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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
