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+#LyX 2.3 created this file. For more info see http://www.lyx.org/
+\lyxformat 544
+\begin_document
+\begin_header
+\save_transient_properties true
+\origin unavailable
+\textclass article
+\use_default_options true
+\maintain_unincluded_children false
+\language slovene
+\language_package default
+\inputencoding utf8
+\fontencoding global
+\font_roman "default" "default"
+\font_sans "default" "default"
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+\graphics default
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+\bibtex_command default
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+\use_hyperref false
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+\use_package amsmath 1
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+\use_package cancel 1
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+\cite_engine basic
+\cite_engine_type default
+\use_bibtopic false
+\use_indices false
+\paperorientation portrait
+\suppress_date false
+\justification true
+\use_refstyle 1
+\use_minted 0
+\index Index
+\shortcut idx
+\color #008000
+\end_index
+\secnumdepth 3
+\tocdepth 3
+\paragraph_separation indent
+\paragraph_indentation default
+\is_math_indent 0
+\math_numbering_side default
+\quotes_style english
+\dynamic_quotes 0
+\papercolumns 1
+\papersides 1
+\paperpagestyle default
+\tracking_changes false
+\output_changes false
+\html_math_output 0
+\html_css_as_file 0
+\html_be_strict false
+\end_header
+
+\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