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\begin_body

\begin_layout Title
Rešen tretji izpit teorije Analize 1 — IŠRM 2023/24
\end_layout

\begin_layout Abstract
Izpit je potekal v petek, 30.
 avgusta 2024 od desete
\begin_inset Foot
status open

\begin_layout Plain Layout
Avtor tega besedila je na izpit zamudil poldrugo uro.
\end_layout

\end_inset

 do dvanajste ure.
 Nosilec predmeta je 
\noun on
Oliver Dragičević
\noun default
.
 Naloge in rešitve sem po spominu spisal 
\noun on
Anton Luka Šijanec
\noun default
.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\left[15\right]$
\end_inset


\begin_inset Newline newline
\end_inset

Podaj natančne definicije naslednjih pojmov:
\end_layout

\begin_deeper
\begin_layout Enumerate
limita zaporedja, stekališče zaporedja
\end_layout

\begin_deeper
\begin_layout Standard
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

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $L$
\end_inset

 je limita 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\sim L=\lim_{n\to\infty}a_{n}\Leftrightarrow\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n>n_{0}:\left|a_{n}-L\right|<\varepsilon$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $L$
\end_inset

 je stekališče 
\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}\Leftrightarrow\forall\varepsilon>0\exists\mathcal{A}\subseteq\mathbb{N},\left|\mathcal{A}\right|=\left|\mathcal{\mathbb{N}}\right|\ni:\left\{ a_{n};n\in\mathcal{A}\right\} \subseteq\left(L-\varepsilon,L+\varepsilon\right)$
\end_inset


\end_layout

\end_deeper
\begin_layout Enumerate
vsota (neskončne) konvergentne vrste
\end_layout

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

 poljubno zaporedje.
 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}\coloneqq\lim_{n\to\infty}\sum_{k=1}^{n}a_{n}$
\end_inset

.
 Če limita obstaja, je vrsta 
\begin_inset Formula $\sum_{n=1}^{\infty}a_{n}$
\end_inset

 konvergentna in njena vsota je enaka tej limiti.
\end_layout

\end_deeper
\begin_layout Enumerate
Cauchyjev pogoj za zaporedja
\end_layout

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

 realno zaporedje.
 Konvergentno je natanko tedaj, ko ustreza Cauchyjevemu pogoju: 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall m,n\geq n_{0}:\left|a_{n}-a_{m}\right|<\varepsilon$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
odprte, zaprte, omejene, kompaktne množice v 
\begin_inset Formula $\mathbb{R}$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
Množica 
\begin_inset Formula $\mathcal{A}$
\end_inset

 je odprta, ko 
\begin_inset Formula $\forall a\in\mathcal{A}\exists\varepsilon>0\ni:\left(a-\varepsilon,a+\varepsilon\right)\subseteq\mathcal{A}$
\end_inset

, ko za vsako točko množice obstaja neka njena okolica, ki je podmnožica
 množice 
\begin_inset Formula $\mathcal{A}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Množica 
\begin_inset Formula $\mathcal{A}$
\end_inset

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

 odprta.
\end_layout

\begin_layout Enumerate
Množica 
\begin_inset Formula $\mathcal{A}$
\end_inset

 je omejena, ko 
\begin_inset Formula $\exists m,M\in\mathbb{R}\forall a\in\mathcal{A}:a\leq M\wedge a\geq m$
\end_inset

.
\end_layout

\begin_layout Enumerate
Množica 
\begin_inset Formula $\mathcal{A}$
\end_inset

 je kompaktna 
\begin_inset Formula $\Leftrightarrow\mathcal{A}$
\end_inset

 zaprta 
\begin_inset Formula $\wedge$
\end_inset

 
\begin_inset Formula $\mathcal{A}$
\end_inset

 omejena.
\end_layout

\end_deeper
\begin_layout Enumerate
limita funkcije v dani točki
\end_layout

\begin_deeper
\begin_layout Standard
Naj bodo 
\begin_inset Formula $a\in\mathbb{R}$
\end_inset

, 
\begin_inset Formula $\mathcal{D}$
\end_inset

 okolica 
\begin_inset Formula $a$
\end_inset

 in 
\begin_inset Formula $f:\mathcal{D}\setminus\left\{ a\right\} \to\mathbb{R}$
\end_inset

 poljubne.
 
\begin_inset Formula $L\in\mathbb{R}$
\end_inset

 je limita 
\begin_inset Formula $f$
\end_inset

 v točki 
\begin_inset Formula $a\sim L=\lim_{x\to a}f\left(x\right)\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}\setminus\left\{ a\right\} :\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
zveznost funkcije
\end_layout

\begin_deeper
\begin_layout Standard
Naj bodo 
\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
\end_inset

, 
\begin_inset Formula $a\in\mathcal{D}$
\end_inset

 in 
\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
\end_inset

 poljubne.
 
\begin_inset Formula $f$
\end_inset

 je zvezna v 
\begin_inset Formula $a\Leftrightarrow\forall\varepsilon>0\exists\delta>0\forall x\in\mathcal{D}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
\end_inset

 .
 
\begin_inset Formula $f$
\end_inset

 je zvezna na množici 
\begin_inset Formula $\mathcal{A}$
\end_inset

, če je zvezna na vsaki točki množice 
\begin_inset Formula $\mathcal{A}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
odvedljivost funkcije
\end_layout

\begin_deeper
\begin_layout Standard
Naj bodo 
\begin_inset Formula $a\in\mathbb{R}$
\end_inset

, 
\begin_inset Formula $\mathcal{D}\subseteq\mathbb{R}$
\end_inset

, 
\begin_inset Formula $f:\mathcal{D}\to\mathbb{R}$
\end_inset

 poljubne.
 
\begin_inset Formula $f$
\end_inset

 je odvedljiva v 
\begin_inset Formula $a\text{\ensuremath{\Leftrightarrow\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}}}\in\mathbb{R}$
\end_inset

, ZDB ko obstaja slednja limita.
 Tedaj definiramo 
\begin_inset Quotes eld
\end_inset

odvod funkcije 
\begin_inset Formula $f$
\end_inset

 v točki 
\begin_inset Formula $a$
\end_inset


\begin_inset Quotes erd
\end_inset

: 
\begin_inset Formula $f'\left(a\right)=\lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}$
\end_inset

.
 
\begin_inset Formula $f$
\end_inset

 je odvedljiva na množici 
\begin_inset Formula $\mathcal{A}$
\end_inset

, če je odvedljiva na vsaki točki množice 
\begin_inset Formula $\mathcal{A}$
\end_inset

.
\end_layout

\end_deeper
\begin_layout Enumerate
določen integral realne funkcije na zaprtem omejenem intervalu.
\end_layout

\begin_deeper
\begin_layout Enumerate
Naj bodo 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

 in 
\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
\end_inset

 poljubne.
\end_layout

\begin_layout Enumerate
Definirajmo pojem delitve 
\begin_inset Formula $\left[a,b\right]$
\end_inset

.
 Delitev so točke 
\begin_inset Formula $t_{0},\dots,t_{n}$
\end_inset

, da velja 
\begin_inset Formula $a=t_{0}<t_{1}<\cdots<t_{n}=b$
\end_inset

 za nek 
\begin_inset Formula $n\in\mathbb{N}$
\end_inset

.
 Točke identificiramo z delilnimi intervali takole: 
\begin_inset Formula $D_{n}=\left[t_{n-1},t_{n}\right]$
\end_inset

.
 Delitev torej identificiramo z množico teh dedlilnih intervalov: 
\begin_inset Formula $D=\left\{ D_{k};\forall k\in\left\{ 1..n\right\} \right\} $
\end_inset

.
 Definiramo tudi velikost delitve: 
\begin_inset Formula $\left|D_{\infty}\right|=\max_{k\in\left\{ 1..n\right\} }\left|D_{k}\right|$
\end_inset

.
\end_layout

\begin_layout Enumerate
Definirajmo pojem izbire za dano delitev.
 Naj bo 
\begin_inset Formula $D$
\end_inset

 delitev.
 Pripadajoča izbira so take izbirne točke 
\begin_inset Formula $\xi_{1},\dots,\xi_{n}$
\end_inset

, da velja 
\begin_inset Formula $\forall k\in\left\{ 1..n\right\} :\xi_{k}\in D_{k}$
\end_inset

.
 Množico teh izbirnih točk označimo z 
\begin_inset Formula $\xi\coloneqq\left\{ \xi_{k};\forall k\in\left\{ 1..n\right\} \right\} $
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset Formula $f$
\end_inset

 je integrabilna na 
\begin_inset Formula $\left[a,b\right]$
\end_inset

, če 
\begin_inset Formula $\exists I\in\mathbb{R}\forall\varepsilon>0\exists\delta>0\forall$
\end_inset

 delitev 
\begin_inset Formula $D\forall$
\end_inset

 izbiro 
\begin_inset Formula $\xi$
\end_inset

, pripadajočo delitvi 
\begin_inset Formula $D:\left|D_{\infty}\right|<\delta\Rightarrow\left|\sum_{k=1}^{n}\left|D_{k}\right|f\left(\xi\right)-I\right|<\varepsilon$
\end_inset

.
 Tedaj pravimo, da je 
\begin_inset Formula $I$
\end_inset

 določen integral 
\begin_inset Formula $f$
\end_inset

 na 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 in pišemo 
\begin_inset Formula $I\eqqcolon\int_{a}^{b}f\left(x\right)dx$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\left[15\right]$
\end_inset


\end_layout

\begin_deeper
\begin_layout Enumerate
Pojasni princip matematične indukcije.
\end_layout

\begin_deeper
\begin_layout Standard
Naj bo 
\begin_inset Formula $\left(P_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 zaporedje logičnih vrednosti/izjav/izrazov.
 Če velja
\end_layout

\begin_layout Enumerate
\begin_inset Formula $P_{1}$
\end_inset

 drži in hkrati
\end_layout

\begin_layout Enumerate
\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
\end_inset

 drži 
\begin_inset Formula $\Rightarrow P_{n+1}$
\end_inset

 drži,
\end_layout

\begin_layout Standard
potem velja 
\begin_inset Formula $\forall n\in\mathbb{N}:P_{n}$
\end_inset

 drži.
\end_layout

\end_deeper
\begin_layout Enumerate
Z matematično indukcijo dokaži
\begin_inset Formula 
\[
\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}
\]

\end_inset


\end_layout

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

: 
\begin_inset Formula $1=\frac{1\left(1+1\right)}{2}$
\end_inset

 Velja.
\end_layout

\begin_layout Enumerate
Indukcijska predpostavka: 
\begin_inset Formula $1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset

.
\end_layout

\begin_layout Enumerate
Korak 
\begin_inset Formula $n\to n+1$
\end_inset

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

\end_inset


\begin_inset Formula 
\[
1+2+\cdots+n\overset{\text{I.P.}}{=}\frac{n\left(n+1\right)}{2}
\]

\end_inset


\end_layout

\begin_layout Enumerate
Sklep: 
\begin_inset Formula $\forall n\in\mathbb{N}:1+2+\cdots+n=\frac{n\left(n+1\right)}{2}$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\left[25\right]$
\end_inset


\begin_inset Newline newline
\end_inset

Naj bosta 
\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

 realni konvergentni zaporedji.
 Dokaži, da je 
\begin_inset Formula $c_{n}\coloneqq a_{n}b_{n}$
\end_inset

 prav tako konvergentno zaporedje.
\end_layout

\begin_deeper
\begin_layout Itemize
Označimo 
\begin_inset Formula $\lim_{n\to\infty}a_{n}\eqqcolon A$
\end_inset

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

.
\end_layout

\begin_layout Itemize
Uganemo, da je 
\begin_inset Formula $\lim_{n\to\infty}a_{n}b_{n}=AB$
\end_inset

.
 To moramo sedaj dokazati.
\end_layout

\begin_layout Itemize
Dokazujemo, da 
\begin_inset Formula $\forall\varepsilon>0\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}b_{n}-AB\right|<\varepsilon\sim\left|a_{n}b_{n}+a_{n}B-a_{n}B-AB\right|=\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|<\varepsilon$
\end_inset


\end_layout

\begin_layout Itemize
Ker po trikotniški neenakosti velja 
\begin_inset Formula $\left|a_{n}\left(b_{n}-B\right)+B\left(a_{n}-A\right)\right|\leq\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|$
\end_inset

, je dovolj za poljuben 
\begin_inset Formula $\varepsilon>0$
\end_inset

 dokazati
\begin_inset Formula 
\[
\exists n_{0}\in\mathbb{N}\forall n\geq n_{0}:\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\varepsilon
\]

\end_inset


\end_layout

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

 konvergentno, 
\begin_inset Formula $\exists n_{1}\in\mathbb{N}\forall n\geq n_{1}:\left|a_{n}-A\right|<\frac{\varepsilon}{2\left|a\right|}$
\end_inset

, kjer je 
\begin_inset Formula $a$
\end_inset

 zgornja meja zaporedja 
\begin_inset Formula $a_{n}$
\end_inset

.
 Slednje je omejeno, ker je konvergentno.
\end_layout

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

 konvergentno, 
\begin_inset Formula $\exists n_{2}\in\mathbb{N}\forall n\geq n_{1}:\left|b_{n}-B\right|<\frac{\varepsilon}{2\left|B\right|}$
\end_inset

.
\end_layout

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

 velja
\begin_inset Formula 
\[
\left|a_{n}\right|\left|b_{n}-B\right|+\left|B\right|\left|a_{n}-A\right|<\frac{\varepsilon\left|a\right|}{2\left|a_{n}\right|}+\frac{\varepsilon}{2}\leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
\]

\end_inset

in izrek je dokazan.
\end_layout

\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\left[?\right]$
\end_inset


\begin_inset Newline newline
\end_inset

Dokaži, da je zvezna realna funkcija na zaprtem intervalu omejena.
 Natančno navedi vse izreke, ki jih pri tem dokazu uporabiš.
\end_layout

\begin_deeper
\begin_layout Standard
Naj bodo 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

 in zvezna 
\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
\end_inset

 poljubne.
\end_layout

\begin_layout Itemize
Dokaz, da je 
\begin_inset Formula $f$
\end_inset

 omejena navzgor.
\end_layout

\begin_deeper
\begin_layout Itemize
PDDRAA 
\begin_inset Formula $f$
\end_inset

 ni navzgor omejena.
 Tedaj 
\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\geq n$
\end_inset

.
\end_layout

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

 na zaprti množici, je omejeno zaporedje, torej ima stekališče.
 Recimo mu 
\begin_inset Formula $s\in\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Itemize
Ker je 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 zaprta, je 
\begin_inset Formula $s\in\left[a,b\right]$
\end_inset

.
\end_layout

\begin_layout Itemize
Ker je 
\begin_inset Formula $f$
\end_inset

 zvezna na 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 in s tem v 
\begin_inset Formula $s$
\end_inset

, velja 
\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
\end_inset

.
\end_layout

\begin_layout Itemize
Po konstrukciji 
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 velja 
\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=\infty$
\end_inset

, torej 
\begin_inset Formula $f\left(s\right)=\infty$
\end_inset

, kar ni mogoče, saj 
\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
\end_inset

 po predpostavki.
 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

.
\end_layout

\begin_layout Standard
Predpostavka 
\begin_inset Quotes eld
\end_inset


\begin_inset Formula $f$
\end_inset

 ni navzgor omejena
\begin_inset Quotes erd
\end_inset

 ne velja, torej smo dokazali, da je 
\begin_inset Formula $f$
\end_inset

 navzgor omejena.
\end_layout

\end_deeper
\begin_layout Itemize
Dokaz, da je 
\begin_inset Formula $f$
\end_inset

 omejena navzdol.
\end_layout

\begin_deeper
\begin_layout Itemize
PDDRAA 
\begin_inset Formula $f$
\end_inset

 ni navzdol omejena.
 Tedaj 
\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in\left[a,b\right]\ni:f\left(x_{n}\right)\leq-n$
\end_inset

.
\end_layout

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

 na zaprti množici, je omejeno zaporedje, torej ima stekališče.
 Recimo mu 
\begin_inset Formula $s\in\mathbb{R}$
\end_inset

.
\end_layout

\begin_layout Itemize
Ker je 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 zaprta, je 
\begin_inset Formula $s\in\left[a,b\right]$
\end_inset

.
\end_layout

\begin_layout Itemize
Ker je 
\begin_inset Formula $f$
\end_inset

 zvezna na 
\begin_inset Formula $\left[a,b\right]$
\end_inset

 in s tem v 
\begin_inset Formula $s$
\end_inset

, velja 
\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=f\left(s\right)$
\end_inset

.
\end_layout

\begin_layout Itemize
Po konstrukciji 
\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
\end_inset

 velja 
\begin_inset Formula $\lim_{n\to\infty}f\left(x_{n}\right)=-\infty$
\end_inset

, torej 
\begin_inset Formula $f\left(s\right)=-\infty$
\end_inset

, kar ni mogoče, saj 
\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
\end_inset

 po predpostavki.
 
\begin_inset Formula $\rightarrow\!\leftarrow$
\end_inset

.
\end_layout

\begin_layout Standard
Predpostavka 
\begin_inset Quotes eld
\end_inset


\begin_inset Formula $f$
\end_inset

 ni navzdol omejena
\begin_inset Quotes erd
\end_inset

 ne velja, torej smo dokazali, da je 
\begin_inset Formula $f$
\end_inset

 navzdol omejena.
\end_layout

\end_deeper
\begin_layout Itemize
Ker je 
\begin_inset Formula $f$
\end_inset

 omejena navzgor in navzdol, je omejena.
\end_layout

\begin_layout Itemize
Uporabljeni izreki.
\end_layout

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

 s členi na kompaktni množici je omejeno.
\end_layout

\begin_layout Itemize
Omejeno zaporedje ima stekališče.
\end_layout

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

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

, obstaja konvergentno podzaporedje 
\begin_inset Formula $\left(a_{n_{k}}\right)_{k\in\mathbb{N}}$
\end_inset

, da je 
\begin_inset Formula $s$
\end_inset

 njegova limita.
\end_layout

\begin_layout Itemize
Množica je kompaktna natanko tedaj, ko vsebuje limite vseh konvergentnih
 zaporedij s členi v njej.
\end_layout

\begin_layout Itemize
Funkcija 
\begin_inset Formula $f$
\end_inset

 je zvezna v 
\begin_inset Formula $s$
\end_inset

, če za vsako k 
\begin_inset Formula $s$
\end_inset

 konvergentno zaporedje velja, da njegovi s 
\begin_inset Formula $f$
\end_inset

 preslikani členi konvergirajo v 
\begin_inset Formula $f\left(s\right)$
\end_inset

.
\end_layout

\end_deeper
\end_deeper
\begin_layout Enumerate
\begin_inset Formula $\left[?\right]$
\end_inset


\begin_inset Newline newline
\end_inset

Za realno funkcijo ene spremenljivke dokaži verižno pravilo.
\end_layout

\begin_deeper
\begin_layout Itemize
Naj bodo 
\begin_inset Formula $\mathcal{D},\mathcal{E},\mathcal{F}\subseteq\mathbb{R}$
\end_inset

, 
\begin_inset Formula $x\in\mathcal{D}$
\end_inset

 in 
\begin_inset Formula $f:\mathcal{D}\to\mathcal{E}$
\end_inset

, 
\begin_inset Formula $g:\mathcal{E}\to\mathcal{F}$
\end_inset

 poljubne.
 Naj bo 
\begin_inset Formula $f$
\end_inset

 odvedljiva v 
\begin_inset Formula $x$
\end_inset

 in 
\begin_inset Formula $g$
\end_inset

 odvedljiva v 
\begin_inset Formula $f\left(x\right)$
\end_inset

.
\end_layout

\begin_layout Itemize
Dokažimo, da je 
\begin_inset Formula $g\circ f$
\end_inset

 odvedljiva v 
\begin_inset Formula $x$
\end_inset

 in da velja 
\begin_inset Formula 
\[
\left(g\circ f\right)'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right).
\]

\end_inset


\end_layout

\begin_layout Itemize
Označimo 
\begin_inset Formula $a\coloneqq f\left(x\right)$
\end_inset

 in 
\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
\end_inset

.
 Potemtakem 
\begin_inset Formula $f\left(x+h\right)=\delta_{h}+a$
\end_inset

.
\begin_inset Formula 
\[
\left(g\circ f\right)'\left(x\right)=\lim_{h\to0}\frac{g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(\delta_{h}+a\right)-g\left(a\right)}{h}=
\]

\end_inset


\begin_inset Formula 
\[
=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=\lim_{h\to0}\frac{g\left(\delta_{h}+a\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
\]

\end_inset

Ker je 
\begin_inset Formula $f$
\end_inset

 v 
\begin_inset Formula $x$
\end_inset

 odvedljiva, je v 
\begin_inset Formula $x$
\end_inset

 zvezna, zato sledi 
\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
\end_inset

.
\begin_inset Formula 
\[
\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
\]

\end_inset


\end_layout

\end_deeper
\end_body
\end_document