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\title[Matematika 1]
{Matematika 1}

\subtitle{5. vaja}

\author[Borut] % 
{B.~Jur\v{c}i\v{c}~Zlobec\inst{1}}

\institute[Univerza v Ljubljani Slovenia] % 
{
  \inst{1}%
    Univerza v Ljubljani,\\Fakulteta za Elektrotehniko\\ 
    1000 Ljubljana, Tr\v{z}a\v{s}ka 25, Slovenija
}


\date[BF 2010] % 
{Matematika FE, Ljubljana, Slovenija 2010}

\subject{Matematika}


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\section{Funkcije}
\subsection{Limita in zveznost funkcije}
\subsubsection{Limita}
\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to x_0}f(x)$}
\formulas{$\displaystyle{f(x)=\frac{x^2-x}{x^2-1},\quad x_0=1}$}.
        \begin{itemize}
        \item Faktoriziramo "stevec in imenovalec: $\lim_{x\to 1}\frac{x(x-1)}{(x-1)(x+1)}$,
        \item kraj"samo skupni faktor in dobimo $\lim_{x\to 1}\frac{x}{x+1}=\frac{1}{2}$. 
        \end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to x_0}f(x)$}
\formulas{$\displaystyle{f(x)=\frac{1-\sqrt{1-x^2}}{x},\quad x_0=0}$}
	\begin{itemize}
	\item Pomno"zimo "stevec in imenovalec v $\lim_{x\to 0}\frac{1-\sqrt{1-x^2}}{x}$ s konjugirano iracionaliteto:\\
	\item $\lim_{x\to 0}\frac{1-(1-x^2)}{x(1+\sqrt{1-x^2})}$,
	\item od tod je $\lim_{x\to 0}\frac{x}{(1+\sqrt{1-x^2})}=0$.
	\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to \infty}f(x)$}
\formulas{$\displaystyle{f(x)=\frac{x^3+x}{x^4-3x+1}}$}
	\begin{itemize}
	\item "Stevec in imenovalec delimo z vodilno potenco $x^4$.
	\item $\lim_{x\to \infty}
	\frac{\frac{1}{x}+\frac{1}{x^3}}{1-\frac{3}{x^3}+\frac{1}{x^4}}$
	\item $\lim_{x\to \infty f(x)=0}$
	\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to \infty}f(x)$}
\formulas{$\displaystyle{f(x)=\frac{x^3+x}{2x^3-3x+1}}$}
	\begin{itemize}
	\item "Stevec in imenovalec delimo z vodilno potenco $x^3$.
	\item $\lim_{x\to \infty}
	\frac{1+\frac{1}{x}}{2-\frac{3}{x^2}+\frac{1}{x^3}}$
	\item $\lim_{x\to \infty f(x)=\frac{1}{2}}$
	\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to x_0}f(x)$}
\formulas{$\displaystyle\frac{1}{1+e^{-\frac{1}{x^2}}}\quad x_0=0 $}.
\begin{itemize}
	\item Velja $\lim_{x\to0}\frac{1}{x^2}=\infty$ in
	\item $\lim_{x\to\infty}e^{-x}=0$,
	\item od tod je $\lim_{x\to0}\frac{1}{1+e^{-\frac{1}{x^2}}}=1$
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to x_0}f(x)$}
\formulas{$\displaystyle x\sin{\frac{1}{x}},\quad x_0=0 $}.
\begin{itemize}
        \item Velja $-1\le \sin\frac{1}{x}\le 1$ za vsak $x\ne 0$,
        \item od tod velja ocena $-|x|\le x\sin\frac{1}{x}\le |x|$.
        \item Torej je $\lim_{x\to0}x\sin\frac{1}{x}=0$
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci limito funkcije $\lim_{x\to \infty}f(x)$}
\formulas{$\displaystyle\arctg\frac{1-x}{1+x}$}.
\begin{itemize}
        \item Velja $\lim_{x\to \infty}\frac{1-x}{1+x}=-1$
        \item in $\arctg(-1)=-\frac{\pi}{4}$.
        \item Torej je $\lim_{x\to\infty}\arctg\frac{1-x}{1+x}=-\frac{\pi}{4}$
\end{itemize}
\end{frame}

\subsubsection{Leva in desna limita funkcije}
\begin{frame}{V to"cki nezveznosti funkcije $f(x)$ izra"cunaj levo in desno limito}
\formulas{$\displaystyle\frac{1}{1+e^{\frac{1}{x}}}$}
\begin{itemize}
	\item Velja $\lim_{x\nearrow0}\frac{1}{x}=-\infty$, $\lim_{x\searrow0}\frac{1}{x}=\infty$,
	\item $\lim_{x\to-\infty} e^x=0$ in $\lim_{x\to\infty} e^x=\infty$. 
	\item Od tod je $\lim_{x\nearrow0}\frac{1}{1+e^{\frac{1}{x}}}=1$ in 
	\item $\lim_{x\searrow0}\frac{1}{1+e^{\frac{1}{x}}}=0$
\end{itemize}
\end{frame}

\begin{frame}{V to"cki nezveznosti funkcije $f(x)$ izra"cunaj levo in desno limito}
\formulas{$\displaystyle\arctg\frac{x}{1+x}$}
\begin{itemize}
	\item Velja $\lim_{x\nearrow-1}\frac{x}{x+1}=\infty$, $\lim_{x\searrow-1}\frac{x}{x+1}=-\infty$,
	\item $\lim_{x\to-\infty} \arctg x=-\frac{\pi}{2}$ in $\lim_{x\to\infty} \arctg x=\frac{\pi}{2}$. 
	\item Od tod je $\lim_{x\nearrow-1}\arctg\frac{x}{1+x}=\frac{\pi}{2}$ in 
	\item $\lim_{x\searrow0}\frac{1}\arctg\frac{x}{1+x}=-\frac{\pi}{2}$
\end{itemize}
\end{frame}

\begin{frame}{Dolo"ci parameter $a$ tako, da bo funkcija $f(x)$ povsod zvezna.}
\formulas{$\displaystyle f(x)=\sign(x - 2)(x^2 + 2 |x + 1|+ a) $}.
\begin{itemize}
	\item Parameter $a$ moramo izbrati tako, da sta leva in desna 
	limita v to"cki $2$ enaki funkcijski vrednosti v tej to"cki.
	\item $f(2)=0$.
	\item $\lim_{x\nearrow2}\sign(x - 2)(x^2 + 2 |x + 1|+ a)=
		\lim_{x\to2}-(x^2 + 2 |x + 1|+ a)=-10-a$
	\item $\lim_{x\searrow2}\sign(x - 2)(x^2 + 2 |x + 1|+ a)=
                \lim_{x\to2}(x^2 + 2 |x + 1|+ a)=10+a$.
	\item Od tod sledi, da je $a=-10$.
\end{itemize}
\end{frame}
\begin{frame}{Graf funkcije $f(x)=\sign(x - 2)(x^2 + 2 |x + 1|-10)$}
\begin{center}
	\includegraphics[scale=0.75]{func-17.pdf}
\end{center}
\end{frame}

\begin{frame}{Dolo"ci parameter $a$ tako, da bo funkcija $f(x)$ povsod zvezna.}
\formulas{$\displaystyle f(x)=\sign(x - 2)(x + \frac{ax}{1 + x^2}) + x $}.
\begin{itemize}
        \item Parameter $a$ moramo izbrati tako, da sta leva in desna
        limita v to"cki $2$ enaki funkcijski vrednosti v tej to"cki.
        \item $f(2)=2$.
        \item $\lim_{x\nearrow2}\sign(x - 2)(x + \frac{ax}{1 + x^2})+x=
                \lim_{x\to2}-(x + \frac{ax}{1 + x^2})+x=-\frac{2a}{5}$
        \item $\lim_{x\searrow2}\sign(x - 2)(x + \frac{ax}{1 + x^2})+x=
                \lim_{x\to2}(x + \frac{ax}{1 + x^2})+x=4+\frac{2a}{5}$.
        \item Od tod sledi, da je $a=-5$.
\end{itemize}
\end{frame}
\begin{frame}{Graf funkcije $f(x)=\sign(x - 2)(x - \frac{5x}{1 + x^2})+x$}
\begin{center}
        \includegraphics[scale=0.75]{func-18.pdf}
\end{center}
\end{frame}
\begin{frame}{Dolo"ci parametra $a$ in $b$ tako, da bo funkcija $f(x)$ povsod zvezna.}
\formulas{
      $f(x)=\begin{cases}
      x^2 + 1, x \le 2\\
      a x^2 + b x + 1, 2 < x < 3\\
      x + 1, x \ge 3
\end{cases}$
}
\begin{itemize}
	\item $\lim_{x\nearrow2}f(x)=\lim_{x\to2} x^2 + 1=5$
	\item $\lim_{x\searrow2}f(x)=\lim_{x\to2}a x^2 + b x + 1=4a+2b+1$
	\item $\lim_{x\nearrow3}f(x)=\lim_{x\to2}a x^2 + b x + 1=9a+3b+1$
 	\item $\lim_{x\searrow3}f(x)=\lim_{x\to2} x + 1=4$
	\item Re"simo sistem ena"cb $4a+2b+1=5$ in $9a+3b+1=4$. 
	\item Re"sitev je $a=-1$ in $b=4$.  
\end{itemize}

\end{frame}
\begin{frame}{Graf funkcije $f(x)$ iz prej"sne naloge.}
\begin{center}
        \includegraphics[scale=0.75]{func-19.pdf}
\end{center}
\end{frame}


\section{Odvod funkcije}
\subsection{Tabela elementarnih odvodov}

\begin{frame}{Pravila za odvajanje}
\begin{enumerate}[I]
	\item $\displaystyle \left(a f(x)+b g(x)\right)'=a f'(x)+b g'(x)$, linearnost.\label{lin}
	\item $\displaystyle \left({f(x)}{g(x)}\right)'={f'(x)g(x)+f(x)g'(x)}$, produkt.\label{prod}
	\item $\displaystyle 
		\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$, kvocient.\label{kvoc}
	\item $\displaystyle \left(f(g(x))\right)'=f'(g(x))g'(x)$, posredna funkcija.\label{posr}
	\item $\displaystyle 
		\left(f^{-1}(x)\right)'=\frac{1}{f'(y)}$, kjer je $x=f(y)$, inverzna funkcija.\label{inv}
\end{enumerate}
\end{frame}

\begin{frame}{Odvodi elementarnih funkcij}
\begin{enumerate}
	\item $\left(x^n\right)'=n\,x^{n-1}$\label{potenca-n},$(n\in\mathbb{Z},\quad x\ne 0)$.
	\item $\left(x^s\right)'=s\,x^{s-1}$\label{potenca-s},$(s\in\mathbb{R},\quad x > 0)$.
	\item $\left(e^x\right)'=e^x$\label{exp}
	\item $\left(\ln |x|\right)'=\frac{1}{x}$\label{loge}
	\item $\left(\sin x\right)'=\cos x$\label{sin}
	\item $\left(\cos x\right)'=-\sin x$\label{cos}
	\item $\left(\arctan x\right)'=\frac{1}{1+x^2}$\label{atan}
	\item $\left(\arcsin x\right)'=\frac{1}{\sqrt{1-x^2}}$\label{asin}
\end{enumerate}
\end{frame}

\subsection{Odvajanje funkcij}
\subsubsection{Algebrske funkcije}
\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=\frac{x}{1+x^2}$}
\begin{itemize}
	\item Uporabimo (\ref{kvoc}) $\rightarrow$  (\ref{lin}) $\rightarrow$(\ref{potenca-n}).
	\item $f'(x)=\frac{x'(1+x^2)-x{(1+x^2)}'}{{(1+x^2)}^2}\rightarrow$
	\item $f'(x)=\frac{x'(1+x^2)-x(1'+{x^2}')}{{(1+x^2)}^2}\rightarrow$
	\item $f'(x)=\frac{(1+x^2)-x(2x)}{{(1+x^2)}^2}=\frac{1-x^2}{{(1+x^2)}^2}$.
\end{itemize}
\end{frame}

\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=x\sqrt{1+x^2}$}.
\begin{itemize}
	\item Uporabimo (\ref{prod})$\rightarrow$(\ref{potenca-s})$\rightarrow$(\ref{posr})
		$\rightarrow$(\ref{lin})$\rightarrow$(\ref{potenca-n}).
	\item $f'(x)=\sqrt{1+x^2}+\frac{x}{2\sqrt{1+x^2}}{(1+x^2)}'\rightarrow$
	\item $f'(x)=\sqrt{1+x^2}+\frac{x}{2\sqrt{1+x^2}}2x\rightarrow$
	\item $f'(x)=\sqrt{1+x^2}+\frac{x^2}{\sqrt{1+x^2}}=\frac{1+2x^2}{\sqrt{1+x^2}}$.
\end{itemize}
\end{frame}

\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=x\sqrt{\frac{1-x^2}{1+x^2}}$}.
\begin{itemize}
	\item Uporabimo (\ref{prod})$\rightarrow$(\ref{potenca-s})$\rightarrow$(\ref{posr})
		$\rightarrow$(\ref{kvoc})$\rightarrow$(\ref{lin})$\rightarrow$(\ref{potenca-n}).
	\item $f'(x)=\sqrt{\frac{1-x^2}{1+x^2}}+x\left(\sqrt{\frac{1-x^2}{1+x^2}}\right)'\rightarrow 
	\sqrt{\frac{1-x^2}{1+x^2}}+
	x\frac{\left(\frac{1-x^2}{1+x^2}\right)'}{2\sqrt{\frac{1-x^2}{1+x^2}}}\rightarrow$
	\item $f'(x)=\sqrt{\frac{1-x^2}{1+x^2}}+
        x\frac{\left(\frac{-2x(1+x^2)-(1-x^2)2x}{(1+x^2)^2}\right)}{2\sqrt{\frac{1-x^2}{1+x^2}}}\rightarrow$
	\item $f'(x)=-\frac{x^4+2x^2-1}{\sqrt{1-x^2}\sqrt{(1+x^2)^3}}$ 
\end{itemize}
\end{frame}

\subsubsection{Eksponentna funkcija in logaritem}
\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=\ln(x+\sqrt{1+x^2})$}
\begin{itemize}
	\item Uporabimo (\ref{loge})$\rightarrow$(\ref{lin})$\rightarrow$(\ref{posr})
		$\rightarrow$(\ref{potenca-s})$\rightarrow$(\ref{lin})$\rightarrow$(\ref{potenca-n}).
	\item $f'(x)=\frac{1}{x+\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)'\rightarrow$
	\item $f'(x)=\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{(1+x^2)'}{2\sqrt{1+x^2}}\right)\rightarrow$
	\item $f'(x)=\frac{1}{x+\sqrt{1+x^2}}\left(1+\frac{2x}{2\sqrt{1+x^2}}\right)=
	\frac{x+\sqrt{1+x^2}}{(x+\sqrt{1+x^2})\sqrt{1+x^2}}\rightarrow$
	\item $f'(x)=\frac{1}{\sqrt{1+x^2}}$.
\end{itemize}
\end{frame}
\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=e^{-\frac{1}{x^2}}$}.
\begin{itemize}
	\item $f'(x)=e^{-\frac{1}{x^2}}\left(-\frac{1}{x^2}\right)'\rightarrow$
	\item $\displaystyle f'(x)=\frac{2}{x^3}e^{-\frac{1}{x^2}}$.
\end{itemize}
\end{frame}

\subsubsection{Trigonometri"cne in ciklometri"cne funkcije}
\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=\frac{\cos x}{2\sin^2x}$}
\begin{itemize}
	\item Uporabimo (\ref{lin})$\rightarrow$
	(\ref{kvoc})$\rightarrow$(\ref{posr})$\rightarrow$(\ref{potenca-n})
	$\rightarrow$(\ref{cos})$\rightarrow$(\ref{sin}).
	\item $\displaystyle f'(x)=\frac{1}{2}\frac{-\sin^3x-2\cos^2x\sin x}{\sin^4x}\rightarrow$
	\item $\displaystyle f'(x)=-\frac{\sin^2x+2\cos^2x}{2\sin^3x}\rightarrow$
	\item $\displaystyle f'(x)=-\frac{1+\cos^2x}{2\sin^3x}$.
\end{itemize}
\end{frame}

\subsubsection{"Se nekaj transcendentnih funkcij}
\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=\left({\left({x^x}\right)^x}\right)^2$}
	\begin{itemize}
		\item Poenostavimo $y=x^{2x^2}$.
		\item Logaritmiramo obe strani $\ln y=2x^2\ln x$.
		\item Odvajamo vsako stran ena"cbe posebej 
		$\frac{y'}{y}=4x\ln x+2x^2\frac{1}{x}=4x \ln x + 2x$.
		\item Pomno"zimo z $y$ in dobimo $y'=y(4x\ln x+2x)\rightarrow$
		\item $\displaystyle f'(x)=x^{2x^2}(4x \ln  x + 2x)$.
	\end{itemize}	
\end{frame}

\begin{frame}{Odvajaj funkcijo $f(x)$.}
\formulas{$\displaystyle f(x)=\left(\frac{x}{1+x}\right)^x$}
        \begin{itemize}
                \item Logaritmiramo obe strani 
			$\ln y=x\ln\left(\frac{x}{x+1}\right)$.
                \item Odvajamo vsako stran ena"cbe posebej\\
                $\frac{y'}{y}=
		\ln\left(\frac{x}{1+x}\right)+
		\frac{x+1}{x}\left(\frac{x}{x+1}\right)'$.
                \item $\frac{y'}{y}=
		\ln\left(\frac{x}{1+x}\right)+
		\frac{x+1}{x}\frac{1}{(x+1)^2}\rightarrow$
                \item $f'(x)=\left(\frac{x}{1+x}\right)^x
		(\ln\left(\frac{x}{1+x}\right)+\frac{1}{x^2+x})$.
        \end{itemize}
\end{frame}

\subsection{L'Hospitalovo pravilo in odvedljivost}
\subsubsection{L'H\^ospitalovo pravilo}
\begin{frame}{L'H\^ospitalovo pravilo}
\formulas{L'H\^ospitalovo pravilo ali tudi Bernoullijevo pravilo:}\\
\begin{enumerate}
 	\item "Ce sta $\lim_{x\to x_0}f(x)$ in 
	$\lim_{x\to x_0}g(x)$ obe enaki $0$ ali $\pm\infty$, kjer je 
		$x_0\in \mathbb{R}$ ali $x_0=\pm\infty$, ter
	\item obstaja limita $\lim_{x\to x_0}\frac{f'(x)}{g'(x)}$, 
	\item potem velja \formulas{$$\lim_{x\to x_0}\frac{f(x)}{g(x)}=
		\lim_{x\to x_0}\frac{f'(x)}{g'(x)}$$}
\end{enumerate}	
\end{frame}
\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito 
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=\frac{\sin x}{x}$ v $x_0=0$}
	\begin{itemize}
		\item $\lim_{x\to0}\sin x=0$ in $\lim_{x\to0}x=0$.
		\item Ker je $\lim_{x\to0}\frac{(\sin x)'}{x'}=
		\frac{\cos x}{1}=1$, potem  
		\item \formulas{$\displaystyle 
			\lim_{x\to 0}\frac{\sin x}{x}=1$}.
	\end{itemize}
\end{frame}

\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=\frac{1-\cos x}{x^2}$ v $x_0=0$}
        \begin{itemize}
                \item $\lim_{x\to0}1-\cos x=0$ in $\lim_{x^2\to0}x=0$.
                \item Ker je $\lim_{x\to0}\frac{(1-\cos x)'}{(x^2)'}=
                \lim_{x\to 0}\frac{\sin x}{2x}=\frac{1}{2}$, potem 
                \item \formulas{$\displaystyle 
			\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$}.
        \end{itemize}
\end{frame}

\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=x\ln x$, $x_0=0$}
        \begin{itemize}
		\item Limito zapi"semo v obliki 
		$\lim_{x\searrow0}\frac{\ln x}{1/x}$
                \item $\lim_{x\searrow0}\ln x=-\infty$ in 
		$\lim_{x\searrow0}\frac{1}{x}=\infty$.
                \item $\lim_{x\to0}\frac{(\ln x)'}{(1/x)'}=
                \lim_{x\searrow0}\frac{1/x}{-1/x^2}=\lim_{x\to0}-x=0$,
                \item \formulas{$\displaystyle 
		\lim_{x\searrow0}x \ln x=0$}.
        \end{itemize}
\end{frame}

\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=x^x$, $x_0=0$}
        \begin{itemize}
                \item Logaritmiramo $\ln f(x)=x \ln x$.
		\item $\lim_{x\searrow0}\ln f(x)=\lim_{x\searrow0}x \ln x=0$ 
                \item \formulas{$\displaystyle 
                \lim_{x\searrow0}x^x=e^0=1$}.
        \end{itemize}
\end{frame}

\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=\frac{1}{x^2} e^{-\frac{1}{x^2}}$, $x_0=0$}
        \begin{itemize}
                \item Nadomestimo $t\rightarrow\frac{1}{x^2}$.
		\item "Ce $x\to 0$,  potem $t\to\infty$.
                \item $\lim_{t\to\infty}te^{-t}=
		\lim_{t\to\infty}\frac{t}{e^t}=
			\lim_{t\to\infty}\frac{1}{e^t}=0$
                \item \formulas{$\displaystyle 
                \lim_{x\to0}x^2e^\frac{1}{x^2}=0$}.
        \end{itemize}
\end{frame}

\begin{frame}{S pomo"cjo L'H\^ospitalovega pravila izra"cunaj limito
funkcije $f(x)$ v to"cki $x_0$.}
\formulas{$\displaystyle f(x)=\frac{1}{x-1}-\frac{1}{\ln x}$, $x_0=1$}
        \begin{itemize}
                \item Skupni imenovalec: $f(x)=\frac{x\ln x-x+1}{(x-1)\ln x}$
                \item $\lim_{x\to1}\frac{x\ln x-x+1}{(x-1)\ln x}\rightarrow$
		\item $\lim_{x\to1}\frac{\ln x}{\frac{x-1}{x}+\ln x}\rightarrow$
		\item $\lim_{x\to1}\frac{x\ln x}{x-1+x\ln x}\rightarrow$
		\item $\lim_{x\to1}\frac{1+\ln x}{2+\ln x}=\frac{1}{2}$ 
                \item \formulas{$\displaystyle 
                \lim_{x\to1}\frac{x\ln x-x+1}{(x-1)\ln x}=\frac{1}{2}$}.
        \end{itemize}
\end{frame}


\subsubsection{Odvedljivost}
\begin{frame}{Za dano funkcijo $f(x)$,} 
\formulas{$f(x)=\begin{cases}
	ax+bx^2,x<\frac{1}{2}\\ 
	\frac{2x}{1+x^2}, x\ge \frac{1}{2}
\end{cases}$},\\
dolo"ci konstanti $a$ in $b$ tako, da bo povsod odvedljiva in
zapi"si ena"cbo tangente v to"cki zlepka.
\begin{itemize}
	\item Izpolnjeni morajo biti naslednji pogoji:
	$\lim_{x\nearrow\frac{1}{2}}f(x)=\lim_{x\searrow\frac{1}{2}} f(x)$,
	$\lim_{x\nearrow\frac{1}{2}}f'(x)=\lim_{x\searrow\frac{1}{2}}f'(x)$.
	\item Ena"cbi: $\frac{a}{2}+\frac{b}{2}=\frac{4}{5}$ in 
		$a+b=\frac{24}{25}$.
	\item Re"sitvi: $a=\frac{56}{25}$, $b=-\frac{32}{25}$.
	\item Ena"cba tangente v to"cki zlepka je 
	$y=\frac{8}{25}+\frac{24}{25}x$. 
\end{itemize}
\end{frame}
\begin{frame}{Grafi\v{c}ni prikaz}
	\begin{center}
		\includegraphics[scale=1]{func-20.pdf}
	\end{center}
\end{frame}

\begin{frame}{Za dano funkcijo $f(x)$,} 
\formulas{$f(x)=\begin{cases}
	ax^2+bx,x<2\\ 
	cx^2+d, x\ge 2
\end{cases},\quad f(1)=1,\quad f(4)=3$},\\
dolo"ci konstante $a$, $b$ $c$ in $d$ tako, da bo povsod odvedljiva in
zapi"si ena"cbo tangente v to"cki zlepka.
\begin{itemize}
	\item Izpolnjeni morajo biti naslednji pogoji:
	$f(1)=1$, $f(4)=3$, 
	$\lim_{x\nearrow2}f(x)=\lim_{x\searrow2} f(x)$,
	$\lim_{x\nearrow2}f'(x)=\lim_{x\searrow2}f'(x)$.
	\item Ena"cbe: $a+b=1$, $16c+d=3$, $4a + 2b = 4c + d$ in $4a+b=4c$.
	\item Re"sitve: $a=-(2/11)$, $b=13/11$, $c=5/44$ in $d=13/11$.
	\item Ena"cba tangente v to"cki zlepka je 
	$y=\frac{8}{11}+\frac{5}{11}x$. 
\end{itemize}
\end{frame}
\begin{frame}{Grafi\v{c}ni prikaz}
	\begin{center}
		\includegraphics[scale=1]{func-21.pdf}
	\end{center}
\end{frame}

\end{document}
