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\def\ce{\mathbb C} 
\def\en{\mathbb N}
\def\zet{\mathbb Z}
\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}
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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, operace s mno\v{z}inami, bin\'arn\'i relace a zobrazen\'i}}\ \\
\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 $n!\leq \left(\tfrac{n+1}{2}\right)^n\quad$(\emph{Hint: p\v{r}i d\r{u}kazu pou\v{z}ijte Bernoulliho nerovnost})
		\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 $\left|\sin\left(\sum_{i=1}^nx_i\right)\right|\leq \sum_{i=1}^n\sin(x_i)$\\[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]
\textbf{4.} Nech\v{t} $p\in \en$. Doka\v{z}te, \v{z}e relace {\it kongruence modulo $p$} definovan\'a p\v{r}edpisem
$$
m\equiv n\ (\Mod p)\quad\iff\quad |m-n|\ \text{je d\v{e}liteln\'e \v{c}\'islem}\ p,
$$
je ekvivalence na $\en$.\\[5pt]
\textbf{5.} Nech\v{t} $A$ je mno\v{z}ina v\v{s}ech funkc\'i z intervalu $[0,1]$ do $\er$ a nech\v{t} $\leq$ je relace definovan\'a p\v{r}edpisem
$$
f\leq g\ \iff\ \forall\, x\in[0,1]:\ f(x)\leq g(x).
$$
Doka\v{z}te, \v{z}e $\leq$ tvo\v{r}\'i na mno\v{z}in\v{e} $A$ \v{c}\'aste\v{c}n\'a uspo\v{r}\'ad\'an\'i, kter\'e nen\'i line\'arn\'i.\\[5pt]
\textbf{6.} Doka\v{z}te, \v{z}e je-li $X$ mno\v{z}ina, ozna\v{c}me $\P(X):=\{A:\;A\subset X\}$. Pak relace
\[
R=\{[A,B]\in \P(X)\times \P(X):\;A\subset B\}
\]
je \v{c}\'aste\v{c}n\'e uspo\v{r}\'ad\'an\'i na $\P(X)$, kter\'e obecn\v{e} nen\'i line\'arn\'i.\\[5pt]
\textbf{7.} Uva\v{z}ujme relaci $\triangle:=\{[A,B]\in \P(\en)\times \P(\en):\;A\cap B\neq \emptyset\}$. Rozhodn\v{e}te, zda je $\triangle$ reflexivn\'i, symetrick\'a, tranzitivn\'i, antisymetrick\'a, slab\v{e} antisymetrick\'a. Sv\'e tvrzen\'i doka\v{z}te.\\[5pt]
\textbf{8.} \nec $f\colon\en\to\en$, $D(f) = \en$. 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}
\textbf{9.} Ur\v{c}ete defini\v{c}n\'i obor a obor hodnot funkc\'i\resetpr \pr $f(x):= x - \sqrt{x^2-1}$ \pr $g(x):=\tfrac{2\sqrt{x}}{4-\sqrt{x}}$\\[3pt]
\newpage
\centerline{\MakeUppercase{iii. axiomy t\v{e}les, uspo\v{r}\'adan\'ych t\v{e}les, suprema, infima}}\ \\
\textbf{1. Doka\v{z}te, \v{z}e pro ka\v{z}d\'e t\v{e}leso $(T,+,\cdot,0,1)$ plat\'i:}\\[5pt]
  a) $\forall x,y,z\in T:\quad x+y = x+z\Rightarrow y=z;\qquad$
  b) Nech\v{t} $x\in T, x\neq 0$. Pak opa\v{c}n\'y prvek $(-x)$ a inverzn\'i prvek $\frac{1}{x}$ jsou
jednozna\v{c}n\v{e} definov\'any.$\qquad$
  c) $\forall x \in T:\ x \cdot 0  = 0;$\qquad
  d) $-0=0;$\\
  e) $\forall a \in \mathbb R:\ -(-a)=a;$\qquad  
  f) $\forall a \in T:\ -a = (-1)\cdot a;$\qquad
  g) $\forall a,b \in T:\ (-a)b=-(ab)=a(-b);$\\
  h) $\forall x,y \in T:\ (-x)(-y)=xy;$\qquad
  i) $\forall a,b \in T:\ ab = 0 \Rightarrow (a = 0 \vee b = 0);$\\
	j) $\forall x,y \in T\setminus\{0\}:\ (xy)^{-1} = y^{-1}x^{-1}.$\\[5pt]
\textbf{2. Doka\v{z}te, \v{z}e pro ka\v{z}d\'e uspo\v{r}\'adan\'e t\v{e}leso $(T,+,\cdot,0,1)$ a pro $x,y\in T$ plat\'i:}\\[5pt]
 a) $0 < 1;$\qquad
 b) $0 < x\Rightarrow 0 < \frac{1}{x};$\qquad
 c) $x < y\Rightarrow x < \frac{x+y}{2} < y,$ kde $2=1+1;$\qquad
 d) $0<x<y\Rightarrow 0 < \frac{1}{y} < \frac{1}{x};$\\
 e) $x\neq 0\Rightarrow x\cdot x > 0;$\qquad  
 f) $x < 0 < y\Rightarrow xy < 0;$\qquad
 g) $0 < x < 1\Rightarrow x\cdot x < x;$\\
 h) $\forall n\in\en:\ 0\leq x < y\Rightarrow x^n < y^n;$\qquad
 i) $x\geq 0\Rightarrow -x\leq 0;$\\
 j) $\forall z,w\in T\setminus\{0\}:\ \tfrac{x}{z}.\tfrac{y}{w} = \tfrac{xy}{zw};$\qquad
 k) $\forall z,w\in T\setminus\{0\}:\ \tfrac{x}{z} + \tfrac{y}{w} = \tfrac{xw + yz}{zw}$\\[5pt]
\textbf{3. 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) $F = \{5^{(-1)^j3^k};\ j \in \zet, \ k \in \zet\};$\\[5pt]
\textbf{4. Nech\v{t} $A,B\subset\er$ jsou nepr\'azdn\'e omezen\'e mno\v{z}iny. Doka\v{z}te, \v{z}e}\\[5pt]
a) $\sup(A\cup B)=\max\{\sup A,\sup B\}$, b) $\sup(-A)=-\inf A\quad$ a $\quad\inf(-A)=-\sup A$\\
c) $\sup(A+B)=\sup A + \sup B$, d) $\sup(A-B)=\sup A - \inf B$.
\newpage
\centerline{\MakeUppercase{iv. limity posloupnost\'i}}\ \\
\textbf{1. Ur\v{c}ete, zda je posloupnost $\{a_n\}_{n=1}^\infty$ monot\'onn\'i a (jednostrann\v{e}) omezen\'a:}\\[5pt]
 a) $\frac{n^2-1}{n}$\quad b) $n^{1-n}$\quad c) $\sin(\frac{n\pi}{4})$\\[5pt]
\textbf{2. 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{3. Doka\v{z}te, \v{z}e posloupnost $\{a_n\}_{n=1}^\infty$ nem\'a vlastn\'i limitu:} a) $a_n=2^{(-1)^nn}\qquad$ b) $a_n=(-1)^n\left(\frac{1}{10} - \frac 1 n\right)$\\[5pt]
\textbf{4. 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{5. 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 $\lp\frac{n^5 + 2^n + 17^n}{n! + n + 3^n}$
\pr $\lp\frac{n+[\sqrt[3]{n}]^3}{n-[\sqrt{n+9}]}$
\pr $\lp\frac{3^n + n^5 + n!}{n(n^6+n!)}$
\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 $\lp\sqrt{\frac{n + \sqrt{n^2 + \sqrt{n^4 + 1}}}{n+1}}$
\pr $\lp\sqrt[n]{\sqrt{3^n + 2\cdot 2^n} - \sqrt{3^n + 2^n}}$
\pr $\lp\frac{\sqrt[n]{2^n + 3^n}}{\sqrt[2n]{4^n + \sqrt{n}}}$\\[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{7. 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 $\lp\left(\tfrac{99}{100} + \tfrac{5}{n}\right)^n$\\[5pt]
\textbf{8. 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 $a_n = \left(1 + \tfrac{1}{n}\right)^n \cos(n\pi)$
\pr $a_n = \tfrac{2n(-1)^n}{n+1} + \sqrt[n]{2}$
\pr $a_n = \sqrt[n]{1 + 2^{n(-1)^n}}$\\[5pt]
\textbf{9. Spo\v{c}t\v{e}te limity posloupnost\'i (jedn\'a se o uk\'azkov\'e p\v{r}\'iklady ke druh\'emu z\'apo\v{c}tov\'emu testu):}\resetpr
\pr $\lp\tfrac{(-1)^n+\sin \tfrac {n\pi}{4}-\cos \tfrac {n\pi}{3}}{\sqrt[n]{n^3}-1}$
\pr $\lp\tfrac{3^n-[\sqrt[3]{(n+1)!}\  ]}{3n^3-[\sqrt[3]{n!}\ ]}$
\pr $\lp\sqrt[n]{[n^4\cos n ]-n^23^n+4^n}$\\
\pr $\lp\tfrac{5+\sqrt[3]{n^3+n^2}-\sqrt{n^2+n}} {\sqrt{n}-\sqrt[4]{n}}$
\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 $\lp\bigl(1-\tfrac{1}{n}\bigr)^n\tfrac{n}{\sqrt[n]{n!}}$
\newpage
\textbf{10. 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 $\lp\frac{\sqrt{n^5 + 17\sqrt[6]{n}} - \sqrt{n^5 -5\sqrt[6]{n}+1}}{\sqrt[3]{n^5 + 18n-16} - \sqrt[3]{n^5-9n}}$\\[200pt]
\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 $\suma \frac{2^k}{k!}$\quad
\pr $\suma \frac{(k!)^2}{(2k)!}$\quad
\pr $\suma \frac{e^{kx}}{k},x\in\er$\quad
\pr $\suma \frac{e^{kx}}{k^2},x\in\er$\quad
%\pr $\suma \frac{1}{\sqrt[k]{\log k}}$\quad
\pr $\suma (\frac{k-1}{k+1})^{k(k-1)}$\\
\pr $\suma (\sqrt{k^2 + 1} - \sqrt{k^2-1})$\quad
\pr $\suma (\sqrt[3]{k^2 + 1} - \sqrt[3]{k^2-1})$\quad
\pr $\suma \frac{2k^2 + 3k + 4}{2k^4 + 3}$\quad
\pr $\suma \frac{2k^2 + 3k + 4}{2k^3}$\\
\pr $\suma \frac{k^8}{3^k + 5^k}$\quad
%\pr $\suma \log(1+A^k),A\geq 1$\quad
\pr $\suma \frac{k^5}{5^k}$\quad
\pr $\suma \frac{\sqrt[9]{k^{10}} + \sqrt[10]{k^9}}{\sqrt[9]{k^{11}} + \sqrt[11]{k^9}}$\\[5pt]
\textbf{2. 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{3. 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{4. 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 $\suma \frac{\log n}{n^2} \cos\left(\frac{\sin n + 8}{n^8}\right)$
\pr $\suma (-1)^n \frac{1}{\log n} \cos\left(\frac{n}{n^2+1}\right)$\\[5pt]
%NA DALSI CVIKO ZBYVA DIRICHLETOVO KRITERIUM a BC PODMINKA PRO RADY (dokazat, ze sin n ma omezene castecne soucty! - a nejake priklady na to; BC podminka .. ted nevim kde se pouzije, neco treba vymyslet)
\textbf{5. Vy\v{s}et\v{r}ete konvergenci (absolutn\'i i neabsolutn\'i) n\'asleduj\'ic\'ich \v{r}ad:}\\ \resetpr
\pr $\suma \frac{\sin^2 n}{n}$
\pr $\suma (-1)^n \frac{n\sin n}{n^2 + 1}$
\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. 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 $\lx[2] \frac{(x^2 - x - 2)^{20}}{(x^3 - 12x + 16)^{10}}$
\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{2. 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{3. 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 $\lx \left(\tfrac{1+\tg x}{1+\sin x}\right)^{\tfrac{1}{\sin^2 x}}$
%PRISTE: \pr $\lx[\frac \pi3] \frac{\sin(x-\frac \pi3)}{1-2\cos x}$% HINT: $\cos(x) = \cos(x-\frac \pi3 + \frac \pi3)$
%PRISTE: \pr $\lx[a] \frac{\sin x - \sin a}{x-a}$% analogicky jako vyse
\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{4. 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 $\lx[0^+] \tfrac{\arccos\tfrac{1-x^2}{1+x^2}}{\sin x}$
\newpage
\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{5. 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 $\lx[\frac \pi4] {\left(\frac{\tg x + \cotg x}{2}\right)^{\tg^2 2x}}$\\[10pt]
\textbf{6. 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{7. 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{8. 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
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. 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 $(e + |x|)^x$\\[10pt]
	\textbf{2. Vy\v{s}et\v{r}ete spojitost (spo\v{c}\'itejte limity v krajn\'ich bodech defini\v{c}n\'iho oboru a pokud lze funkci spojit\v{e} dodefinovat, ud\v{e}lejte to a pracujte s dodefinovanou funkc\'i) a najd\v{e}te derivaci funkce }$f(x) = $\\ \resetpr  
  \pr $\left(\frac1x\right)^{\frac1x}$
  \pr $\left(\sin x\right)^{\cos x}$
  \pr $\log\arccos x$
  \pr $x\arcsin\sqrt{\frac{x}{x+1}}+\arctg\sqrt x-\sqrt x$
  \pr $\arctg e^x-\log\sqrt{\frac{e^{2x}}{e^{2x}+1}}$\\[10pt]
		
	\centerline{\MakeUppercase{viii. aplikace derivac\'i}}\ \\ 
	\textbf{1. Nalezn\v{e}te glob\'aln\'i extr\'emy funkce }$f(x) = x^3 - 3x + 2$,\quad$x\in[-3,2]$\\
	\textbf{2. 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{3. 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{4. Doka\v{z}te, \v{z}e }$\arctg x = \arcsin \tfrac{x}{\sqrt{1+x^2}}$\\
	\textbf{5. 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 $|(1-x^2)e^{-x}|$	
\end{document}