.
 Številu 
\begin_inset Formula $s=\lim_{n\to\infty}s_{n}$
\end_inset

 tedaj pravimo vsota vrste in pišemo 
\begin_inset Formula $s\eqqcolon\sum_{j=1}^{\infty}a_{j}$
\end_inset

.
 Pojem neskončne vsote torej prevedemo na pojem limite pridruženega zaporedja delnih vsot.
 Včasih vrsto (kot operacijo) enačimo z njeno vsoto (izidom operacije).
\end_layout

\begin_layout Definition*
Če vrsta ni konvergentna,
 rečemo,
 da je divergentna.
 Enako,
 če je 
\begin_inset Formula $s\in\left\{ \pm\infty\right\} $
\end_inset

.
\end_layout

\begin_layout Example*
Primeri vrst.
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $a_{n}=\frac{1}{2^{n}}$
\end_inset

,
 torej zaporedje 
\begin_inset Formula $\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots$
\end_inset

.
 Ali se sešteje v 1?
 Velja 
\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=1}^{n}a_{j}$
\end_inset

.
 Pišimo 
\begin_inset Formula $q=\frac{1}{2}$
\end_inset

,
 tedaj 
\begin_inset Formula $a_{n}=q^{n}$
\end_inset

 in 
\begin_inset Formula 
\[
s_{n}=q+q^{2}+q^{3}+\cdots+q^{n}=q\left(1+q+q^{2}+\cdots+q^{n-1}\right)=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)\left(1-q\right)}{1-q}=
\]

\end_inset


\begin_inset Formula 
\[
=q\frac{\left(1+q+q^{2}+\cdots+q^{n-1}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n}\right)}{1-q}=q\frac{1-q^{n}}{1-q}=\frac{q}{1-q}\left(1-q^{n}\right)
\]

\end_inset

Izračunajmo 
\begin_inset Formula $\lim_{n\to\infty}s_{n}=\lim_{n\to\infty}\frac{q}{1-q}\left(1-\cancelto{0}{q^{n}}\right)=\frac{q}{1-q}$
\end_inset

 (velja,
 ker 
\begin_inset Formula $q\in\left(-1,1\right)$
\end_inset

),
 torej je 
\begin_inset Formula $s=\sum_{n=1}^{\infty}q^{n}=\frac{q}{1-q}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Geometrijska vrsta (splošno).
 Naj bo 
\begin_inset Formula $q\in\mathbb{R}$
\end_inset

.
 Vrsta 
\begin_inset Formula $\sum_{j=0}^{\infty}q^{j}$
\end_inset

 se imenuje geometrijska vrsta.
 Velja 
\begin_inset Formula $s=\lim_{n\to\infty}\sum_{j=0}^{n}q^{j}$
\end_inset

 in 
\begin_inset Formula $s_{n}=1+q+q^{2}+q^{3}+\cdots+q^{n}$
\end_inset

.
 Če je 
\begin_inset Formula $q=1$
\end_inset

,
 je 
\begin_inset Formula $s_{n}=n+1$
\end_inset

,
 sicer množimo izraz z 
\begin_inset Formula $\left(1-q\right)$
\end_inset

:
\begin_inset Formula 
\[
\left(1+q+q^{2}+\cdots+q^{n}\right)\left(1-q\right)=\left(1+q+q^{2}+\cdots+q^{n}\right)-\left(q+q^{2}+q^{3}+\cdots+q^{n+1}\right)=1-q^{n+1}
\]

\end_inset

torej 
\begin_inset Formula $s_{n}=\frac{1-q^{n+1}}{1-q}$
\end_inset

 in vrsta konvergira 
\begin_inset Formula $\Leftrightarrow q\not=1$
\end_inset

 in 
\begin_inset Formula $\lim_{n\to\infty}\frac{1-q^{n+1}}{1-q}\exists$
\end_inset

 v 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
 To pa se zgodi natanko za 
\begin_inset Formula $q\in\left(-1,1\right)$
\end_inset

,
 takrat je 
\begin_inset Formula $\lim_{n\to\infty}\frac{1-\cancelto{0}{q^{n+1}}}{1-q}=\frac{1}{1-q}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Harmonična vrsta.
 Je vrsta 
\begin_inset Formula $\sum_{j=1}^{\infty}\frac{1}{j}$
\end_inset

.
 Velja 
\begin_inset Formula $\frac{1}{j}\underset{j\to\infty}{\longrightarrow}0$
\end_inset

,
 toda vrsta divergira.
 Dokaz sledi kmalu malce spodaj.
\end_layout

\end_deeper
\begin_layout Question*
Kako lahko enostavno določimo,
 ali dana vrsta konvergira?
\end_layout

\begin_layout Subsection
Konvergenčni kriteriji
\end_layout

\begin_layout Theorem*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hypertarget{cauchyvrste}{Cauchyjev pogoj}
\end_layout

\end_inset

.
 Vrsta 
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset

 je konvergentna 
\begin_inset Formula $\Leftrightarrow$
\end_inset

 delne vrste ustrezajo Cauchyjevemu pogoju;
 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n,m\in\mathbb{N}:n,m\geq n_{0}\Rightarrow\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|<\varepsilon$
\end_inset

.
\end_layout

\begin_layout Corollary*
\begin_inset Formula $\sum_{j=1}^{\infty}a_{j}$
\end_inset

 konvergira 
\begin_inset Formula $\Rightarrow\lim_{j\to\infty}a_{j}=0$
\end_inset

.
\end_layout

\begin_layout Proof
Uporabimo izrek zgoraj za 
\begin_inset Formula $n=m-1$
\end_inset

:
 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|s_{n}-s_{n+1}\right|=\left|a_{n}\right|<\varepsilon$
\end_inset

.
\end_layout

\begin_layout Example*
Vrsti 
\begin_inset Formula $\sum_{j=1}^{\infty}\cos n$
\end_inset

 in 
\begin_inset Formula $\sum_{j=1}^{\infty}\sin n$
\end_inset

 divergirata,
 saj smo videli,
 da členi ne ene ne druge ne konvergirajo nikamor,
 torej tudi ne proti 0,
 kar je potreben pogoj za konvergenco vrste.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
Harmonična vrsta divergira.
 Protiprimer Cauchyjevega pogoja:
 Naj bo 
\begin_inset Formula $\varepsilon=\frac{1}{4}$
\end_inset

.
 Tedaj ne glede na izbiro 
\begin_inset Formula $n_{0}$
\end_inset

 najdemo:
\begin_inset Formula 
\[
s_{2n}-s_{n}=\sum_{j=n+1}^{2n}\frac{1}{j}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}
\]

\end_inset

Dokaz divergence brez Cauchyjevega pogoja:
 
\begin_inset Formula $s_{2^{n}}=a_{1}+\sum_{j=1}^{n}\left(s_{2^{j}}-s_{s^{j-1}}\right)>1+\frac{n}{2}$
\end_inset

 in 
\begin_inset Formula $\lim_{n\to\infty}1+\frac{n}{2}=\infty$
\end_inset

.
\begin_inset Note Note
status open

\begin_layout Plain Layout
Geometrični argument za divergenco:
 TODO XXX FIXME DODAJ
\end_layout

\end_inset


\end_layout

\begin_layout Theorem*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hypertarget{pk}{Primerjalni kriterij}
\end_layout

\end_inset

.
 Naj bosta 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 in 
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset

 vrsti z nenegativnimi členi.
 Naj bo 
\begin_inset Formula $\forall k\geq k_{0}:a_{k}\leq b_{k}$
\end_inset

 (od nekod naprej) —
 pravimo,
 da je 
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset

 majoranta za 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 od nekod naprej.
\end_layout

\begin_deeper
\begin_layout Itemize
Če 
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}$
\end_inset

 konvergira,
 tedaj tudi 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 konvergira.
\end_layout

\begin_layout Itemize
Če 
\begin_inset Formula $\text{\ensuremath{\sum_{n=1}^{\infty}a_{n}=\infty}}$
\end_inset

,
 tedaj tudi 
\begin_inset Formula $\sum_{n=1}^{\infty}b_{n}=\infty$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Example*
Videli smo,
 da 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k}$
\end_inset

 divergira.
 Kaj pa 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
\end_inset

?
 Preverimo naslednje in uporabimo primerjalni kriterij:
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall k\in\mathbb{N}:\frac{1}{k^{2}}\leq\frac{2}{k\left(k+1\right)}$
\end_inset

?
 Računajmo 
\begin_inset Formula $k^{2}\geq\frac{k\left(k+1\right)}{2}\sim k\geq\frac{k+1}{2}\sim\frac{k}{2}\geq\frac{1}{2}$
\end_inset

.
 Velja,
 ker 
\begin_inset Formula $k\in\mathbb{N}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Vrsta 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}$
\end_inset

 konvergira?
 Opazimo 
\begin_inset Formula $\frac{1}{k}-\frac{1}{k+1}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k\left(k+1\right)}$
\end_inset

.
 Za delne vsote vrste 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)}$
\end_inset

 velja:
\begin_inset Formula 
\[
\sum_{k=1}^{n}\frac{1}{k\left(k+1\right)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}\underset{n\to\infty}{\longrightarrow}1,
\]

\end_inset

torej 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{2}{k\left(k+1\right)}=2$
\end_inset

.
 Posledično po primerjalnem kriteriju tudi 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$
\end_inset

 konvergira.
 Izkaže se 
\begin_inset Formula $\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}\approx1,645$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Theorem*
Kvocientni oz.
 d'Alembertov kriterij.
 Za vrsto s pozitivnimi členi 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 definirajmo 
\begin_inset Formula $D_{n}\coloneqq\frac{a_{n+1}}{a_{n}}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n\geq n_{0}:D_{n}\leq q\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
\end_inset

 (vrsta konvergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:D_{n}\geq1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
\end_inset

 (vrsta divergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\exists D=\lim_{n\to\infty}D_{n}\in\mathbb{R}\Longrightarrow$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:kvocientni3a"

\end_inset


\begin_inset Formula $D<1\Longrightarrow\sum_{n=1}^{\infty}a_{n}<\infty$
\end_inset

 (vrsta konvergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $D>1\Longrightarrow\sum_{n=1}^{\infty}a_{n}=\infty$
\end_inset

 (vrsta divergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $D=1\Longrightarrow$
\end_inset

 s tem kriterijem ne moremo določiti konvergence.
\end_layout

\end_deeper
\end_deeper
\begin_layout Proof
Razlaga.
 
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\forall n>n_{0}:D_{n}\leq q$
\end_inset

,
 torej 
\begin_inset Formula $\frac{a_{n+1}}{a_{n}}\leq q\sim a_{n+1}\leq qa_{n}$
\end_inset

 in hkrati 
\begin_inset Formula $\text{\ensuremath{\frac{a_{n+2}}{a_{n+1}}\leq q\sim a_{n+2}\leq qa_{n+1}}}$
\end_inset

,
 torej skupaj 
\begin_inset Formula $a_{n+2}\leq qa_{n+1}\leq qqa_{n}=q^{2}a_{n}$
\end_inset

,
 sledi 
\begin_inset Formula $q_{n+2}\leq q^{2}a_{n}$
\end_inset

 in 
\begin_inset Formula $\forall k\in\mathbb{N}:q_{n+k}\leq q^{k}a_{n}$
\end_inset

.
 Vrsto smo majorizirali z geometrijsko vrsto,
 ki ob 
\begin_inset Formula $q\in\left(0,1\right)$
\end_inset

 konvergira po primerjalnem kriteriju,
 zato tudi naša vrsta konvergira.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall n>n_{0}:\frac{a_{n+1}}{a_{n}}\geq D\geq1$
\end_inset

,
 torej 
\begin_inset Formula $a_{n+1}\geq a_{n}$
\end_inset

 in hkrati 
\begin_inset Formula $a_{n+2}\geq a_{n+1}$
\end_inset

,
 torej skupaj 
\begin_inset Formula $a_{n+2}\geq a_{n}$
\end_inset

,
 sledi 
\begin_inset Formula $\forall k\in\mathbb{N}:a_{n+k}\geq a_{n}$
\end_inset

.
 Naša vrsta torej majorizira konstantno vrsto,
 ki očitno divergira;
 
\begin_inset Formula $\sum_{k=n_{0}}^{\infty}a_{k}\geq\sum_{k=n_{0}}^{\infty}a_{n}=0$
\end_inset

.
 Potemtakem tudi naša vrsta divergira.
 Poleg tega niti ne velja 
\begin_inset Formula $a_{k}\underset{k\to\infty}{\longrightarrow}0$
\end_inset

,
 torej vrsta gotovo divergira.
\end_layout

\begin_layout Enumerate
Enako kot 1 in 2.
\end_layout

\end_deeper
\begin_layout Example*
Za 
\begin_inset Formula $x>0$
\end_inset

 definiramo 
\begin_inset Formula $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$
\end_inset

.
 Vrsta res konvergira po točki 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:kvocientni3a"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\begin_inset Formula 
\[
D_{n}=\frac{\frac{x^{n+1}}{\left(n+1\right)!}}{\frac{x^{n}}{n!}}=\frac{x^{n+1}n!}{x^{n}\left(n+1\right)!}=\frac{x}{n+1}\underset{n\to\infty}{\longrightarrow}0
\]

\end_inset


\end_layout

\begin_layout Theorem*
Korenski oz.
 Cauchyjev kriterij.
 Naj bo 
\begin_inset Formula $\sum_{k=1}^{\infty}a_{k}$
\end_inset

 vrsta z nenegativnimi členi.
 Naj bo 
\begin_inset Formula $c_{n}\coloneqq\sqrt[n]{a_{n}}$
\end_inset

.ž
\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N},q\in\left(0,1\right)\forall n>n_{0}:c_{n}\leq q\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
\end_inset

 (vrsta konvergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\exists n_{0}\in\mathbb{N}\forall n>n_{0}:c_{n}\geq1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
\end_inset

 (vrsta divergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\exists c=\lim_{n\to\infty}c_{n}\in\mathbb{R}\Longrightarrow$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
\begin_inset Formula $c<1\Longrightarrow\sum_{k=1}^{\infty}a_{k}<\infty$
\end_inset

 (vrsta konvergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $c>1\Longrightarrow\sum_{k=1}^{\infty}a_{k}=\infty$
\end_inset

 (vrsta divergira)
\end_layout

\begin_layout Enumerate
\begin_inset Formula $c=1\Longrightarrow$
\end_inset

 s tem kriterijem ne moremo določiti konvergence.
\end_layout

\end_deeper
\end_deeper
\begin_layout Proof
Skica dokazov.
\end_layout

\begin_deeper
\begin_layout Enumerate
Velja 
\begin_inset Formula $\forall n>n_{0}:c_{n}\leq q$
\end_inset

.
 To pomeni 
\begin_inset Formula $\sqrt[n]{a_{n}}\leq q$
\end_inset

,
 torej 
\begin_inset Formula $a_{n}\leq q^{n}$
\end_inset

 in 
\begin_inset Formula $a_{n+1}\leq q^{n+1}$
\end_inset

,
 torej je vrsta majorizirana z geometrijsko vrsto 
\begin_inset Formula $\sum_{n=1}^{\infty}q^{n}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Velja 
\begin_inset Formula $\forall n>n_{0}:c_{n}\geq1$
\end_inset

.
 To pomeni 
\begin_inset Formula $\sqrt[n]{a_{n}}\geq1$
\end_inset

,
 torej 
\begin_inset Formula $a_{n}\geq1$
\end_inset

,
 torej je vrsta majorizirana s konstantno in zato divergentno vrsto 
\begin_inset Formula $\sum_{n=1}^{\infty}1$
\end_inset

.
\end_layout

\begin_layout Enumerate
Enako kot 1 in 2.
\end_layout

\end_deeper
\begin_layout Subsection
Alternirajoče vrste
\end_layout

\begin_layout Definition*
Vrsta je alternirajoča,
 če je predznak naslednjega člena nasproten predznaku tega člena.
 ZDB 
\begin_inset Formula $\forall n\in\mathbb{N}:\sgn a_{n+1}=-\sgn a_{n}$
\end_inset

,
 kjer je 
\begin_inset Formula $\sgn:\mathbb{R}\to\left\{ -1,0,1\right\} $
\end_inset

 s predpisom 
\begin_inset Formula $\sgn a=\begin{cases}
-1 & ;a<0\\
1 & ;a>0\\
0 & ;a=0
\end{cases}$
\end_inset

.
 ZDB 
\begin_inset Formula $\forall n\in\mathbb{N}:a_{n+1}a_{n}\leq0$
\end_inset

.
\end_layout

\begin_layout Theorem*
Leibnizov konvergenčni kriterij.
 Naj bo 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 padajoče zaporedje in 
\begin_inset Formula $\lim_{n\to\infty}a_{n}=0$
\end_inset

.
 Tedaj vrsta 
\begin_inset Formula $\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset

 konvergira.
 Če je 
\begin_inset Formula $s\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset

 in 
\begin_inset Formula $s_{n}\coloneqq\sum_{k=1}^{\infty}\left(-1\right)^{k}a_{k}$
\end_inset

,
 tedaj 
\begin_inset Formula $\left|s-s_{k}\right|\leq a_{n+1}$
\end_inset

.
\end_layout

\begin_layout Proof
Skica dokaza.
 Vidimo,
 da delne vsote 
\begin_inset Formula $s_{2n}$
\end_inset

 padajo k 
\begin_inset Formula $s''$
\end_inset

 in delne vsote 
\begin_inset Formula $s_{2n-1}$
\end_inset

 naraščajo k 
\begin_inset Formula $s'$
\end_inset

.
 Toda ker 
\begin_inset Formula $s_{2n}-s_{2n-1}=a_{2n}$
\end_inset

,
 velja 
\begin_inset Formula $s'=s''$
\end_inset

.
 Limita razlike dveh zaporedij je razlika limit teh dveh zaporedij,
 torej 
\begin_inset Formula $s'=s''=s$
\end_inset

.
 
\begin_inset Formula $s$
\end_inset

 je supremum lihih in infimum sodih vsot.
 
\begin_inset Formula $\left|s-s_{n}\right|\leq\left|s_{n+1}-s_{n}\right|=a_{n+1}$
\end_inset

.
\end_layout

\begin_layout Example*
Harmonična vrsta 
\begin_inset Formula $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\to\infty$
\end_inset

,
 toda alternirajoča harmonična vrsta 
\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\to\log2$
\end_inset

.
\end_layout

\begin_layout Subsection
Absolutno konvergentne vrste
\end_layout

\begin_layout Definition*
Vrsta 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 je absolutno konvergentna,
 če je 
\begin_inset Formula $\sum_{n=1}^{\infty}\left|a_{n}\right|$
\end_inset

 konvergentna.
\end_layout

\begin_layout Theorem*
Absolutna konvergenca 
\begin_inset Formula $\Rightarrow$
\end_inset

 konvergenca.
\end_layout

\begin_layout Proof
Uporabimo 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{cauchyvrste}{Cauchyjev pogoj za konvergenco vrst}
\end_layout

\end_inset

 in trikotniško neenakost.
\begin_inset Formula 
\[
\left|s_{m}-s_{n}\right|=\left|\sum_{j=n+1}^{m}a_{j}\right|\leq\sum_{j=n+1}^{m}\left|a_{j}\right|<\varepsilon
\]

\end_inset

za 
\begin_inset Formula $m,n\geq n_{0}$
\end_inset

 za nek 
\begin_inset Formula $n_{0}$
\end_inset

.
\end_layout

\begin_layout Remark*
Obrat ne velja,
 protiprimer je alternirajoča harmonična vrsta.
\end_layout

\begin_layout Subsection
Pogojno konvergentne vrste
\end_layout

\begin_layout Standard
\begin_inset Formula $\sum_{k=0}^{\infty}2-\sum_{k=0}^{\infty}1\not=\sum_{k=0}^{\infty}\left(2-1\right)$
\end_inset

,
 temveč 
\begin_inset Formula $\infty-\infty=$
\end_inset

 nedefinirano.
\end_layout

\begin_layout Question*
Ross-Littlewoodov paradoks.
 Ali smemo zamenjati vrstni red seštevanja,
 če imamo neskončno mnogo sumandov?
\end_layout

\begin_layout Standard
Najprej vprašanje natančneje opredelimo in vpeljimo orodja za njegovo obravnavo.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\mathcal{M}\subset\mathbb{N}$
\end_inset

.
 Permutacija 
\begin_inset Formula $\mathcal{M}$
\end_inset

 je vsaka bijektivna preslikava 
\begin_inset Formula $\pi:\mathcal{M}\to\mathcal{M}$
\end_inset

.
 Če je 
\begin_inset Formula $\mathcal{M}=\left\{ a_{1},\dots,a_{n}\right\} $
\end_inset

 končna množica,
 tedaj 
\begin_inset Formula $\pi$
\end_inset

 označimo s tabelo:
\begin_inset Formula 
\[
\left(\begin{array}{ccc}
a_{1} & \cdots & a_{n}\\
\pi\left(a_{1}\right) & \cdots & \pi\left(a_{n}\right)
\end{array}\right)
\]

\end_inset


\end_layout

\begin_layout Example*
\begin_inset Formula 
\[
\pi=\left(\begin{array}{ccccc}
1 & 2 & 3 & 4 & 5\\
5 & 3 & 1 & 4 & 2
\end{array}\right)
\]

\end_inset


\end_layout

\begin_layout Definition*
Vrsta 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 je brezpogojno konvergentna,
 če za vsako permutacijo 
\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}$
\end_inset

 vrsta 
\begin_inset Formula $\sum_{n=1}^{\infty}\pi\left(a_{n}\right)$
\end_inset

 konvergira in vsota ni odvisna od 
\begin_inset Formula $\pi$
\end_inset

.
 Vrsta je pogojno konvergentna,
 če je konvergentna,
 toda ne brezpogojno.
\end_layout

\begin_layout Example*
\begin_inset Formula $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots$
\end_inset

 je pogojno konvergentna,
 ker pri seštevanju z vrstnim redom,
 pri katerem tisočim pozitivnim členom sledi en negativen in njemu zopet tisoč pozitivnih itd.,
 vrsta ne konvergira.
\end_layout

\begin_layout Theorem*
Absolutna konvergenca 
\begin_inset Formula $\Leftrightarrow$
\end_inset

 Brezpogojna konvergenca
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Theorem*
Riemannov sumacijski izrek.
 Če je vrsta pogojno konvergentna,
 tedaj 
\begin_inset Formula $\forall x\in\mathbb{R}\cup\left\{ \pm\infty\right\} \exists$
\end_inset

 permutacija 
\begin_inset Formula $\pi:\mathbb{N}\to\mathbb{N}\ni:\sum_{n=1}^{\infty}a_{\pi\left(n\right)}=x$
\end_inset

.
 ZDB Končna vsota je lahko karkoli,
 če lahko poljubno spremenimo vrstni red seštevanja.
 Prav tako obstaja taka permutacija 
\begin_inset Formula $\pi$
\end_inset

,
 pri kateri 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{\pi\left(n\right)}$
\end_inset

 nima vsote ZDB delne vsotee ne konvergirajo.
\end_layout

\begin_layout Example*
\begin_inset Formula $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}$
\end_inset

.
\end_layout

\begin_layout Section
Funkcijske vrste
\end_layout

\begin_layout Standard
Tokrat poskušamo seštevati funkcije.
 V prejšnjem razdelku seštevamo le realna števila.
 Funkcijska vrsta,
 če je 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 zaporedije funkcij 
\begin_inset Formula $X\to\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $x$
\end_inset

 zunanja konstanta,
 izgleda takole:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\sum_{n=1}^{\infty}a_{n}\left(x\right)
\]

\end_inset


\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $X$
\end_inset

 neka množica in 
\begin_inset Formula $\Phi=\left\{ \varphi_{n}:X\to\mathbb{R},n\in\mathbb{N}\right\} $
\end_inset

 družina funkcij.
\end_layout

\begin_layout Definition*
Pravimo,
 da funkcije 
\begin_inset Formula $\varphi_{n}$
\end_inset

 konvergirajo po točkah na 
\begin_inset Formula $X$
\end_inset

,
 če je 
\begin_inset Formula $\forall x\in X$
\end_inset

 zaporedje 
\begin_inset Formula $\left(\varphi_{n}\left(x\right)\right)_{n\in\mathbb{N}}$
\end_inset

 konvergentno.
\end_layout

\begin_layout Definition*
Označimo limito s 
\begin_inset Formula $\varphi\left(x\right)$
\end_inset

.
 ZDB to pomeni,
 da 
\begin_inset Formula 
\[
\forall\varepsilon>0,x\in X:\exists n_{0}=n_{0}\left(\varepsilon,x\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon.
\]

\end_inset


\end_layout

\begin_layout Definition*
Pravimo,
 da funkcije 
\begin_inset Formula $\varphi_{n}$
\end_inset

 konvergirajo enakomerno na 
\begin_inset Formula $X$
\end_inset

,
 če 
\begin_inset Formula 
\[
\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall x\in X,n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon
\]

\end_inset

 oziroma ZDB 
\begin_inset Formula 
\[
\forall\varepsilon>0\exists n_{0}=n_{0}\left(\varepsilon\right)\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\sup_{x\in X}\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|\leq\varepsilon.
\]

\end_inset


\end_layout

\begin_layout Definition*
Poudariti je treba,
 da je pri konvergenci po točkah 
\begin_inset Formula $n_{0}$
\end_inset

 lahko odvisen od 
\begin_inset Formula $\varepsilon$
\end_inset

 in 
\begin_inset Formula $x$
\end_inset

,
 pri enakomerni konvergenci pa le od 
\begin_inset Formula $\varepsilon$
\end_inset

.
\end_layout

\begin_layout Note*
Očitno enakomerna konvergenca implicira konvergenco po točkah,
 obratno pa ne velja.
\end_layout

\begin_layout Example*
Za 
\begin_inset Formula $n\in\mathbb{N}$
\end_inset

 definiramo 
\begin_inset Formula $\varphi_{n}:\left[0,1\right]\to\left[0,1\right]$
\end_inset

 s predpisom 
\begin_inset Formula $\varphi_{n}\left(x\right)=x^{n}$
\end_inset

.
 Tedaj obstaja 
\begin_inset Formula $\varphi\left(x\right)\coloneqq\lim_{n\to\infty}\varphi_{n}\left(x\right)=\begin{cases}
0 & ;x\in[0,1)\\
1 & ;x=1
\end{cases}$
\end_inset

.
 Torej po definiciji velja 
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset

 po točkah,
 toda ne velja 
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset

 enakomerno.
 Za poljubno velik pas okoli 
\begin_inset Formula $\varphi\left(x\right)$
\end_inset

 bodo še tako pozne funkcijske vrednosti 
\begin_inset Formula $\varphi_{n}\left(x\right)$
\end_inset

 od nekega 
\begin_inset Formula $x$
\end_inset

 dalje izven tega pasu.
 Če bi 
\begin_inset Formula $\varphi_{n}\to\varphi$
\end_inset

 enakomerno,
 tedaj bi za poljuben 
\begin_inset Formula $\varepsilon\in\left(0,1\right)$
\end_inset

 in dovolj pozne 
\begin_inset Formula $n$
\end_inset

 (večje od nekega 
\begin_inset Formula $n_{0}\in\mathbb{N}$
\end_inset

) veljalo 
\begin_inset Formula $\forall x\in\left[0,1\right]:\left|\varphi_{n}\left(x\right)-\varphi\left(x\right)\right|<\varepsilon$
\end_inset

.
 To je ekvivalentno 
\begin_inset Formula $\forall x\in\left(0,1\right):\left|x^{n}\right|<\varepsilon\Leftrightarrow n\log x<\log\varepsilon\Leftrightarrow n>\frac{\log\varepsilon}{\log x}$
\end_inset

.
 Toda 
\begin_inset Formula $\lim_{x\nearrow1}\frac{\log\varepsilon}{\log x}=\infty$
\end_inset

,
 zato tak 
\begin_inset Formula $n$
\end_inset

 ne obstaja.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $X$
\end_inset

 neka množica in 
\begin_inset Formula $\left(f_{j}:X\to\mathbb{R}\right)_{j\in\mathbb{N}}$
\end_inset

 dano zaporedje funkcij.
 Pravimo,
 da funkcijska vrsta 
\begin_inset Formula $\sum_{j=1}^{\infty}f_{j}$
\end_inset

 konvergira po točkah na 
\begin_inset Formula $X$
\end_inset

,
 če 
\begin_inset Formula $\forall x\in X:\sum_{j=1}^{\infty}f_{j}\left(x\right)<0$
\end_inset

 (številska vrsta je konvergentna).
 ZDB to pomeni,
 da funkcijsko zaporedje delnih vsot 
\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
\end_inset

 konvergira po točkah na 
\begin_inset Formula $X$
\end_inset

.
\end_layout

\begin_layout Definition*
Funkcijska vrsta 
\begin_inset Formula $s=\sum_{j=1}^{\infty}$
\end_inset

 konvergira enakomerno na 
\begin_inset Formula $X$
\end_inset

,
 če funkcijsko zaporedje delnih vsot 
\begin_inset Formula $s_{n}\coloneqq\sum_{j=1}^{n}f_{j}$
\end_inset

 konvergira enakomerno na 
\begin_inset Formula $X$
\end_inset

.
\end_layout

\begin_layout Definition*
Funkcija oblike 
\begin_inset Formula $x\mapsto\sum_{j=1}^{\infty}f_{j}\left(x\right)$
\end_inset

 se imenuje funkcijska vrsta.
\end_layout

\begin_layout Exercise*
Dokaži,
 da 
\begin_inset Formula $\sum_{n=1}^{\infty}x^{n}$
\end_inset

 ne konvergira enakomerno!
 Vrsta konvergira po točkah le na intervalu 
\begin_inset Formula $x\in\left(0,1\right)$
\end_inset

,
 za druge 
\begin_inset Formula $x$
\end_inset

 divergira.
 Ko fiksiramo zunanjo konstanto,
 gre za geometrijsko vrsto.
 Delna vsota 
\begin_inset Formula $\sum_{j=1}^{n}x^{j}=\frac{x\left(1-x^{n}\right)}{1-x}$
\end_inset

.
 Velja 
\begin_inset Formula $\lim_{n\to\infty}\frac{x\left(1-x^{n}\right)}{1-x}=x\lim_{n\to\infty}\frac{1-\cancelto{0}{x^{n}}}{1-x}=\frac{x}{1-x}$
\end_inset

.
 Sedaj prevedimo,
 ali 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall x\in\left(-1,1\right),n\geq n_{0}:\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|<\varepsilon$
\end_inset

.
 Za začetekk si oglejmo le 
\begin_inset Formula $x>0$
\end_inset

.
 Ker je tedaj 
\begin_inset Formula $\frac{x\left(1-x^{n}\right)}{1-x}<\frac{x}{1-x}$
\end_inset

,
 je 
\begin_inset Formula $\left|\frac{x\left(1-x^{n}\right)}{1-x}-\frac{x}{1-x}\right|=\frac{x}{1-x}-\frac{x\left(1-x^{n}\right)}{1-x}=\frac{\cancel{x-x+}x^{n+1}}{1-x}$
\end_inset

.
 Računajmo sedaj 
\begin_inset Formula $\frac{x^{n+1}}{1-x}<\varepsilon\sim x^{n+1}<\varepsilon\left(1-x\right)\sim\left(n+1\right)\log x<\log\left(\varepsilon\left(1-x\right)\right)\sim n+1>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}\sim n>\frac{\log\left(\varepsilon\left(1-x\right)\right)}{\log x}-1$
\end_inset

.
 Ker je 
\begin_inset Formula $n$
\end_inset

 odvisen od 
\begin_inset Formula $x$
\end_inset

,
 vsota ni enakomerno konvergentna.
\end_layout

\begin_layout Standard
Poseben primer funkcijskih vrst so funkcijske vrste funkcij oblike 
\begin_inset Formula $f_{j}=b_{j}\cdot x^{j}$
\end_inset

,
 torej potence (monomi).
\end_layout

\begin_layout Definition*
Potenčna vrsta je funkcijska vrsta oblike 
\begin_inset Formula $\sum_{j=1}^{\infty}b_{j}\cdot x^{j}$
\end_inset

,
 kjer so a 
\begin_inset Formula $\left(b_{j}\right)_{j\in\mathbb{N}}$
\end_inset

 dana realna števila.
\end_layout

\begin_layout Theorem*
Cauchy-Hadamard.
 Za vsako potenčno vrsto obstaja konvergenčni radij 
\begin_inset Formula $R\in\left[0,\infty\right]\ni:$
\end_inset


\end_layout

\begin_deeper
\begin_layout Itemize
vrsta absolutno konvergira za 
\begin_inset Formula $\left|x\right|<R$
\end_inset

,
\end_layout

\begin_layout Itemize
vrsta divergira za 
\begin_inset Formula $\left|x\right|>R$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Theorem*
Velja 
\begin_inset Formula $\text{\ensuremath{\frac{1}{R}=\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}}$
\end_inset

,
 kjer vzamemo 
\begin_inset Formula $\frac{1}{0}\coloneqq\infty$
\end_inset

 in 
\begin_inset Formula $\frac{1}{\infty}\coloneqq0$
\end_inset

.
\end_layout

\begin_layout Proof
Rezultat že poznamo za zelo poseben primer 
\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
\end_inset

 (geometrijska vrsta).
 Ideja dokaza je,
 da konvergenco vsake potenčne vrste opišemo s pomočjo geometrijske vrste.
\end_layout

\begin_deeper
\begin_layout Itemize
Konvergenca:
 Za 
\begin_inset Formula $x=0$
\end_inset

 vrsta očitno konvergira,
 zato privzamemo 
\begin_inset Formula $x\not=0$
\end_inset

.
 Definirajmo 
\begin_inset Formula $R$
\end_inset

 s formulo iz definicije (
\begin_inset Formula $R=\frac{1}{\limsup_{k\to\infty}\sqrt[k]{\left|b_{k}\right|}}$
\end_inset

).
 Naj bo 
\begin_inset Formula $x$
\end_inset

 tak,
 da 
\begin_inset Formula $\left|x\right|<R\leq\infty$
\end_inset

 (sledi 
\begin_inset Formula $R>0$
\end_inset

).
 Naj bo 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Tedaj po definiciji 
\begin_inset Formula $R$
\end_inset

 velja 
\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\leq\frac{1}{R}+\varepsilon$
\end_inset

 za vse dovolj velike 
\begin_inset Formula $k$
\end_inset

.
 Za take 
\begin_inset Formula $k$
\end_inset

 sledi
\begin_inset Formula 
\[
\left|b_{k}\right|\left|x\right|^{k}\leq\left(\left(\frac{1}{R}+\varepsilon\right)\left|x\right|\right)^{k}.
\]

\end_inset

Opazimo,
 da je desna stran neenačbe člen geometrijske vrste,
 s katero majoriziramo vrsto iz absolutnih vrednosti členov naše vrste.
 Preverimo,
 da desna stran konvergira.
 Konvergira,
 kadar 
\begin_inset Formula $\left(\frac{1}{R}+\varepsilon\right)\left|x\right|<1$
\end_inset

 oziroma 
\begin_inset Formula $\varepsilon<\frac{1}{\left|x\right|}-\frac{1}{R}$
\end_inset

.
 Po 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{pk}{primerjalnem kriteriju}
\end_layout

\end_inset

 torej naša vrsta absolutno konvergira.
\end_layout

\begin_layout Itemize
Divergenca:
 Vzemimo poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

.
 Po definciji 
\begin_inset Formula $R$
\end_inset

 sledi,
 da je 
\begin_inset Formula $\sqrt[k]{\left|b_{k}\right|}\geq\frac{1}{R}-\varepsilon$
\end_inset

 za vse dovolj velike 
\begin_inset Formula $k$
\end_inset

.
 Za take 
\begin_inset Formula $k$
\end_inset

 sledi
\begin_inset Formula 
\[
\left|b_{k}\right|\left|x\right|^{k}\geq\left(\left(\frac{1}{R}-\varepsilon\right)\left|x\right|\right)^{k}.
\]

\end_inset

Opazimo,
 da je desna stran neenačbe člen geometrijske vrste,
 ki je majorizirana z vrsto iz absolutnih vrednosti členov naše vrste.
 Desna stran divergira,
 ko 
\begin_inset Formula $\left(\frac{1}{R}-\varepsilon\right)\left|x\right|=1$
\end_inset

 oziroma 
\begin_inset Formula $\varepsilon=\frac{1}{R}-\frac{1}{\left|x\right|}$
\end_inset

,
 zato tudi naša vrsta divergira.
\end_layout

\end_deeper
\begin_layout Example*
Primer konvergenčnega radija potenčne vrste od prej:
 
\begin_inset Formula $\sum_{j=1}^{\infty}x^{j}$
\end_inset

.
 Velja 
\begin_inset Formula $\forall j\in\mathbb{N}:b_{j}=1$
\end_inset

,
 torej 
\begin_inset Formula $R=\frac{1}{\limsup_{j\to\infty}\sqrt[k]{\left|b_{k}\right|}}=1$
\end_inset

,
 torej po zgornjem izreku vrsta konvergira za 
\begin_inset Formula $x\in\left(-1,1\right)$
\end_inset

 in divergira za 
\begin_inset Formula $x\not\in\left[-1,1\right]$
\end_inset

.
 Ročno lahko še preverimo,
 da divergira tudi v 
\begin_inset Formula $\left\{ -1,1\right\} $
\end_inset

.
\end_layout

\begin_layout Section
Zveznost
\begin_inset Note Note
status open

\begin_layout Plain Layout
TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
 recimo dodaj dokaz zveznosti x^2
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Ideja:
 Izdelati želimo formulacijo,
 s katero preverimo,
 le lahko z dovolj majhno spremembo 
\begin_inset Formula $x$
\end_inset

 povzročimo majhno spremembo funkcijske vrednosti.
\end_layout

\begin_layout Example*
Primer nezvezne funkcije je 
\begin_inset Formula $f\left(x\right)=\begin{cases}
0 & ;0\leq x<1\\
1 & ;x=1
\end{cases}$
\end_inset

.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $D\subseteq\mathbb{R},a\in D$
\end_inset

 in 
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset

.
 Pravimo,
 da je 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

,
 če 
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

.
 
\begin_inset Formula $f$
\end_inset

 je zvezna na množici 
\begin_inset Formula $x\subseteq D$
\end_inset

,
 če je zvezna na vsaki točki v 
\begin_inset Formula $D$
\end_inset

.
\end_layout

\begin_layout Theorem*
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hypertarget{kzzz}{Karakterizacija zveznosti z zaporedji}
\end_layout

\end_inset

.
 Naj bodo 
\begin_inset Formula $D,a,f$
\end_inset

 kot prej.
 Velja:
 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a\Leftrightarrow\forall\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D:\lim_{n\to\infty}a_{n}=a\Rightarrow\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
\end_inset

 ZDB 
\begin_inset Formula $f$
\end_inset

 je zvezna v 
\begin_inset Formula $a$
\end_inset

,
 če za vsako k 
\begin_inset Formula $a$
\end_inset

 konvergentno zaporedje na domeni velja,
 da funkcijske vrednosti členov zaporedja konvergirajo k funkcijski vrednosti 
\begin_inset Formula $a$
\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Predpostavimo,
 da je 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

,
 torej 
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

.
 Naj bo 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 poljubno zaporedje na 
\begin_inset Formula $D$
\end_inset

,
 ki konvergira k 
\begin_inset Formula $a$
\end_inset

,
 se pravi 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\in\mathbb{N}:n\geq n_{0}\Rightarrow\left|a-a_{n}\right|<\varepsilon$
\end_inset

.
 Naj bo 
\begin_inset Formula $\varepsilon$
\end_inset

 poljuben.
 Vsled zveznosti 
\begin_inset Formula $f$
\end_inset

 velja,
 da je 
\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

 za vse take 
\begin_inset Formula $a_{n}$
\end_inset

,
 da velja 
\begin_inset Formula $\left|a_{n}-a\right|<\delta$
\end_inset

 za neko 
\begin_inset Formula $\delta\in\mathbb{R}$
\end_inset

.
 Ker je zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 konvergentno k 
\begin_inset Formula $a$
\end_inset

,
 so vsi členi po nekem 
\begin_inset Formula $n_{0}$
\end_inset

 v 
\begin_inset Formula $\delta-$
\end_inset

okolici 
\begin_inset Formula $a$
\end_inset

,
 torej velja pogoj 
\begin_inset Formula $\left|a_{n}-a\right|<\delta$
\end_inset

,
 torej velja 
\begin_inset Formula $\left|f\left(a_{n}\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

 za vse 
\begin_inset Formula $n\geq n_{0}$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 PDDRAA 
\begin_inset Formula $f$
\end_inset

 ni zvezna v 
\begin_inset Formula $a$
\end_inset

.
 Da pridemo do protislovja,
 moramo dokazati,
 da 
\begin_inset Formula $\exists\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in D\ni:\lim_{n\to\infty}a_{n}=a$
\end_inset

,
 a vendar 
\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)\not=f\left(a\right)$
\end_inset

.
 Ker 
\begin_inset Formula $f$
\end_inset

 ni zvezna,
 velja,
 da 
\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x\in D\ni:\left|x-a\right|<\delta\wedge\left|f\left(x\right)-f\left(a\right)\right|\geq\varepsilon$
\end_inset

.
 Izberimo 
\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
\end_inset

.
 Tedaj 
\begin_inset Formula $\forall n\in\mathbb{N}\exists\varepsilon>0,x\in D\eqqcolon x_{n}\ni:\left|x_{n}-a\right|<\frac{1}{n}\wedge\left|f\left(x_{n}\right)-f\left(a\right)\right|\geq\varepsilon$
\end_inset

.
 S prvim argumentom konjunkcije smo poskrbeli za to,
 da je naše konstruiramo zaporedje 
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 konvergentno k 
\begin_inset Formula $a$
\end_inset

.
 Konstruirali smo zaporedje,
 pri katerem so funkcijske vrednosti za vsak 
\begin_inset Formula $\varepsilon$
\end_inset

 izven 
\begin_inset Formula $\varepsilon-$
\end_inset

okolice 
\begin_inset Formula $f\left(a\right)$
\end_inset

,
 torej zaporedje ne konvergira k 
\begin_inset Formula $f\left(a\right)$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Theorem*
Naj bo 
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset

.
 
\begin_inset Formula $f$
\end_inset

 je zvezna na 
\begin_inset Formula $D\Leftrightarrow$
\end_inset

 za vsako odprto množico 
\begin_inset Formula $V\subset\mathbb{R}$
\end_inset

 je 
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset

 spet odprta množica
\begin_inset Foot
status open

\begin_layout Plain Layout
Za funkcijo 
\begin_inset Formula $f:D\to V$
\end_inset

 za 
\begin_inset Formula $X\subseteq V$
\end_inset

 definiramo 
\begin_inset Formula $f^{-1}\left(X\right)\coloneqq\left\{ x\in D;f\left(x\right)\in V\right\} \subseteq D$
\end_inset

.
\end_layout

\end_inset

.
\end_layout

\begin_layout Proof
Dokazujemo ekvivalenco.
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Leftarrow\right)$
\end_inset

 Predpostavimo,
 da za vsako odprto množico 
\begin_inset Formula $V\subset\mathbb{R}$
\end_inset

 je 
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset

 spet odprta množica.
 Dokazujemo,
 da je 
\begin_inset Formula $f$
\end_inset

 zvezna na 
\begin_inset Formula $D$
\end_inset

.
 Naj bosta 
\begin_inset Formula $a\in D,\varepsilon>0$
\end_inset

 poljubna.
 Naj bo 
\begin_inset Formula $V\coloneqq\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)$
\end_inset

 odprta množica.
 Po predpostavki sledi,
 da je 
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset

 spet odprta.
 Ker je 
\begin_inset Formula $a\in f^{-1}\left(V\right)$
\end_inset

,
 je 
\begin_inset Formula $a\in V$
\end_inset

.
 Ker je 
\begin_inset Formula $V$
\end_inset

 odprta,
 
\begin_inset Formula $\exists\delta>0\ni:\left(a-\delta,a+\delta\right)\in V$
\end_inset

.
 Torej 
\begin_inset Formula $\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

,
 torej je 
\begin_inset Formula $f$
\end_inset

 zvezna na 
\begin_inset Formula $D$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $\left(\Rightarrow\right)$
\end_inset

 Predpostavimo,
 da je 
\begin_inset Formula $f$
\end_inset

 zvezna na 
\begin_inset Formula $D$
\end_inset

,
 to pomeni 
\begin_inset Formula $\forall a\in D\forall\varepsilon>0\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

.
 Naj bo 
\begin_inset Formula $V$
\end_inset

 poljubna odprta podmnožica 
\begin_inset Formula $\mathbb{R}$
\end_inset

 in naj bo 
\begin_inset Formula $a\in f^{-1}\left(V\right)$
\end_inset

 poljuben (torej 
\begin_inset Formula $f\left(a\right)\in V$
\end_inset

).
 Ker je 
\begin_inset Formula $f\left(a\right)\in V$
\end_inset

,
 ki je odprta,
 
\begin_inset Formula $\exists\varepsilon>0\ni:\left(f\left(a\right)-\varepsilon,f\left(a\right)+\varepsilon\right)\subseteq V$
\end_inset

.
 Ker je 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

,
 
\begin_inset Formula $\exists\delta>0\forall x\in D:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

,
 torej je tudi neka odprta okolica 
\begin_inset Formula $f\left(a\right)$
\end_inset

 v 
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset

.
 Ker je bil 
\begin_inset Formula $a$
\end_inset

 poljuben,
 je 
\begin_inset Formula $f^{-1}\left(V\right)$
\end_inset

 odprta,
 ker je bila 
\begin_inset Formula $V$
\end_inset

 poljubna,
 je izrek dokazan.
\end_layout

\end_deeper
\begin_layout Theorem*
Naj bosta 
\begin_inset Formula $f,g:D\to\mathbb{R}$
\end_inset

 zvezni v 
\begin_inset Formula $a\in D$
\end_inset

.
 Tedaj so v 
\begin_inset Formula $a$
\end_inset

 zvezne tudi funkcije 
\begin_inset Formula $f+g,f-g,f\cdot g$
\end_inset

 in 
\begin_inset Formula $f/g$
\end_inset

,
 slednja le,
 če je 
\begin_inset Formula $g\left(a\right)\not=0$
\end_inset

.
\end_layout

\begin_layout Proof
Ker je 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

 po 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{kzzz}{izreku o karakterizaciji zveznosti z zaporedji}
\end_layout

\end_inset

 velja za vsako 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},\forall n\in\mathbb{N}:a_{n}\subset D,\lim_{n\to\infty}a_{n}=a$
\end_inset

 tudi 
\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(a\right)$
\end_inset

.
 Po 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{pmkdlim}{izreku iz poglavja o zaporedjih}
\end_layout

\end_inset

 velja,
 da 
\begin_inset Formula $f\left(a_{n}\right)*g\left(a_{n}\right)\to\left(f*g\right)\left(a_{n}\right)$
\end_inset

 za 
\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
\end_inset

.
 Zopet uporabimo izrek o karakterizaciji zveznosti z zaporedji,
 ki pove,
 da so tudi 
\begin_inset Formula $f*g$
\end_inset

 za 
\begin_inset Formula $*\in\left\{ +,-,\cdot,/\right\} $
\end_inset

 zvezne v 
\begin_inset Formula $a$
\end_inset

.
 Pri deljenju velja omejitev 
\begin_inset Formula $f\left(a\right)\not=0$
\end_inset

.
\end_layout

\begin_layout Theorem*
Če sta 
\begin_inset Formula $D,E\subseteq\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $f:D\to E$
\end_inset

 in 
\begin_inset Formula $g:E\to\mathbb{R}$
\end_inset

,
 je 
\begin_inset Formula $g\circ f:D\to\mathbb{R}$
\end_inset

.
 Hkrati pa,
 če je 
\begin_inset Formula $f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

 in 
\begin_inset Formula $g$
\end_inset

 zvezna v 
\begin_inset Formula $f\left(a\right)$
\end_inset

,
 je 
\begin_inset Formula $g\circ f$
\end_inset

 zvezna v 
\begin_inset Formula $a$
\end_inset

.
\begin_inset Foot
status open

\begin_layout Plain Layout
Velja 
\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
\end_inset

.
\end_layout

\end_inset


\end_layout

\begin_layout Proof
Vzemimo poljubno 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\subseteq D$
\end_inset

,
 da 
\begin_inset Formula $a_{n}\to a\in D$
\end_inset

.
 Zopet uporabimo 
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
hyperlink{kzzz}{izrek o karakterizaciji zveznosti z zaporedji}
\end_layout

\end_inset

:
 ker je 
\begin_inset Formula $f$
\end_inset

 zvezna,
 velja 
\begin_inset Formula $f\left(a_{n}\right)\to f\left(a\right)$
\end_inset

 in ker je 
\begin_inset Formula $g$
\end_inset

 zvezna,
 velja 
\begin_inset Formula $g\left(f\left(a_{n}\right)\right)\to g\left(f\left(a\right)\right)$
\end_inset

.
 Potemtakem 
\begin_inset Formula $\left(g\circ f\right)\left(a_{n}\right)\to\left(g\circ f\right)\left(a\right)$
\end_inset

 in po istem izreku je 
\begin_inset Formula $g\circ f$
\end_inset

 zvezna na 
\begin_inset Formula $D$
\end_inset

.
\end_layout

\begin_layout Theorem*
Vsi polinomi so zvezni na 
\begin_inset Formula $\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Proof
Vzemimo 
\begin_inset Formula $p\left(x\right)=\sum_{k=0}^{n}a_{k}k^{k}$
\end_inset

.
 Uporabimo prejšnji izrek.
 Polinom je sestavljen iz vsote konstantne funkcije,
 zmnožene z identiteto,
 ki je s seboj 
\begin_inset Formula $n-$
\end_inset

krat množena.
 Ker vsota in množenje ohranjata zveznost,
 je treba dokazati le,
 da je 
\begin_inset Formula $f\left(x\right)=x$
\end_inset

 zvezna in da so 
\begin_inset Formula $\forall c\in\mathbb{R}:f\left(x\right)=c$
\end_inset

 zvezne.
\end_layout

\begin_deeper
\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=x$
\end_inset

 Ali velja 
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

?
 Da,
 velja.
 Vzamemo lahko katerokoli 
\begin_inset Formula $\delta\in(0,\varepsilon]$
\end_inset

.
\end_layout

\begin_layout Labeling
\labelwidthstring 00.00.0000
\begin_inset Formula $f\left(x\right)=c$
\end_inset

 Naj bo 
\begin_inset Formula $c\in\mathbb{R}$
\end_inset

 poljuben.
 Tu je 
\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|c-c\right|=0$
\end_inset

,
 torej je desna stran implikacije vedno resnična,
 torej je implikacija vedno resnična.
\end_layout

\end_deeper
\begin_layout Theorem*
Vse elementarne funkcije so na njihovih definicijskih območjih povsod zvezne.
 To so:
 polinomi,
 potence,
 racionalne funkcije,
 koreni,
 eksponentne funkcije,
 logaritmi,
 trigonometrične,
 ciklometrične in kombinacije neskončno mnogo naštetih,
 spojenih s 
\begin_inset Formula $+,-,\cdot,/,\circ$
\end_inset

.
\end_layout

\begin_layout Proof
Tega izreka ne bomo dokazali.
\end_layout

\begin_layout Example*
\begin_inset Formula $f\left(x\right)\coloneqq\log\left(\sin^{3}x+\frac{1}{8}\right)+\frac{1}{\sqrt[4]{x-7}}$
\end_inset

 je zvezna povsod,
 kjer je definirana.
\end_layout

\begin_layout Definition*
Naj bo 
\begin_inset Formula $\varepsilon>0,a\in\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $f:\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \to\mathbb{R}$
\end_inset

.
 Pravimo,
 da je 
\begin_inset Formula $L\in\mathbb{R}$
\end_inset

 limita 
\begin_inset Formula $f$
\end_inset

 v točki 
\begin_inset Formula $a$
\end_inset

 (zapišemo 
\begin_inset Formula $L=\lim_{x\to a}f\left(x\right)$
\end_inset

),
 če za vsako zaporedje 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}},a_{n}\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} $
\end_inset

,
 za katero velja 
\begin_inset Formula $a_{n}\to a$
\end_inset

,
 velja 
\begin_inset Formula $f\left(a_{n}\right)\to L$
\end_inset


\end_layout

\begin_layout Definition*
ZDB če 
\begin_inset Formula $\forall\varepsilon>0\exists\delta>0\ni:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
\end_inset


\end_layout

\begin_layout Definition*
ZDB če za 
\begin_inset Formula $\overline{f}:\left(a-\varepsilon,a+\varepsilon\right)\to\mathbb{R}$
\end_inset

 s predpisom 
\begin_inset Formula $\overline{f}\left(x\right)\coloneqq\begin{cases}
f\left(x\right) & ;x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} \\
L & ;x\in a
\end{cases}$
\end_inset

 velja,
 da je zvezna v 
\begin_inset Formula $a$
\end_inset

.
\end_layout

\begin_layout Note*
Vrednost 
\begin_inset Formula $f\left(a\right)$
\end_inset

,
 če sploh obstaja,
 nima vloge pri vrednosti limite.
\end_layout

\begin_layout Corollary*
Naj bo 
\begin_inset Formula $a\in D\subseteq\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $f:D\to\mathbb{R}$
\end_inset

.
 
\begin_inset Formula $f$
\end_inset

 je zvezna v 
\begin_inset Formula $a\Leftrightarrow\lim_{x\to a}f\left(x\right)=f\left(a\right)$
\end_inset

.
\end_layout

\begin_layout Example*
Naj bo 
\begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
\end_inset

 s predpisom 
\begin_inset Formula $x\mapsto\frac{\sin x}{x}$
\end_inset

.
 Zanima nas,
 ali obstaja 
\begin_inset Formula $\lim_{x\to0}f\left(x\right)$
\end_inset

.
 Grafični dokaz.
\end_layout

\begin_layout Example*
\begin_inset Float figure
placement document
alignment document
wide false
sideways false
status open

\begin_layout Plain Layout
TODO XXX FIXME SKICA S TKZ EUCLID,
 glej ZVZ III/ANA1P1120/str.8
\end_layout

\begin_layout Plain Layout
\begin_inset Caption Standard

\begin_layout Plain Layout
Skica.
\end_layout

\end_inset


\end_layout

\end_inset

Očitno velja 
\begin_inset Formula $\triangle ABD\subset$
\end_inset

 krožni izsek 
\begin_inset Formula $DAB\subset\triangle ABC$
\end_inset

,
 torej za njihove ploščine velja 
\begin_inset Formula 
\[
\frac{\sin x}{2}\leq\frac{x}{2\pi}\cdot x=\frac{x}{2}\leq\frac{\tan x}{2}\quad\quad\quad\quad/\cdot\frac{2}{\sin x}
\]

\end_inset


\begin_inset Formula 
\[
1\leq\frac{x}{\sin x}\leq\frac{1}{\cos x}\quad\quad\quad\quad/\lim_{x\to0}
\]

\end_inset


\begin_inset Formula 
\[
\lim_{x\to0}1\leq\lim_{x\to0}\frac{x}{\sin x}\leq\lim_{x\to0}\frac{1}{\cos x}
\]

\end_inset


\begin_inset Formula 
\[
1\leq\lim_{x\to0}\frac{x}{\sin x}\leq1
\]

\end_inset


\begin_inset Formula 
\[
\lim_{x\to0}\frac{x}{\sin x}=1
\]

\end_inset

Da naš sklep res potrdimo,
 je potreben spodnji izrek.
\end_layout

\begin_layout Theorem*
Če za 
\begin_inset Formula $f,g,h:D\to\mathbb{R}$
\end_inset

 velja za 
\begin_inset Formula $a\in D$
\end_inset

:
\end_layout

\begin_deeper
\begin_layout Itemize
\begin_inset Formula $\exists\varepsilon>0\forall x\in\left(a-\varepsilon,a+\varepsilon\right)\setminus\left\{ a\right\} :f\left(x\right)\leq g\left(x\right)\leq h\left(x\right)$
\end_inset

 in hkrati
\end_layout

\begin_layout Itemize
\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right),\lim_{x\to a}h\left(x\right)$
\end_inset

 in 
\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)=1$
\end_inset

,
 tedaj tudi 
\begin_inset Formula $\exists\lim_{x\to a}g\left(x\right)$
\end_inset

 in 
\begin_inset Formula $\lim_{x\to a}g\left(x\right)=1$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Proof
TODO XXX FIXME DOKAZ V SKRIPTi
\end_layout

\begin_layout Example*
\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
\end_inset

 ne obstaja.
 Zakaj?
 Izračunajmo levo in desno limito:
\begin_inset Formula 
\[
\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
\]

\end_inset

Toda 
\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
\end_inset

.
\end_layout

\begin_layout Definition*
Označimo 
\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
\end_inset

.
 Če 
\begin_inset Formula $\exists f\left(a+0\right)$
\end_inset

 in 
\begin_inset Formula $\exists f\left(a-0\right)$
\end_inset

,
 vendar 
\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
\end_inset

,
 pravimo,
 da ima 
\begin_inset Formula $f$
\end_inset

 v točki 
\begin_inset Formula $a$
\end_inset

 
\begin_inset Quotes gld
\end_inset

skok
\begin_inset Quotes grd
\end_inset

.
\end_layout

\begin_layout Corollary*
sssssssssss 
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\begin_layout Corollary*
sssssssssss
\end_layout

\end_body
\end_document