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\def\ce{\mathbb C} 
\def\en{\mathbb N}
\def\zet{\mathbb Z}
\def\qe{\mathbb Q} 
\def\er{\mathbb R}
\def\R{\er}
\def\N{\en}
\def\nec{Nech\v t\ }
\def\eqdef{\stackrel{\text{\tiny def}}{=}}
\def\Mod{\operatorname{mod}}
\def\lp{\displaystyle\lim_{n\to\infty}}
\newcommand\lx[1][0]{\displaystyle\lim_{x\to #1}}
\def \cotg {\operatorname{cotg}}
\def \tg {\operatorname{tg}}
\def \sgn {\operatorname{sgn}}
\def \arctg {\operatorname{arctg}}
\def \arccotg {\operatorname{arccotg}}
\def \suma {\sum_{n=1}^\infty}
\def \sumad {\sum_{n=2}^\infty}
\def \dx {\, \mathrm{d} x}
\def \P {\mathcal{P}}
\newtheorem*{thm}{V\v{e}ta}

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\title{Cvi\v{c}en\'i - Matematick\'a anal\'yza 1, MFF}  % tyto dvě položky jsou zde v podstatě formálně, ve skutečnosti nejsou nikde 
\author{Marek C\'uth}

\begin{document}		
\centerline{\MakeUppercase{i. opakov\'an\'i st\v{r}edo\v{s}kolsk\'e l\'atky, v\'yroky}}\ \\
\textbf{1. \v{R}e\v{s}te n\'asleduj\'ic\'i nerovnosti a rovnosti v $\er$:}\\[5pt]
  a) $\frac{x-2}{2x-8} \geq 1;\qquad$
  b) $\frac{x+2}{x+3} > \frac{2x+3}{x+6};\qquad$
  c) $\log {}_{\frac 18}(x^2 - 3x + 3)\geq 0;$\qquad
  d) $\log{(x^2+1)} =2 \log{(3-x)};$\\
  e) $\sin{2x} < \cos{x};$\qquad  
  f) $x\leq\left|\frac{x+2}{x-3}\right|;$\qquad
  g) $|x-|x+1||\leq 2x;$\\
  h) $|x - |x-1|| = 1 - |x|;$\qquad
  i) $\frac{x^2 + 2}{x+7} < 2(x-7);$\qquad
  j) $\log {}_{7}(49x^2) = 4\cdot(\log {}_{7}x)^2.$\\[5pt]
\textbf{2. Doka\v{z}te:}\\[5pt]
 a) Pro ka\v{z}d\'e $a$, $b \in \mathbb R$:\quad $||a| - |b|| \leq |a-b|$;\qquad
 b) Pro ka\v{z}d\'e $a$, $b \in \mathbb R$:\quad $|a + b| \leq |a| + |b|$.\\[5pt]
\textbf{3. Na\v{c}rtn\v{e}te graf funkce}\\[5pt]
 a) $f(x) = 1 - |\cos{\tfrac{x}{2}}|;$\qquad
 b) $g(x) = \tfrac{1}{2} + \cos(\tfrac{\pi}{2} + |x|);$\qquad
 c) $h(x) = ||||x|-1|-1|-1|$;\\
 d) $i(x) = \left|\frac{3x + 3}{2x - 4}\right|$;\qquad
 e) $j(x) = |\log|1-x||$.\\[5pt]
\textbf{4. }V z\'avislosti na parametru $c\in\er$ ur\v{c}ete, pro kter\'a $x\in\er$ plat\'i:\\[5pt]
a) $cx^2 + x + 1 > 0$\qquad b) $ce^x\in (-1,0]$\qquad c) $|\cos x| - c > 0$\qquad d) $\log|x| + c\in (-\pi/2,\pi/2)$\\ \\
\textbf{5. }Nech\v{t} $M$ zna\v{c}\'i mno\v{z}inu v\v{s}ech mu\v{z}\r{u} a $Z$ mno\v{z}inu v\v{s}ech \v{z}en. Uva\v{z}ujme n\'asleduj\'ic\'i v\'yrokov\'e formy:

  $S(m,z)$: ,,Mu\v{z} $m$ je man\v{z}elem \v{z}eny $z$.'';
  
  $L_1(m,z)$: ,,Mu\v{z} $m$ miluje \v{z}enu $z$.'';\quad$L_2(m,z)$: ,,\v{Z}ena $z$ miluje mu\v{z}e $m$.''.\\
Pomoc\'{\i} kvantifik\'ator\r{u}, logick\'ych spojek a forem $S$, $L_1$ a
$L_2$ zapi\v{s}te n\'asleduj\'{\i}c\'{\i} v\'yroky:\\
a) Ka\v{z}d\'y \v{z}enat\'y mu\v{z} miluje svou man\v{z}elku.\quad
b) Ka\v{z}dou \v{z}enu miluje n\v{e}jak\'y mu\v{z}.\\
c) Ka\v{z}d\'a \v{z}ena m\'a nejv\'y\v{s} jednoho man\v{z}ela.\\
d) Ka\v{z}d\'y mu\v{z} m\'a nejv\'y\v{s} jednu man\v{z}elku. (\v{R}\'{\i}k\'a tento v\'yrok tot\'e\v{z},
co c)?)\\
e) Existuje vdan\'a \v{z}ena.\quad
f) Existuje \v{z}enat\'y mu\v{z}. (\v{R}\'{\i}k\'a tento v\'yrok tot\'e\v{z},
co e)?)\\
g) Existuj\'{\i} nev\v{e}rn\'e man\v{z}elky. (Man\v{z}elku prohl\'as\'{\i}me za nev\v{e}rnou,
pokud miluje jin\'eho mu\v{z}e ne\v{z} sv\'eho man\v{z}ela.)\\N\'asleduj\'{\i}c\'{\i} v\'yroky p\v{r}elo\v{z}te do \v{c}e\v{s}tiny.\\
h) $\exists m\in M\forall z\in Z (\neg S(m,z))$;\\
i) $\exists z\in Z\forall m\in M (L_1(m,z)\Rightarrow \neg L_2(m,z))$;\\
j) $\exists z\in Z\forall m\in M (L_2(m,z)\Rightarrow \neg L_1(m,z))$;\\
k) $\forall z\in Z ( (\exists m\in M: L_2(m,z))\Rightarrow (\exists m\in M: L_1(m,z) \ \& \neg L_2(m,z) )$.\\ \\
\textbf{6. }Uva\v{z}me n\'asleduj\'ic\'i v\'yroky:
\begin{itemize}
	\item[(i)] $\forall x\in M\;\exists y\in M\; \exists z\in M\quad x = y+z$\qquad\qquad (ii)\quad $\exists y\in M\;\forall x\in M\; \exists z\in M\quad x = y+z$
	\item[(iii)] $\exists y\in M\;\exists z\in M\; \forall x\in M\quad x = y+z$
\end{itemize}
Kter\'e z nich jsou pravdiv\'e a kter\'e nepravdiv\'e, je--li\\
a) $M = \en$\quad b) $M = \en\cup\{0\}$\quad c) $M = (0,1)$\quad d) $M = \{0\}$?\\[5pt]
\textbf{7. } Rozhodn\v{e}te o spr\'avnosti n\'asleduj\'{\i}c\'{\i}ch v\'yrok\r{u} a napi\v{s}te
jejich negace.\\
a) $\forall x \in \en \ \exists y \in \en \ \forall z \in\en \, (z > x \Rightarrow y < z)$;\quad
b) $\exists y\in\en\ \forall x\in\en\ \forall z\in\en\ (z>x\Rightarrow y<z)$;\\
c) $\exists x \in \en \ \forall y \in \en \ \forall z \in\en \, (z > x \Rightarrow y < z)$;\quad
d) $\exists x \in \en \ \forall y \in \en \ \forall z \in\en \, (z < x \Rightarrow y > z)$;\\
e) $\forall x\in\er \ \forall \varepsilon>0\ \exists\delta>0\ \forall y\in\er\, (|y-x|<\delta \Rightarrow y<x+\frac \varepsilon3)$.\\[5pt]
\textbf{8. } Vyj\'ad\v{r}ete co nejjednodu\v{s}\v{s}eji:\\
a) $\forall\varepsilon>0\ \forall y\in\er : |y-7|<5\Rightarrow |f(y)-15|<\varepsilon$;\\
b) $\forall x\in\er\ \exists \delta>1 \forall y\in\er\ \forall n\in\en : |y-x|<\delta \Rightarrow
|f(x)-f(y)|<\frac1n$.
\newpage
\centerline{\MakeUppercase{ii. d\r{u}kazov\'a technika, mno\v{z}iny, zobrazen\'i a mohutnost}}\ \\
\textbf{1.} (Bernoulli) Nech\v{t} $x \in \R$, $x \geq -1$, $n \in \N$. Pak plat\'i Bernoulliova nerovnost
$$(1+x)^n \geq 1 + nx.$$ Doka\v{z}te matematickou indukc\'i. Indukc\'i s krokem ,,$n \to n{+}2$'' d\'ale uka\v{z}te, \v{z}e uveden\'a nerovnost plat\'i i pro $x \geq -2$. Uka\v{z}te na p\v{r}\'iklad\v{e}, \v{z}e pro $x<-2$ a obecn\'e $n\in \N$ ji\v{z} Bernoulliova nerovnost neplat\'i.\\[5pt]
\textbf{2.} Doka\v{z}te indukc\'i:\resetpr
		\pr $\sum_{k=1}^n k = \tfrac{1}{2}n(n+1)$
		\pr $\sum_{k=1}^n k^3 = (1+2+\ldots+n)^2$
		\pr $\sum_{k=0}^n q^k = \tfrac{1-q^{n+1}}{1-q},q\neq 1$
		\pr $\sum_{k=1}^n \tfrac{1}{(2k-1)(2k+1)} = \tfrac{n}{2n+1}$
		\pr Pro $x_1,\ldots,x_n\in[0,\pi]$ plat\'i, \v{z}e $\left|\sin\left(\sum_{i=1}^nx_i\right)\right|\leq \sum_{i=1}^n\sin(x_i)$\\
		\pr $n!\leq \left(\tfrac{n+1}{2}\right)^n\quad$(\emph{Hint: p\v{r}i d\r{u}kazu pou\v{z}ijte Bernoulliho nerovnost})\\[5pt]
\textbf{3.} Plat\'i pro libovoln\'e mno\v{z}iny $A$, $B$ a $C$ n\'asleduj\'c\'i vztahy? Sv\'e tvrzen\'i doka\v{z}te.\\[3pt]
\resetpr \pr $A\cup B = B\Leftrightarrow A\cap B = A$ \pr $A\subset B\Leftrightarrow A\cup B = B$ \pr $(A\cap B)\setminus (B\cap C) = (C\cap B)\setminus (A\cap C)$ \pr $(A\cap B)\setminus (B\cap C) = (A\cap B)\setminus (A\cap C)$ \pr $(A\cup B)\setminus C = A\setminus (C\cup B)$\\[5pt]
\noindent\textbf{4.} Uvažujme zobrazení $f:X \to Y$ a množiny $A \subset X$, $B \subset X$. Doka\v{z}te n\'asleduj\'ic\'i rovnosti.
	\begin{multicols}{2}\begin{enumerate}[\upshape (i)]
		\item $f(A \cup B) = f(A) \cup f(B)$,
		\item $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$,
		\item $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$,
		\item $f^{-1}(A \setminus B) = f^{-1}(A) \setminus f^{-1}(B)$.
	\end{enumerate}\end{multicols}
\noindent\textbf{5.} Uva\v{z}ujme zobrazen\'i $f: X \to Y$ a mno\v{z}iny $A \subset X$, $B \subset X$. Uka\v{z}te, \v{z}e n\'asleduj\'ic\'i vztahy obecn\v{e} neplat\'i.
	\begin{multicols}{2}\begin{enumerate}[\upshape (i)]
		\item $f(A \cap B) = f(A) \cap f(B)$,
		\item $f(A \setminus B) = f(A) \setminus f(B)$.
	\end{enumerate}\end{multicols}
\noindent\textbf{6.} \nec $f\colon\en\to\en$. Rozhodn\v{e}te o platnosti n\'asleduj\'ic\'ich tvrzen\'i:
	\begin{multicols}{2}\begin{enumerate}[\upshape (i)]
		\item Je-li $f(\en)$ kone\v{c}n\'a, nen\'i $f$ prost\'e.
		\item Je-li $f(\en)$ nekone\v{c}n\'a, je $f$ prost\'e.
		\item Je-li $\en\setminus f(\en)=\emptyset$, je $f$ prost\'e.
		\item Je-li $f$ prost\'e, je $\en\setminus f(\en)=\emptyset$.
	\end{enumerate}\end{multicols}
\noindent\textbf{7.}Rozhodn\v{e}te, zda jsou n\'asleduj\'ic\'i mno\v{z}iny spo\v{c}etn\'e. Sv\'e tvrzen\'i doka\v{z}te:
	\begin{multicols}{2}\begin{enumerate}[\upshape (i)]
		\item $\zet$ (mno\v{z}ina cel\'ych \v{c}\'isel)
		\item $\qe$ (mno\v{z}ina racion\'aln\'ich \v{c}\'isel)
		\item $\er$		
		\item mno\v{z}ina v\v{s}ech zobrazen\'i z $\en$ do $\{0,1\}$
		\item mno\v{z}ina v\v{s}ech iracion\'aln\'ich \v{c}\'isel
		\item mno\v{z}ina v\v{s}ech otev\v{r}en\'ych interval\r{u} na $\er$
		\item mno\v{z}ina v\v{s}ech kone\v{c}n\'ych podmno\v{z}in $\en$
		\item mno\v{z}ina v\v{s}ech nekone\v{c}n\'ych podmno\v{z}in $\en$
	\end{enumerate}\end{multicols}
\centerline{\MakeUppercase{iii. suprema, infima}}\ \\
\textbf{1. Najd\v{e}te suprema a infima n\'asleduj\'ic\'ich mno\v{z}in (pokud existuj\'i). Existuj\'i maxima a minima?}\\[5pt]
 a) $A = \{-\frac{1}{n};\ n \in \en\};$\qquad
 b) $B = \{\frac{n+ (-1)^n}{n};\ n \in \en\};$\\
 c) $C_1 = \{n^2 - m^2;\ n,m \in \en\},\quad C_2 = \{n^2 - m^2;\ \ n,m \in \en, n>m\},\quad C_3 = \{n^2 - m^2;\ \ n,m \in \en, n\leq m\};$\\
 d) $D_1 = \{2^{-n} + 3^{-n};\ n \in \en\},\quad D_2 = \{2^{-n} + 3^{-n};\ n \in \zet\};$\\
 e) $E = \{n^{(-1)^n};\ n \in \en\};$\qquad
 f) $F = \{5^{(-1)^j3^k};\ j \in \zet, \ k \in \zet\};$\newpage
\centerline{\MakeUppercase{iv. limity posloupnost\'i}}\ \\
\textbf{1. Vypo\v{c}t\v{e}te p\v{r}\'imo z definice limity:}\\[5pt]
 a) $\lp\frac{1}{n^2}$\qquad b) $\lp\frac{n^2 + n}{n^3 + 3}$\quad c) $\lp(-1)^n\frac{2n}{n^3+1} = 0\qquad$ d) $\lp\frac{\sin(n\frac \pi7)}{n^2} = 0$ \\[5pt]
\textbf{2. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i:}\resetpr \pr $\lp\frac{2n^2+n-3}{n^3-1}$
\pr $\lp\frac{2n^3+6n}{n^3-7n+7}$\\
\pr $\lp\frac{2n^5+3n-2}{n^5-3n^3+1}$
\pr $\lp\left(\sqrt{n+1}-\sqrt n\right)$
\pr $\lp\left(\root 3\of{n+11}-\root 3\of n\right)$\\
\pr $\lp(-1)^n\sqrt{n}(\sqrt{n+1}-\sqrt{n})$
\pr $\lp\frac{(n+4)^{100} - (n+3)^{100}}{(n+2)^{100} - n^{100}}$
\pr $\lp\frac{n^2 + (n+1)^2}{(n+3)^2 + (n+4)^2}$\\
\pr $\lp\frac{(n+4)^{27} - (n+1)^{27}}{(2n^2 + 5)^{13} - (n^2-1)^{13}}$
\pr $\lp\frac{(n^2-1)(n+2)^3 - n^2}{3n(n^2+5)(n^2+1)}$
\pr $\lp(\sqrt{n+\sqrt{n}} - \sqrt{n})$\\
\pr $\lp(\sqrt[3]{n+2} - \sqrt[3]{n-1})$
\pr $\lp(\sqrt{n^2+n+1} - \sqrt[3]{n^3+1})$
\pr $\lp\frac{\sqrt[4]{n+2} - \sqrt[4]{n+1}}{\sqrt[3]{n+3} - \sqrt[3]{n}}$\\
\pr $\lp\frac{\sqrt{n^2+n+1} - n-1}{\sqrt{n}}$
\pr $\lp\sqrt{\frac{n+\sqrt{n+\sqrt{n+1}}}{n+1}}$\\[5pt]
\textbf{3. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i}($[\cdot]$ zna\v{c}\'i celou \v{c}\'ast):\\ \resetpr
\pr $\lp\frac{3^n + 5^n + n^3}{1 + 5n^2 + 3\cdot5^n}$
\pr $\lp\frac{(n^7+1)^8 - (n^7 + 2)^8}{(n+1)^{50} - (n+2)^{50}}$
\pr $\lp\sqrt{\frac{n + \sqrt{n^2 + \sqrt{n^4 + 1}}}{n+1}}$\\
\pr $\lp(-1)^n\cdot n\cdot (\sqrt[3]{n^3+1} - \sqrt{n^2+5})$
\pr $\lp\left(\sqrt{n + \sqrt{n}} - \sqrt{n+6}\right)\frac{3 + (-5)^n + n! + 4^n}{n^n + 8 - 6n!}$\\
\pr $\lp\frac{(n^6+n)^5 - (n^6+7n)^5}{(n^5+1)^6 - (n^5+6)^6}$
%\pr $\lp\frac{n^5 + 2^n + 17^n}{n! + n + 3^n}$
\pr $\lp\frac{n+[\sqrt[3]{n}]^3}{n-[\sqrt{n+9}]}$
\pr $\lp\frac{3^n + n^5 + (n+1)!}{n(n^6+n!)}$
%\pr $\lp\frac{4^n + n^8 - 4n!}{5n! - 3^n + n^9}$\\
\pr $\lp\sqrt[n]{5^n + 4^n + 3^n + 2^n}$\\
\pr $\lp \frac{(-2)^n + 3^n}{(-2)^{n+1} + 3^{n+1}}$
%\pr $\lp{\frac{1}{n^2}\sum\limits_{k=1}^n [kx]},\,x\in\er$\\[3pt]
\pr $\lp\sqrt[n]{\sqrt{3^n + 2\cdot 2^n} - \sqrt{3^n + 2^n}}$
\pr $\lp\frac{\sqrt[n]{2^n + 3^n}}{\sqrt[2n]{4^n + \sqrt{n}}}$\\
\pr $\lp\sin\left(\frac{n\pi}{4}\right)\sqrt[n]{5 + n}$
\pr $\lp\frac{3^n - [\sqrt[3]{n!}]}{3n^3 - [\sqrt[3]{(n+1)!}]}$
\pr $\lp\sqrt[n]{[n^4\cos n ]-n^23^n+4^n}$\\
\pr $\lp\bigl(\tfrac{2\sqrt{n}-\sqrt[3]{n}}{\sqrt{n+1}}-\cos \tfrac{n\pi}{4}\big)\tfrac{2n}{1-\sqrt[n]{2n}}$
\pr $\lp\tfrac{\sqrt[6]{n^5+1}-\sqrt[5]{n^4-n^3}+\sqrt[n]{n^2}}{\sqrt{n-\sqrt{n}}}$
\pr $\lp\tfrac{5+\sqrt[3]{n^3+n^2}-\sqrt{n^2+n}} {\sqrt{n}-\sqrt[4]{n}}$\\[5pt]
\textbf{4. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i (p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na FSV):} \\[3pt] \resetpr
  \pr $\lp((n^{10}+n^3)^7 - (n^7+1)^{10})\cdot\left(\sqrt[3]{(1+\frac{1}{n^9})^7} - 1\right)^7$\\[3pt]
  \pr $\lp\frac{(n+\frac 2n)^{30} - (n+\frac 1n)^{30}}{\sqrt{(2+n^7)^8 - 2^8}}$\qquad\qquad\qquad\qquad
  \pr $\lp\frac{n\sqrt{n}\sqrt[n]{(n+1)^n+n^{n+1}}}{[\sqrt{n}]+[2\sqrt{n}]+\cdots+[n\sqrt{n}]}$\\[3pt]
  \pr $\lp\left(\left(25+\frac 1n\right)^6 - \left(5+\frac 1n\right)^{12}\right)\cdot\sqrt[6]{(n+2)^7 - (n-1)^7}$\\[3pt]
  \pr $\lp((n^5+2)^{25} - (n+5)^{125})\cdot\left(\left(1+\frac{1}{25n^4}\right)^{125} - 1\right)^{31}$
  \pr $\lp\frac{\sqrt{n^5 + 17\sqrt[6]{n}} - \sqrt{n^5 -5\sqrt[6]{n}+1}}{\sqrt[3]{n^5 + 18n-16} - \sqrt[3]{n^5-9n}}$\\[5pt]
%\textbf{6. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i pomoc\'i Stolzovy v\v{e}ty:}\resetpr
%\pr $\lp \tfrac{1^k + 2^k + \ldots + n^k}{n^{k+1}},\; k\in\en$\\
%\pr $\lp \tfrac{1^k + 3^k + \ldots + (2n-1)^k}{n^{k+1}},\; k\in\en$
%\pr $\lp \left(\tfrac{1^k + 2^k + \ldots + n^k}{n^k} - \tfrac{n}{k+1}\right),\; k\in\en$
%\pr $\lp \tfrac{\log n}{n}$\\[5pt]
\textbf{5. Uka\v{z}te, \v{z}e rekurentn\v{e} zadan\'a posloupnost $\{a_n\}_{n=1}^\infty$ m\'a limitu a ur\v{c}ete j\'i:}\\ \resetpr \pr $a_{n+1} = \tfrac{a_n + 3}{4},\, a_1 = 0$ \pr $a_{n+1} = \sqrt{2 + a_n},\, a_1 = \sqrt{2}$ \pr $a_{n+1} = a_n + \tfrac{1}{2}(x - a_n^2),\, a_1 = 0,\, 0\leq x\leq 1$\\ \pr $a_{n+1} = \tfrac{1}{2}(a_n + \tfrac{1}{a_n}),\, a_1 > 0$ \pr $a_{n+1} = \tfrac{a_n(a_n^2 + 3x)}{3a_n^2 + x},\, a_1 > 0,\, x\geq 0$ \pr $a_{n+1} = \tfrac{1}{3}(2a_n + \tfrac{x}{a_n^2}),\, a_1 >0,\, x > 0$\\[5pt]
\textbf{6. Spo\v{c}t\v{e}te limity posloupnost\'i}:\resetpr
\pr $\lp\left(1 + \tfrac{1}{n}\right)^{3n}$
\pr $\lp\left(1 - \tfrac{1}{100+2n}\right)^n$
\pr $\lp\left(1 + \tfrac{1}{n^2}\right)^n$\\
\pr $\lp \left(\tfrac{99}{100} + \tfrac{1}{5n}\right)^n$
\pr $\sqrt[n]{n! + 3^n + 4n}$\\[5pt]
\noindent\textbf{7. Pro posloupnost $(a_n)_{n=1}^\infty$ ur\v{c}ete hodnoty $\limsup_{n\to \infty} a_n$ a $\liminf_{n\to \infty} a_n$:}\\[3pt] \resetpr
\pr $a_n = \tfrac{(-1)^n}{n} + \tfrac{1 + (-1)^n}{2}$
\pr $a_n = (-1)^n\tfrac{n^2 + 1}{n + 1}$
\pr $a_n = 1 + \tfrac{n}{n+1}\cos\left(\tfrac{n\pi}{2}\right)$
\pr $a_n = \left(1 + \tfrac{1}{n}\right)^n \cos(n\pi)$
\pr $a_n = n^{(-1)^n}$
\pr $a_n = \tfrac{n - 1}{n + 1}\cos\left(\tfrac{2n\pi}{3}\right)$
\pr $a_n = \sqrt[n]{1 + 2^{n(-1)^n}}$\\[5pt]
\textbf{8. Pro posloupnost $(a_n)_{n=1}^\infty$ najd\v{e}te mno\v{z}inu hromadn\'ych bod\r{u} $H(\{a_n\})$:}\\[3pt] \resetpr
\pr $a_n = n\sin\left(\frac{n\pi}{4}\right)$
\pr $a_n = 1 + 2(-1)^{n+1} + 3(-1)^{\frac{n(n-1)}{2}}$
\pr $a_n = 1 + n\sin\left(\frac{n\pi}{2}\right)$
\pr $a_n = (1+\frac{1}{n})^n(-1)^n + \sin\left(\frac{n\pi}{4}\right)$
\pr $a_n = (-1)^n\frac{n+4}{3n+1} + \sin\left(\frac{n\pi}{2}\right)\sqrt[n]{\frac{4^n + 6n}{3^n + 5n}}$\\[5pt]
\textbf{9. Pro posloupnost $(a_n)_{n=1}^\infty$ nalezn\v{e}te co nejjednodu\v{s}\v{s}\'i posloupnost $(b_n)_{n=1}^\infty$ kter\'a se chov\'a jako $(a_n)_{n=1}^\infty$ ve smyslu $\lp \frac{a_n}{b_n}\in (0,\infty)$:}\\[3pt] \resetpr
\pr $a_n = \frac{2n^2 + 4}{n^3 + 4n + 6}$
\pr $a_n = \frac{4n^2 + 6n + 2}{2^n + 3^n}$
\pr $a_n = (\sqrt[3]{n+2} - \sqrt[3]{n+1})$
\pr $a_n = \frac{\sqrt{n^3 + 2} - \sqrt[3]{n^2 + 4}}{(n+2)^{20} - n^{20}}$\\
\pr $a_n = (1+\frac{1}{n})^n\frac{\sqrt[9]{n^{10}} + \sqrt[8]{n^7}}{\sqrt[4]{n^{5}} + \sqrt[3]{n^2}}$
\pr $a_n = \frac{[(n+3)^2]}{\sqrt[n]{n^n + 5n! + 4^n}}$
\pr $a_n = \sqrt[n]{\frac{1}{((n+1)^3 - n^3 - 3n^2)^{n-1}}}$\\[5pt]
\textbf{10. Pomoc\'i Bolzano-Cauchyovy podm\'inky a vztahu $|\cos x - \cos y|\leq |x - y|,\;x,y\in\er$ doka\v{z}te, \v{z}e posloupnost $(a_n)_{n=1}^\infty$ pro kterou plat\'i $a_1 = 1$, $a_{n+1} = \cos\left(\frac{a_n}{2}\right)$ je konvergentn\'i.}\\
\rule{\textwidth}{1pt} \\[5pt]
\begin{center}\textbf{Uk\'azkov\'e p\v{r}\'iklady k prvn\'imu z\'apo\v{c}tov\'emu testu}\end{center}
V n\' asleduj{\'\i}c{\'\i}ch p\v r{\'\i}kladech ur\v cete 
$$\lim_{n\rightarrow \infty} a_n,$$ 
nebo doka\v zte, \v ze tato limita neexistuje.\resetpr
\begin{align*}\text{\pr} a_n = \frac{1}{n(\sqrt{n^2 + 2} - \sqrt[3]{n^3 + 1})}\qquad & \text{\pr} a_n = \frac{n^n - (n+1)^n}{(n+1)^n - (n+2)^n}\\
\text{\pr} a_n = (-1)^n\frac{2^n + 4n + 6}{2^n(\sqrt[n]{5} - 1)}\qquad & \text{\pr} a_n = \frac{(n+2)^{711} - (n+8)^{711}}{(n^{11}+3)^{71} - (n^{11}+4)^{71}}\\
\text{\pr} a_n = \sqrt[n]{\left(1+\frac{1}{n}\right)^n + 3^n}\qquad & \text{\pr} a_n = \frac{2^n + 3n^2 + [n!]}{3^n + 4n^2 + [2n!]}
\end{align*}

\pr Posloupnost zadan\'a rekurentn\v{e}: $a_1 = 4,\quad a_{n+1} = (a_n)^2 - 6$

\pr Ur\v{c}ete hodnoty $\limsup_{n\to \infty} a_n$ a $\liminf_{n\to \infty} a_n$ pro posloupnost $a_n = (1+\frac{1}{n})^n(-1)^n + \sin\left(\frac{n\pi}{4}\right)$
\newpage
\centerline{\MakeUppercase{v. \v{r}ady}}\ \\
\textbf{1. Ur\v{c}ete, zda konverguj\'i n\'asleduj\'ic\'i \v{r}ady:}\\[5pt]\resetpr
\pr $\suma (-1)^n\frac{n^2 + 3n + 4}{n^2 + 5}$
\pr $\suma \frac{n^2}{n^3 + 1}$
\pr $\suma \frac{1}{\sqrt{2n+1}\sqrt{2n+3}}$
\pr $\suma \frac{3}{2^n - 2n}$
\pr $\suma \frac{2^n + (-1)^nn}{3^n + (-1)^nn}$\\
\pr $\suma (\sqrt{n^3 + 1} - \sqrt{n^3 - 1})$
\pr $\suma \frac{3^n + 4^n}{4^n + 5^n}$
\pr $\suma \frac{\sqrt[3]{n^2 + 4} - \sqrt[3]{n^2 + 1}}{\sqrt[3]{n}}$
\pr $\suma \frac{1}{(2 + \frac{1}{n})^n}$
\pr $\suma \frac{1}{n\sqrt[n]{n}}$\\
\pr $\suma \frac{n^{n-1}}{\sqrt{(n^2 - n + 1)^{n+1}}}$
\pr $\suma \frac{\sqrt{n^3 + 2} - \sqrt[3]{n^2 + 4}}{(n+2)^{20} - n^{20}}$
\pr $\suma \frac{\sqrt{n + 6} - \sqrt{n+1}}{\sqrt[n]{n^n + 3n! + 4^n}}$
\pr $\suma \frac{n}{3^n}$
\pr $\suma \frac{n^3 + 4n + 1}{4^n + 3n}$\\[5pt]
\textbf{2. Ur\v{c}ete, zda konverguj\'i n\'asleduj\'ic\'i \v{r}ady:}\\[5pt]\resetpr
\pr $\suma \frac{2^n}{n!}$\quad
\pr $\suma \frac{(n!)^2}{(2n)!}$\quad
\pr $\suma \frac{e^{nx}}{n},x\in\er$\quad
\pr $\suma \frac{e^{nx}}{n^2},x\in\er$\quad
%%\pr $\suma \frac{1}{\sqrt[n]{\log n}}$\quad
\pr $\suma (\frac{n-1}{n+1})^{n(n-1)}$\\
\pr $\suma (\sqrt{n^2 + 1} - \sqrt{n^2-1})$\quad
\pr $\suma (\sqrt[3]{n^2 + 1} - \sqrt[3]{n^2-1})$\quad
\pr $\suma \frac{2n^2 + 3n + 4}{2n^4 + 3}$\quad
\pr $\suma \frac{2n^2 + 3n + 4}{2n^3}$\\
\pr $\suma \frac{n^8}{3^n + 5^n}$\quad
%%\pr $\suma \log(1+A^k),A\geq 1$\quad
\pr $\suma \frac{n^5}{5^n}$\quad
\pr $\suma \frac{\sqrt[9]{n^{10}} + \sqrt[10]{n^9}}{\sqrt[9]{n^{11}} + \sqrt[11]{n^9}}$
\pr $\suma \frac{n^{n + \frac{1}{n}}}{(n + \frac{1}{n})^n}$
\pr $\suma \frac{n^3(\sqrt{2} + (-1)^n)^n}{3^n}$\\[5pt]
\textbf{3. Za pomoci Raabeho krit\'eria (viz. n\'i\v{z}e) ur\v{c}ete, zda konverguj\'i n\'asleduj\'ic\'i \v{r}ady:}
\begin{thm}[Raabeovo krit\'erium]Nech\v{t} $\suma a_n$ je \v{r}ada s kladn\'ymi \v{c}leny.
\begin{itemize}
	\item[(i)] Je-li $\lim n(\frac{a_n}{a_{n+1}} - 1) > 1$, pak je $\suma a_n$ konvergentn\'i.
	\item[(ii)] Je-li $\lim n(\frac{a_n}{a_{n+1}} - 1) < 1$, pak je $\suma a_n$ divergentn\'i.
\end{itemize}
\end{thm}\resetpr
\pr $\suma \frac{4^n(n!)^2}{(2n+1)!}$
\pr $\suma \frac{\sqrt{n!}}{(2+\sqrt{1})\cdots(2+\sqrt{n})}$
\pr $\suma \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots (2n)}$
\pr $\suma \frac{1\cdot 4\cdot 7\cdots (3n-2)}{3\cdot 6\cdot 9\cdots (3n)}$\\[5pt]
\textbf{4. Pomoc\'i kondenza\v{c}n\'iho krit\'eria ur\v{c}ete konvergenci n\'asleduj\'ic\'ich \v{r}ad:}\\ \resetpr
\pr $\suma \frac{1}{n\log n}$ (obecn\v{e}ji pak $\suma \frac{1}{n(\log n)^a}$ pro $a > 0$)
\pr $\suma \frac{\log n}{n^2}$ (obecn\v{e}ji pak $\suma \frac{\log n}{n^a}$ pro $a > 0$; m\r{u}\v{z}ete bez d\r{u}kazu pou\v{z}\'it fakt \v{z}e pro ka\v{z}d\'e $a > 0$ existuje $n_0$ takov\'e, \v{z}e posloupnost $(\frac{\log n}{n^a})_{n=n_0}^\infty$ je nerostouc\'i)\\[5pt]
\textbf{5. Ur\v{c}ete, zda n\'asleduj\'ic\'i \v{r}ady konverguj\'i absolutn\v{e}:}\\ \resetpr
\pr $\suma (-1)^n\frac{1}{n\sqrt[n]{n}}$
\pr $\suma \sin(n^2 + 3n)\frac{4^n + 3^n}{5^n + 2^n}$
\pr $\suma (-1)^n\frac{\sqrt[9]{n^{10}} + \sqrt[10]{n^9}}{\sqrt[9]{n^{11}} + \sqrt[11]{n^9}}$\\[5pt]
\textbf{6. Vy\v{s}et\v{r}ete konvergenci (absolutn\'i i neabsolutn\'i) n\'asleduj\'ic\'ich \v{r}ad:}\\ \resetpr
\pr $\suma (-1)^n \tfrac{1}{n \log n}$
\pr $\suma (-1)^n \frac{n}{n+1}$
\pr $\suma (-1)^n \frac{n^2}{n^3 + 1}$
\pr $\suma (-1)^n (\sqrt{n + 1} - \sqrt{n})$\\
\pr $\suma (-1)^n (\sqrt{n^2 + n} - n)$
\pr $\suma (-1)^n \frac{n}{n^2 + 6}$\\
\pr $\suma (-1)^n (\sqrt{n^4 + 1} - n^2)\log n$ \emph{(pou\v{z}ijte v\'ysledek p\v{r}\'ikladu 4b)}
\pr $\suma (-1)^n \frac{n}{(n^2+1)\log n}$\\[5pt]
\textbf{7. Vy\v{s}et\v{r}ete konvergenci (absolutn\'i i neabsolutn\'i) n\'asleduj\'ic\'ich \v{r}ad:}\\ \resetpr
%%\pr $\suma (-1)^n \frac{\sqrt[n]{n^9}}{\log(\log n)}$
\pr $\suma (-1)^n \frac{1}{\sqrt{n}\log n}$
\pr $\suma \cos(n^2)(\sqrt{n^6 + n} - n^3) $
\pr $\suma (-1)^n \frac{n}{n+1}\cos(\frac 1n) $\\
\pr $\suma (-1)^n \frac{n}{n^2+1}\cos(\frac 1n)$
\pr $\suma \frac{1}{\sqrt[n]{\log n}}$\\
\pr $\suma \frac{\log n}{n^2} \cos\left(\frac{\sin n + 8}{n^8}\right)$ \emph{(pou\v{z}ijte v\'ysledek p\v{r}\'ikladu 4b)}
\pr $\suma (-1)^n \frac{1}{\log n} \cos\left(\frac{n}{n^2+1}\right)$\\[5pt]
\pr $\suma \frac{\sin^2 n}{n}$
\pr $\suma (-1)^n \frac{n\sin n}{n^2 + 1}$
\pr $\suma \frac{\sin(n)\cos(\tfrac{1}{n})}{n}$
\pr $\suma \frac{\sin(n + \tfrac{1}{n})}{\log(\log n)}$
\pr $\suma \frac{\cos(2n)}{\log n}$\\
\pr $\suma (-1)^n \frac{2n^2 + 3n + 4}{2n^4 + 3}$
\newpage
\centerline{\MakeUppercase{vi. limita funkc\'i}}\ \\
\textbf{1. Uka\v{z}te, \v{z}e n\'asleduj\'ic\'i v\'yroky jsou ekvivalentn\'i v\'yroku $\lim_{x\to c} f(x) = A$ (v ka\v{z}d\'e \'uloze je uvedeno, pro jak\'a $A,c$ m\'ate ekvivalenci dokazovat).}\resetpr\\[5pt]
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x\in P(c,\delta):\quad f(x)\in B(A,5\varepsilon)\qquad (\text{uva\v{z}ujte }A\in\er, c\in\er)$\\
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x\in P(c,8\delta):\quad f(x)\in B(A,\varepsilon)\qquad (\text{uva\v{z}ujte }A = +\infty, c\in\er)$\\
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x\in P(c,\delta):\quad f(x)\in [A - \varepsilon, A + \varepsilon]\qquad (\text{uva\v{z}ujte }A\in\er, c = - \infty)$\\[5pt]
\textbf{2. Spo\v{c}t\v{e}te z definice (nebo z definice doka\v{z}te \v{z}e limita neexistuje):}\resetpr\\
\pr $\lim_{x\to 5}x$
\pr $\lim_{x\to 2^+}\frac{1}{x-2}$
\pr $\lim_{x\to 0} \max\{4 - |x-4|,0\}$
\pr $\lim_{x\to 0} x\sin\frac{1}{x}$
\pr $\lim_{x\to 0}\sin\frac{1}{x}$\\
\pr $\lim_{x\to 2^+}f(x)$, $\lim_{x\to 2^-}f(x)$ a $\lim_{x\to 2}f(x)$, kde funkce $f$ je d\'ana p\v{r}edpisem $f(x) = \begin{cases}1 & x\in(0,2)\\
												0 & \text{jinak.}\end{cases}$\\[5pt]
\textbf{3. Pro ka\v{z}d\'y z n\'asleduj\'ic\'ich v\'yrok\r{u} nalezn\v{e}te funkci $f:\er\to\er$ a \v{c}\'isla $A,c\in\er$, \v{z}e tvrzen\'i ``$\lim_{x\to c}f(x) = A$'' nen\'i tomuto v\'yroku ekvivalentn\'i.}\resetpr\\
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x\in P(c,\delta):\quad f(x) = A$\\
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x:\quad \big(|x-c| = \delta\implies f(x)\in B(A,\varepsilon)\big)$\\
\pr $\forall \varepsilon > 0\;\exists\delta > 0\;\forall x\in B(c,\delta):\quad f(x)\in B(A,\varepsilon)$\\[5pt]
\textbf{4. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje}\\ \resetpr
\pr $\lx[3] \frac{x^2 - 5x + 6}{x^2 - 8x + 15}$
\pr $\lx[1] \frac{x^3 - 3x + 2}{x^4 - 4x + 3}$
\pr $\lx[1] \frac{x^4 - 3x + 2}{x^5 - 4x + 3}$
\pr $\lx[2] \frac{(x^2 - x - 2)^{20}}{(x^3 - 12x + 16)^{10}}$
\pr $\lx[1] \frac{x^{100} - 2x + 1}{x^{50} - 2x + 1}$\\[5pt]
\textbf{5. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje (p\v{r}\'iklady na pod\'ily polynom\r{u}):}\\ \resetpr
\pr $\lx[1] \frac{x^2 + 2x + 1}{x^2 - 2x + 1}$
\pr $\lx[2] \left(\frac{1}{x(x-2)^2} - \frac{1}{x^2 - 3x + 2}\right)$
\pr $\lx[1] \left(\frac{x+2}{x^2 - 5x + 4} + \frac{x-4}{3(x^2 - 3x + 2)}\right)$\\[5pt]
\textbf{6. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje (p\v{r}\'iklady na v\'yrazy s odmocninami):}\\ \resetpr
\pr $\lx \frac{\sqrt{1+x} - \sqrt{1-x}}{x^n},\;n\in\zet$
\pr $\lx[1] \frac{\sqrt[3]{x^2 + 7} - \sqrt[3]{x^3 + 7}}{(x-1)^n},\;n\in\zet$
\pr $\lx[8] \frac{\sqrt{9+2x}-5}{\sqrt[3]{x} - 2}$\\
\pr $\lx[3] \frac{\sqrt{x+13}-2\sqrt{1+x}}{x^2 - 9}$\\[5pt]
\textbf{7. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje (p\v{r}\'iklady na pou\v{z}it\'i zn\'am\'ych limit $\tfrac{\sin x}{x}\to 1$, $\tfrac{\log x}{x-1}\to 1$, $\frac{e^x-1}{x}\to 1$ a VOLSF):}\\ \resetpr
\pr $\lx \frac{\sin 5x - \sin 3x}{\sin x}$% na tabuli
\pr $\lx \frac{1-\cos x}{x^2}$ (tuto limitu d\'ale lze pova\v{z}ovat za ,,zn\'amou``)% na tabuli
\pr $\lx \frac{\log(\cos x)}{x^2}$\\% na tabuli
\pr $\lx \frac{\log(1+x)}{x}$ (tuto limitu d\'ale lze pova\v{z}ovat za ,,zn\'amou``)
\pr $\lx \frac{e^{\sin^2 x} - \cos x}{x^2}$
\pr $\lx \frac{1-\cos x}{\sin^3 x}$\\
\pr $\lx \frac{\tg x - \sin x}{\sin^3 x}$
\pr $\lx[\infty] \left(\tfrac{x+2}{2x+1}\right)^{x^2}$
\pr $\lx[\infty] \left(\tfrac{x^2+1}{x^2-2}\right)^{x^2}$
\pr $\lx \left(1+x^2\right)^{\cotg^2 x}$
\pr $\lx \left(\tfrac{1+\tg x}{1+\sin x}\right)^{\tfrac{1}{\sin^2 x}}$\\
\pr $\lx \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}$
\pr $\lx \frac{\sqrt{1+\tg x} - \sqrt{1+\sin x}}{x^3}$\\
\pr $\lx \frac{x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$
%%PRISTE:
\pr $\lx[\tfrac{\pi}{6}] \frac{2\sin^2x + \sin x - 1}{2\sin^2x - 3\sin x + 1}$
\pr $\lx[\frac\pi4] (\tg x)^{\tg 2x}$
\pr $\lx \left(\tfrac{1+x\cdot 2^x}{1+x\cdot 3^x}\right)^{\tfrac{1}{x^2}}$
\pr $\lx (x + e^x)^{\frac{1}{x^2}}$\\[10pt]
\textbf{8. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje:}\\ \resetpr
\pr $\lx[\infty] \tfrac{\log\left(x^2-x+1\right)}{\log\left( x^{10}+x+1\right)}$
\pr $\lx[\frac \pi3] \frac{\sin(x-\frac \pi3)}{1-2\cos x}$% HINT: $\cos(x) = \cos(x-\frac \pi3 + \frac \pi3)$
\pr $\lx[\infty] (\sin \sqrt{x+1} - \sin \sqrt{x})$
\pr $\lx \tfrac{\arcsin x}{x}$\\
%%to budu pocitat ja - a ted oni analogicke:
\pr $\lx[\infty] \tfrac{\log\left(1+\sqrt x +\root 3\of x\right)}{\log\left(1+\root 3\of x+\root 4\of x\right)}$
\pr $\lx[a] \frac{\sin x - \sin a}{x-a}$
\pr $\lx[\infty] (\cos \sqrt{x+1} - \cos \sqrt{x-1})$
\pr $\lx[1^-] \frac{\arccos x}{\sqrt{1-x}}$\\
%%a dale jiz smesice vseho - at poznaji jak na kterou limitu
\pr $\lx[\frac\pi4] \tfrac{\sin x -\cos x}{\left(x-\frac\pi4\right)^2}$
\pr $\lx[\infty] \log(1+2^x)\cdot\log\left(1+\frac3x\right)$
\pr $\lx \tfrac{\arcsin x}{\sin(\sin x)}$
\pr $\lx \tfrac{(\cos x)^x-\sqrt{1+\sin^3 x}}{x^3}$\\
\pr $\lx[\frac \pi4] \tg 2x\cdot\tg\left(\frac\pi4-x\right)$
\pr $\lx[\infty] x\left(\frac{\pi}{2} - \arcsin \frac{x}{\sqrt{x^2 + 1}}\right)$
%%\pr $\lx[0] \frac{\arctg(1+x) - \arctg(1-x)}{x}$\\ je treba trigonometricka formule pro arctg a to je asi blbost se ucit
\pr $\lx[0^+] \tfrac{\arccos\tfrac{1-x^2}{1+x^2}}{\sin x}$\\[10pt]
\begin{thm}[Heineova v\v{e}ta]
Nech\v{t} $c \in \er^*$, $A \in \er^*$ a funkce $f$ je definov\'ana na n\v{e}jak\'em prstencov\'em okol\'i bodu $c$. Pak jsou n\'asleduj\'ic\'i dva v\'yroky ekvivalentn\'i.
\begin{itemize}
\item[(i)] Plat\'i $\lim_{x \to c} f(x) = A$.

\item[(ii)] Pro ka\v{z}dou posloupnost $\{x_n\}$ spl\v{n}uj\'ic\'i $x_n \in D_f$, $x_n \neq c$ pro v\v{s}echna $n \in \N$ a $\lim_{n \to \infty}x_n = c$,
plat\'i $\lim_{n \to \infty}f(x_n) = A$.
\end{itemize}\end{thm}\ \\
\textbf{9. Pro limity spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje. Pro \v{r}ady vy{s}et\v{r}ete jejich absolutn\'i konvergenci.}\\ \resetpr
\pr $\lp \left(\frac{n+1}{n-1}\right)^{\frac{1-\sqrt{n}}{1-n}}$
\pr $\lx[\infty] \frac{\log(e^{2x} + x^3)}{\log(e^x + x^4)}$
\pr $\suma \frac{1}{\sqrt{n}}\log\frac{n}{n + 1}$\\
\pr $\lp n(\sqrt[n]{2} - 1)$
\pr $\lx[\infty] \frac{\sqrt[x]{1 + x^2 + 3^x + x^x}}{2x}$
\pr $\suma (\sqrt{n+2} - \sqrt{n})\arctg\frac{1}{\sqrt{n} + n}$
\\[10pt]
\textbf{10. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje (jedn\'a se o p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na FSV):}\\ \resetpr
  \pr $\lx {\left(\frac{(1+\cos x)^x + 1}{2}\right)^\frac 1x}$
  \pr $\lx[\infty] {\frac{\log(1 + x^3 + 3^x + x^x)}{\log(1 + x^3 + 3^x)\log(1 + x^3)}}$
  \pr $\lx[-\frac \pi2] {\frac{\tg x\cdot \log(\sin^2x)}{\sqrt{1 + \sin x}}}$\\
  \pr $\lp {\sqrt[n]{n^n + n^{n+1} + \ldots + n^{2n}}\left(1 - \cos \tfrac 3n\right)}$
  \pr $\lx {\frac{\sqrt{1 + \cos x\cdot \log(\cos x)} - \sqrt{1 + \log(\cos x)}}{\sqrt[3]{\tg x} - \sqrt[3]{\sin x}}\cdot\frac{\sqrt[3]{x}}{\sin^2 x}}$\\
  \pr $\lp {\frac{\sqrt{n^3 + \sin n} - \sqrt{n^3 + 3n}}{\sqrt[4]{n^2 + n} - \sqrt[4]{n^2 + \arctg n}}}$
  \pr $\lx[\frac{\pi}{4}]{\left(\frac{\tg x + \cotg x}{2}\right)^{\tg^2 2x}}$\\[10pt]
\textbf{11. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i, nebo uka\v{z}te \v{z}e limita neexistuje (jedn\'a se o p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na MFF):}\\
a) $\lp \frac{(1+\frac 1n)^{120} - (1+\frac{2}{n})^{80}}{(1-\frac{1}{n})^{100} + (1+\frac 3n)^{100} - 2}$\quad
b) $\lp \frac{\log(n^2 + n + 1) - \log(n^2 + n)}{\sqrt{n^4 + 2} - \sqrt{n^4 + 1}}$\\[10pt]
\textbf{12. Vy\v{s}et\v{r}ete konvergenci a absolutn\'i konvergenci n\'asleduj\'ic\'ich \v{r}ad (jedn\'a se o p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na MFF):}\\ \resetpr
\pr $\suma \arcsin(\sqrt{n^2 + 3} - \sqrt{n^2 - 1})\sqrt{\sin(\frac 1n)}$
\pr $\suma \arctg(\frac 1n - \frac{1}{\sqrt{n^2 + 1}})\frac{\sqrt[3]{n+1} - \sqrt[3]{n}}{(\sin(\frac 1n))^{\frac{10}{3}}}$\\
\pr $\sum_{n=2}^\infty \frac{1}{\log n}\arctg((-1)^n\frac{1}{\sqrt{n}})$
\pr $\suma \frac{\arccotg(2n)}{\sqrt[3]{n+4}}$
\pr $\suma \cos(\pi n)\arccos(\log(e - \frac{1}{n}))$ (zde AK nevy\v{s}et\v{r}ujte)\\[10pt]
\textbf{13. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje (jedn\'a se o p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na MFF):}\\
%a) $\lx \frac{\cos(\sin x) - 1}{\log(\sqrt{1+x^2})}$\quad
a) $\lx[0^+] \frac{\log(1 + x^3 + 3^x + x^x)}{\log(1 + x^3 + 3^x)\log(1 + x^3)}$
b) $\lx \left(\frac{1^x + 2^x + 3^x}{3}\right)^{1/x}$\quad
c) $\lx[0^+] \frac{\arccos(e^{-x})}{\sqrt{x}}$\quad
d) $\lx \frac{e^{\sin 2x} - e^{\arcsin x}}{\tg x}$
\newpage
\centerline{\MakeUppercase{vii. spojitosti a derivace}}\ \\ 
\textbf{1. Zjist\v{e}te, ve kter\'ych bodech defini\v{c}n\'iho oboru je funkce spojit\'a (pokud nen\'i v n\v{e}jak\'em bod\v{e} spojit\'a, ur\v{c}ete jestli je spojit\'a zleva a zprava).}\\[5pt] \resetpr
\pr $\max\{2x, x^2\}$ \pr $[x]\cdot \log x$ \pr $f(x) = \begin{cases}\left(\sin x\right)^{\cos x} & x\in (0,\pi),\\
	0 & x = 0,\\
	1 & x = \pi.
	\end{cases}$\\
\textbf{2. Najd\v{e}te $A\in\er$, aby pro ka\v{z}d\'e $x\in(0,\infty)$ platil vztah}
$$\left(\log(\cos^2x + \sqrt{1+\cos^4x}) + \arctg x + \arcsin\left(\frac{1}{\sqrt{1+x^2}}\right)\right)' = A\frac{\sin 2x}{\sqrt{1+\cos^4 x}}.$$
\textbf{3. Spo\v{c}t\v{e}te derivaci (resp. jednostrann\'e derivace) n\'asleduj\'ic\'ich funkc\'i ve v\v{s}ech bodech, kde existuje.}\\[5pt] \resetpr
\pr $\arccos(\frac{1-x^2}{1+x^2})$ \pr $x^2\exp(-|x-1|)$ \pr $\sqrt{1-e^{-x^2}}$ \pr $x\arcsin\sqrt{\frac{x}{x+1}}+\arctg\sqrt x-\sqrt x$
  \pr $\log \arccos x$\\ \pr $f(x) = \begin{cases}\left(\sin x\right)^{\cos x} & x\in (0,\pi),\\
	0 & x = 0.
	\end{cases}$\\\\[5pt]
\textbf{4. Vy\v{s}et\v{r}ete spojitost a najd\v{e}te derivaci funkce }$f(x) = $\\[5pt] \resetpr
	\textbf{Ze zkou\v{s}kov\'ych p\'isemek na FSV:}
	\pr $\max\{5x - 4, x^2\}$
	\pr $[x]\cdot \sqrt[3]{x^2 - 9}$
	\pr $|\cos 2x|\cdot (\tg x - 1)$\\[5pt]
	\textbf{Ze zkou\v{s}kov\'ych p\'isemek na MFF:}
	\pr $f(x) = \begin{cases}x^2(\sin\tfrac{1}{x} + \cos\tfrac{1}{x}) & x\neq 0,\\
	0 & x = 0.
	\end{cases}$\\
	\pr $\cos x \cdot [\sin x]$
	\pr $(e + |x|)^x$\\[20pt]
\centerline{\MakeUppercase{viii. aplikace derivac\'i}}\ \\
\textbf{1. Spo\v{c}t\v{e}te limity (m\r{u}\v{z}ete pou\v{z}\'ivat l'Hospitalovo pravidlo)}\\[5pt]
\resetpr \pr $\lx \frac{\tg x -x }{x - \sin x}$ % 2
\pr $\lx \frac{3\tg 4x - 12 \tg x}{3\sin 4x-12\sin x}$ % -2
\pr $\lx \frac{x\cotg x-1}{x^2}$ % -1/3
\pr $\lx[\tfrac{\pi}{4}] \frac{\sqrt[3]{\tg x} - 1}{2\sin^2x - 1}$\\ % 1/3
\pr $\lx \frac{(1+x)^{1/x} - e}{x}$ % -e/2
\pr $\lx[1] \frac{x^x-x}{\ln x-x+1}$ % -2
\pr $\lx \left(\frac{1}{x} - \frac{1}{e^x - 1}\right)$ % 1/2
\pr $\lx \left(\frac{\arcsin x}{x}\right)^{1/x^2}$\\ % e^{1/6}
\pr $\lx[a] \frac{a^x - x^a}{x-a},\; a>0$ % a^a (\ln a - 1)
\\[10pt]
\textbf{2. Nalezn\v{e}te glob\'aln\'i extr\'emy funkce }$f(x) = x^3 - 3x + 2$,\quad$x\in[-3,2]$\\
\textbf{3. Nalezn\v{e}te extr\'emy (lok\'aln\'i i glob\'aln\'i) funkc\'i }\\ \resetpr \pr $f(x) = \tfrac{x^2 - 3x + 2}{x^2 + 2x + 1}$,\quad$x\in\er\setminus\{-1\}$
\pr $g(x) = |x|e^{-|x-1|}$,\quad$x\in\er$\\
\textbf{4. Doka\v{z}te n\'asleduj\'ic\'i nerovnosti:}\resetpr \pr $\forall x>0: x - \tfrac{x^3}{6} < \sin x < x$ \pr $\forall x>0: x^e\leq e^x$\\
\textbf{5. Doka\v{z}te, \v{z}e }$\arctg x = \arcsin \tfrac{x}{\sqrt{1+x^2}}$\\
\textbf{6. Vy\v{s}et\v{r}ete pr\r{u}b\v{e}h funkce }$f(x) = $\\[5pt] \resetpr
\textbf{Ze zkou\v{s}kov\'ych p\'isemek na FSV:}
\pr $\left(3^{x+2|x|} - 9\right)^2$
\pr $\arctg{\frac{x^2}{x^2 - 6}}$
\pr $(x + \log 2)\cdot 2^{-\frac 5x}$\\[5pt]
\textbf{Ze zkou\v{s}kov\'ych p\'isemek na MFF:}
\pr $\tfrac{x^3}{\sqrt{|x^4 - 1|}}$
\pr $|(1-x^2)e^{-x}|$	
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