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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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\newcommand{\pr}{\quad\alph{Priklad})\;\addtocounter{Priklad}{1}}
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\newcounter{cviko}
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\newcommand{\dalsiCviko}{ \begin{center}
	-----------------\textbf{\arabic{cviko}. cvi\v{c}en\'i}-----------------
\end{center}\addtocounter{cviko}{1}}

\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}	
\setcounter{cviko}{1}
\dalsiCviko
\centerline{\MakeUppercase{i. \'uvod}}\ \\
\textbf{1.} 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{2. 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{3. Plat\'i pro libovoln\'e mno\v{z}iny $A$, $B$ a $C$ n\'asleduj\'ic\'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. Doka\v{z}te, \v{z}e pro ka\v{z}d\'e t\v{e}leso $(T,+,\cdot,0,1)$ plat\'i:}\\
(pozn: $(T,+,\cdot,0,1)$ je t\v{e}leso pokud tato p\v{e}tice apl\v{n}uje sadu I. axiom\r{u} re\'aln\'ych \v{c}\'isel z p\v{r}edn\'a\v{s}ky)\\[5pt]\resetpr
  \pr $\forall x,y,z\in T:\quad x+y = x+z\Rightarrow y=z$\\
  \pr 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.\\
  \pr $\forall x \in T:\ x \cdot 0  = 0$
  \pr $-0=0$
  \pr $\forall a \in \mathbb R:\ -(-a)=a$
  \pr $\forall a \in T:\ -a = (-1)\cdot a$\\
  \pr $\forall a,b \in T:\ (-a)b=-(ab)=a(-b)$
  \pr $\forall x,y \in T:\ (-x)(-y)=xy$\\
  \pr $\forall a,b \in T:\ ab = 0 \Rightarrow (a = 0 \vee b = 0)$\\[5pt]
	\dalsiCviko
\textbf{5. 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{i. \'uvod - v\'ysledky}}\ \\	
\textbf{1.} \resetpr \pr (i), (ii), (iii) neplat\'i \pr (i), (ii) plat\'i; (iii) neplat\'i \pr (i) plat\'i; (ii), (iii) neplat\'i \pr (i), (ii), (iii) plat\'i\\
\textbf{2.} \resetpr \pr Plat\'{\i}, negace $\exists x \in \en \ \forall y \in \en \
\exists z \in \en \, (z > x \ \&\ y \ge z)$;\\
 \pr Plat\'{\i}, negace $\forall y\in\en\ \exists x\in\en\ \exists z\in\en\ (z>x\ \&\ y\ge<z)$;\\
 \pr Neplat\'{\i}, negace $\forall x \in \en \ \exists y \in \en \ \exists z \in \en \, (z > x \ \&\ y \ge z)$;\\
 \pr Plat\'{\i}, negace $\forall x \in \en \ \exists y \in \en \ \exists z \in \en \, (z < x \ \&\ y \le z)$;\\
 \pr Plat\'{\i}, negace $\exists x\in\er \ \exists \varepsilon>0\ \forall\delta>0\ \exists y\in\er\, (|y-x|<\delta\ \&\ y\geq x+\frac \varepsilon3)$.\\
\textbf{3.} \resetpr \pr ano \pr ano \pr ne \pr ano \pr ne\\
\textbf{5.} \resetpr
\pr $\sup A = 0, \min A = -1$
\pr $\max B = \frac 32, \min B = 0$
\pr $C_1$ nem\'a supremum ani infimum (nen\'i zdola, ani shora omezen\'a), $C_2$ nen\'i shora omezen\'a, $\min C_2 = 3$, $\max C_3 = 0$, $C_3$ nen\'i zdola omezen\'a
\pr $\max D_1 = \frac 56$, $\inf D_1 = 0$, $D_2$ nen\'i shora omezen\'a, $\inf D_2 = 0$
\pr $E$ nen\'i shora omezen\'a, $\inf E = 0$
\pr $F$ nen\'i shora omezen\'a, $\inf F = 0$.
\newpage
\dalsiCviko
\centerline{\MakeUppercase{ii. limity posloupnost\'i}}\ \\
\textbf{1. 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]
\dalsiCviko
\textbf{2. 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 $\lp\frac{n+[\sqrt[3]{n}]^3}{n-[\sqrt{n+9}]}$
\pr $\lp\frac{3^n + n^5 + (n+1)!}{n(n^6+n!)}$
\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 $\lp\sqrt[n]{\sqrt{3^n + 2\cdot 2^n} - \sqrt{3^n + 2^n}}$
\pr $\lp\frac{\sqrt[n]{2^n + 3^n}}{\sqrt[2n]{4^n + \sqrt{n}}}$\\
\pr $\lp\frac{3^n - [\sqrt[3]{n!}]}{3n^3 - [\sqrt[3]{(n+1)!}]}$
\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{3. 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 $\lp((n^{10}+n^3)^7 - (n^7+1)^{10})\cdot\left(\sqrt[3]{(1+\frac{1}{n^9})^7} - 1\right)^7$
  \pr $\lp\frac{(n+\frac 2n)^{30} - (n+\frac 1n)^{30}}{\sqrt{(2+n^7)^8 - 2^8}}$\\[3pt]
  \pr $\lp\left(\left(25+\frac 1n\right)^6 - \left(5+\frac 1n\right)^{12}\right)\cdot\sqrt[6]{(n+2)^7 - (n-1)^7}$
  \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]	
\noindent\textbf{4. 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 $a_n = n^{(-1)^n}$\\[5pt]
\textbf{5. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity posloupnost\'i (uk\'azkov\'e p\v{r}\'iklady k prvn\'imu z\'apo\v{c}tov\'emu testu):} \\[3pt] \resetpr
\pr $\lp \frac{1}{n(\sqrt{n^2 + 2} - \sqrt[3]{n^3 + 1})}$
\pr $\lp (-1)^n\frac{2^n + 4n + 6}{2^n(\sqrt[n]{5} - 1)}$
\pr $\lp \frac{(n+2)^{711} - (n+8)^{711}}{(n^{11}+3)^{71} - (n^{11}+4)^{71}}$\\
\pr $\lp \frac{2^n + 3n^2 + [n!]}{3^n + 4n^2 + [2n!]}$
\newpage
\centerline{\MakeUppercase{ii. limity posloupnost\'i - v\'ysledky}}\ \\
\textbf{1.} \resetpr \pr 0 \pr 2 \pr 2 \pr 0 \pr 0 \pr Nem\'a limitu \pr 1/2 \pr 1 \pr $\tfrac{81}{2^{13} - 1}$ \pr $\tfrac{1}{3}$ \pr $\tfrac{1}{2}$ \pr 0 \pr $\tfrac{1}{2}$ \pr 0 \pr 0 \pr 1\\
\textbf{2.} \resetpr \resetpr \pr $\tfrac{1}{3}$ \pr $\tfrac{8}{50}$ \pr $\sqrt{1+\sqrt{2}}$ \pr Nem\'a limitu \pr 0 \pr 1 \pr 2 \pr 1 \pr 5 \pr $\tfrac{1}{3}$ \pr $\tfrac{2}{\sqrt{3}}$ \pr $\tfrac{3}{2}$ \pr 0 \pr $+\infty$ \pr 0\\
\textbf{3.}\resetpr \pr $-\frac{7^7}{3^6}$ \pr 30 \pr $-54\cdot 5^{10}\cdot\sqrt[6]{21}$ \pr $\frac{11}{9}$\\
\textbf{4.}\resetpr \pr $\limsup_{n\to \infty} a_n = 1$, $\liminf_{n\to \infty} a_n = 0$
\pr $\limsup_{n\to \infty} a_n = \infty$, $\liminf_{n\to \infty} a_n = -\infty$
\pr $\limsup_{n\to \infty} a_n = 2$, $\liminf_{n\to \infty} a_n = 0$
\pr $\limsup_{n\to \infty} a_n = e$, $\liminf_{n\to \infty} a_n = -e$
\pr $\limsup_{n\to \infty} a_n = \infty$, $\liminf_{n\to \infty} a_n = 0$\\
\textbf{5.}\resetpr \pr 1 \pr limita neexistuje \pr 0 \pr $\tfrac{1}{2}$
\newpage
\dalsiCviko
\centerline{\MakeUppercase{iii. limita a spojitost funkce}}\ \\
\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. 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) $j(x) = |\log|1-x||$.
\dalsiCviko
\textbf{3. 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{4. 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{5. 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{6. 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 $\lx \frac{e^{\sin^2 x} - \cos x}{x^2}$
\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 $\lx \left(\tfrac{1+\tg x}{1+\sin x}\right)^{\tfrac{1}{\sin^2 x}}$\\
\dalsiCviko
\textbf{7. 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 $\lx \tfrac{\arcsin x}{x}$ (tuto limitu d\'ale lze pova\v{z}ovat za ,,zn\'amou``)
\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 $\lx[\frac\pi4] \tfrac{\sin x -\cos x}{\left(x-\frac\pi4\right)^2}$
\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 $\lx \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}$
\pr $\lx \frac{x^2}{\sqrt{1+x\sin x} - \sqrt{\cos x}}$
\pr $\lx \left(\tfrac{1+x\cdot 2^x}{1+x\cdot 3^x}\right)^{\tfrac{1}{x^2}}$\\[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{8. Spo\v{c}t\v{e}te limitu, nebo uka\v{z}te \v{z}e limita neexistuje.}\\ \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 $\lp n(\sqrt[n]{2} - 1)$
\pr $\lx[\infty] \frac{\sqrt[x]{1 + x^2 + 3^x + x^x}}{2x}$
\\[10pt]
\textbf{9. 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 $f(x) = \begin{cases}\left(\sin x\right)^{\cos x} & x\in (0,\pi),\\
	0 & x = 0,\\
	1 & x = \pi.
	\end{cases}$\\
	\dalsiCviko
\textbf{9. 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 $\lp {\frac{\sqrt{n^3 + \sin n} - \sqrt{n^3 + 3n}}{\sqrt[4]{n^2 + n} - \sqrt[4]{n^2 + \arctg n}}}$
  \pr $\lx[\frac{\pi}{4}]{\left(\frac{\tg x + \cotg x}{2}\right)^{\tg^2 2x}}$\\[10pt]
\textbf{10. 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{11. 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) $\lx \left(\frac{1^x + 2^x + 3^x}{3}\right)^{1/x}$\\[10pt]
\textbf{12. Spo\v{c}t\v{e}te n\'asleduj\'ic\'i limity, nebo doka\v{z}te \v{z}e limita neexistuje (uk\'azkov\'e p\v{r}\'iklady k druh\'emu z\'apo\v{c}tov\'emu testu)}\\ \resetpr \pr $\lx[+\infty] \frac{\log(1 + x^2 + 2^x)}{\log(1 + x^3 + 3^x)}$ \pr $\lx[0] \left(\frac{x^2 + 3x + 4}{x + 4}\right)^{1/\sin x}$ \pr $\lx[4] \frac{(x-4)\sin x}{x^2 - 8x + 16}$ \pr $\lx[0] \frac{\sqrt{1 + \sin^4x} - \sqrt{1 + \sin^5x}}{x^4}$\\ \pr $\lx[0] \frac{\left(1 + x\right)^{x} - \cos x}{x^2}$ \pr $\lp \frac{(1 + \tfrac{5}{n})^{60} - 1}{(1 + \tfrac{3}{n})^{50} - 1}$
\newpage
\centerline{\MakeUppercase{iii. limita a spojitost funkce - v\'ysledky}}\ \\
\textbf{1.} a) $(4;6]$\quad b) $(-6;-3)\cup(\tfrac{-1-\sqrt{13}}{2},\tfrac{-1+\sqrt{13}}{2})$\quad c) $[1;2]$\quad d) $\tfrac{4}{3}$\\ e) $\bigcup_{k\in\zet}{((-\frac \pi2 + 2k\pi; \frac \pi6 + 2k\pi)\cup(\frac \pi2 + 2k\pi; \frac {5\pi}{6} + 2k\pi))}$\\
 f) $(-\infty,3)\cup(3,2+\sqrt{6}]$\quad g) $[\tfrac{1}{2},\infty)$\\
 h) $\{0,\tfrac{2}{3}\}$\quad i) $(-10;-7)\cup(10;\infty)$\quad j) $\{7,\tfrac{\sqrt{7}}{7}\}$\\[5pt]
\textbf{3.}\resetpr \pr $-\frac{1}{2}$ \pr $\frac{1}{2}$ \pr $1$ \pr $\frac{3^{10}}{2^{10}}$ \pr $\frac{49}{24}$
\textbf{4.}\resetpr \pr $+\infty$ \pr $\infty$ \pr 0\\
\textbf{5.}\resetpr \pr 0 pro $n<1$, 1 pro $n=1$, pro $n>1$ sud\'e limita neexistuje a pro $n>1$ lich\'e je limita $\infty$ \pr 0 pro $n<1$, $-\frac{1}{12}$ pro $n=1$, pro $n>1$ sud\'e limita neexistuje a pro $n>1$ lich\'e je limita $-\infty$ \pr $\frac{12}{5}$ \pr $-\frac{1}{16}$\\
\textbf{6.}\resetpr \pr 2 \pr $\frac{1}{2}$ \pr $-\frac 12$ \pr $\frac 32$ \pr neexistuje \pr $\frac 12$ \pr 0 \pr $e^3$ \pr $e$ \pr 1\\
\textbf{7.}\resetpr \pr $\frac 15$ \pr 1 \pr $\frac 32$ \pr neexistuje \pr $3\log 2$ \pr 1 \pr -1 \pr $\frac 12$ \pr $-\frac {1}{12}$ \pr $\frac 43$ \pr $\frac{2}{3}$\\
\textbf{8.}\resetpr \pr $1$ \pr $2$ \pr $\log 2$ \pr $\frac{1}{2}$\\
\textbf{9.}\resetpr \pr spojit\'a na $\er$ \pr spojit\'a na $(0,\infty)\setminus\en$ a v bod\v{e} $x = 1$, v bodech $x\in \en\setminus\{1\}$ spojit\'a zprava a nespojit\'a zleva \pr spojit\'a v $(0,\pi)$, v bod\v{e} $x=0$ spojit\'a zprava (spojitost zleva nem\'a smysl vy\v{s}et\v{r}ovat), v bod\v{e} $x=\pi$ nen\'i spojit\'a zleva (spojitost zprava nem\'a smysl vy\v{s}et\v{r}ovat)\\
\textbf{9.}\resetpr \pr $\sqrt{2}$ \pr $\frac {1}{3\log 3}$ \pr limita neexistuje (zprava $\sqrt {2}$, zleva $-\sqrt{2}$) \pr -6 \pr $\sqrt{e}$\\
 \textbf{10.}\resetpr \pr $-\frac 15$ \pr 2\\
  \textbf{11.}\resetpr \pr $\frac{1}{3\log 3}$ \pr $\sqrt[3]{6}$ \\
	\textbf{12.}\resetpr \pr $\frac{\log 2}{\log 3}$ \pr $\sqrt{e}$ \pr limita neexistuje \pr $\frac{1}{2}$ \pr $\frac{3}{2}$ \pr 2
\newpage
\centerline{\MakeUppercase{iv. derivace}}\ \\ 
\textbf{1. Spo\v{c}t\v{e}te derivaci n\'asleduj\'ic\'ich funkc\'i ve v\v{s}ech bodech, kde existuje (uk\'azkov\'e p\v{r}\'iklady k druh\'emu z\'apo\v{c}tov\'emu testu)}\\
\resetpr
\pr $(1 + \sin^2 x)^{\arctg x}$
\pr $(x^3 + 4)e^{\sin x}$\\[5pt]
\textbf{2. Najd\v{e}te $A\in\er$, aby pro ka\v{z}d\'e $x\in(0,\infty)$ platil vztah}
$$\left(\log(\cos^2x + \sqrt{1+\cos^4x}) + \arctg x + \arcsin\left(\frac{1}{\sqrt{1+x^2}}\right)\right)' = A\frac{\sin 2x}{\sqrt{1+\cos^4 x}}.$$
\dalsiCviko
\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 $(e + |x|)^x$\\[10pt]
\textbf{5. 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)
\dalsiCviko
\textbf{6. Nalezn\v{e}te glob\'aln\'i extr\'emy funkce }$f(x) = x^3 - 3x + 2$,\quad$x\in[-3,2]$\\
\textbf{7. 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{8. 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 $|(1-x^2)e^{-x}|$	
\newpage
\centerline{\MakeUppercase{iv. derivace - v\'ysledky}}\ \\ 
\textbf{1.}\resetpr \pr $(1 + \sin^2 x)^{\arctg x}\cdot \left(\frac{\log(1 + \sin^2 x)}{1+x^2} + \frac{2\sin x\cos x\arctg x}{1 + \sin^2 x}\right),\;x\in\er$
\pr $3x^2e^{\sin x} + (x^3 + 4)e^{\sin x}\cos x,\;x\in\er$\\
\textbf{2.} $A = - 1$\\
\textbf{3.}\resetpr \pr $f'(x) = \frac{2\sgn(x)}{1+x^2}$ pro $x\in\er\setminus\{0\}$, $f'_+(0) = 2$, $f'_-(0) = -2$.\\
\pr $f'(x) = 2x\exp(-|x-1|) - x^2\exp(-|x-1|)\sgn(x-1)$ pro $x\in\er\setminus\{1\}$, $f'_+(1) = 1$, $f'_-(1) = 3$.\\
\pr $f'(x) = \frac{xe^{-x^2}}{\sqrt{1-e^{-x^2}}}$ pro $x\in\er\setminus\{0\}$, $f'_+(0) = 1$, $f'_-(0) = -1$.\\
\pr $f'(x)=\arcsin\sqrt{x\over x+1}$ pro $x>0$, $f'_+(0)=0$.\\
\pr $f'(x)=-{1\over\arccos x\cdot\sqrt{1-x^2}}$ pro $x\in(-1,1)$, $f'_+(-1)=-\infty$.\\
\pr $f'(x)=(\sin x)^{\cos x}\cdot({\cos^2 x\over \sin x}-\sin x\log\sin x)$ pro $x\in (0,\pi)$, $f'_+(0)=1$.\\
\textbf{4.}\resetpr \pr $f$ je definov\'ana a spojit\'a na $\er$, $f'(x)=5$ pro $x\in (1,4)$; $f'(x)=2x$ pro $x\in (-\infty,1)\cup (4,\infty)$; $f'_+(1)=f'_-(4)=5$, $f'_-(1)=2$, $f'_+(4)=8$.\\
	\pr $D_f = \er$, $f$ je spojit\'a v ka\v{z}d\'em bod\v{e} $\er\setminus(\zet\setminus\{-3, 3\})$, v bodech $\zet\setminus\{-3, 3\}$ je spojit\'a zprava a nespojit\'a zleva.
	$f'(x) = \tfrac{2}{3}x[x](x^2-9)^{-2/3}$ pro $x\in \er\setminus\zet$; $f'(-3)=f'(3)=\infty$; pro $k\in\zet\setminus\{-3,3\}$ je $f'_+(k)=\tfrac{2}{3}k^2(k^2-9)^{-2/3}$ a $f'_-(k)=\begin{cases}\infty,\quad& |k|>3,\\ -\infty,& |k|<3.\end{cases}$\\
	\pr $D_f = \bigcup_{k\in\zet}(-\tfrac{\pi}{2} + k\pi,\tfrac{\pi}{2} + k\pi)$, $f$ je spojit\'a v ka\v{z}d\'em bod\v{e} $D_f$. $f'(x) = -2\sin 2x\cdot(\tg x - 1)\cdot\sgn(\cos 2x) + |\cos 2x|\tfrac{1}{\cos^2x}$ pro $x\in\bigcup_{k\in\zet}(-\tfrac{\pi}{2} + k\pi,-\tfrac{\pi}{4} + k\pi)\cup(-\tfrac{\pi}{4} + k\pi, \tfrac{\pi}{4} + k\pi)\cup (\tfrac{\pi}{4} + k\pi,\tfrac{\pi}{2} + k\pi)$; $f'(\tfrac{\pi}{4} + k\pi) = 0$, $f'_+(-\tfrac{\pi}{4} + k\pi)=-4$, $f'_-(-\tfrac{\pi}{4} + k\pi)=4$ pro $k\in\zet$.\\
	\pr $f'(x)=\begin{cases}2x(\sin\tfrac{1}{x} + \cos\tfrac{1}{x}) +  \sin\tfrac{1}{x} - \cos\tfrac{1}{x}& x\neq 0,\\
	0 & x = 0.\end{cases}$\\
	\pr $D_f = \er$, $f$ spojit\'a v ka\v{z}d\'em bod\v{e} $\er\setminus\{k\pi:\; k\in\zet\}$, v bodech $\{2k\pi:\; k\in\zet\}$ je spojit\'a zprava a nespojit\'a zleva, v bodech $\{(2k+1)\pi:\; k\in\zet\}$ je spojit\'a zleva a nespojit\'a zprava. Pro $k\in\zet$ m\'ame: $f'(x) = 0$ pro $x\in(2k\pi,(2k+1)\pi)$; $f'(x) = \sin x$ pro $x\in((2k-1)\pi,2k\pi)$; $f'_+(2k\pi) = f'_-((2k+1)\pi)=0$; $f'_-(2k\pi) = f'_+((2k+1)\pi)=+\infty$.\\
	\pr $f$ je definov\'ana a spojit\'a na $\er$, $f'(x) = (e + |x|)^x\cdot (\log(e+|x|) + \tfrac{|x|}{e+|x|})$ pro $x\in\er$\\(pozor, $f'(0) = 1$ je pot\v{r}eba spo\v{c}\'itat zvl\'a\v{s}\v{t})\\	
	\textbf{5. }\resetpr \pr 2 \pr -2 \pr $-\tfrac 13$ \pr $\tfrac 13$ \pr $-\tfrac{e}{2}$ \pr -2 \pr $\tfrac 12$ \pr $e^{1/6}$ \pr $a^a(\ln a - 1)$\\
	\textbf{6.} $\max f(x) = f(2) = 4$, $\min f(x) = f(-3) = -16$\\
	\textbf{7.} \resetpr \pr maximum neexistuje (ani lok\'aln\'i), glob\'aln\'i minimum v bod\v{e} $\tfrac{7}{5}$ \pr glob\'aln\'i maximum v bod\v{e} 1, glob\'aln\'i minimum v bod\v{e} 0, lok\'aln\'i maximum v bod\v{e} -1\\
	\textbf{8.} \resetpr \pr, \pr, \pr: viz. v\'ysledky zkou\v{s}kov\'e p\'isemky z roku 2008/2009, varianty E, D, B\\ \url{http://www.karlin.mff.cuni.cz/~kalenda/pis-fsv/0809/pis.htm}\\
	\pr, \pr: viz. sepsan\'e \'ulohy od Petra Po\v{s}ty (seps\'ano v roce 2010/11)\\ \url{http://www.karlin.mff.cuni.cz/~pposta/priklady/PrubFR.pdf}
\newpage
\dalsiCviko
\centerline{\MakeUppercase{v. \v{r}ady}}\ \\
\textbf{1. 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{2. 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{3. 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{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 $\suma (-1)^n \frac{n}{(n^2+1)\log n}$
\dalsiCviko
\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 \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 $\suma (-1)^n \frac{1}{\log n} \cos\left(\frac{n}{n^2+1}\right)$
\pr $\suma (-1)^n \frac{n\sin n}{n^2 + 1}$
\pr $\suma \frac{\sin(n)\cos(\tfrac{1}{n})}{n}$
\pr $\suma (-1)^n \frac{2n^2 + 3n + 4}{2n^4 + 3}$\\
\textbf{8. 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]
\newpage
\centerline{\MakeUppercase{v. \v{r}ady - v\'ysledky}}\ \\
\textbf{1.}\resetpr \pr $\frac{1}{n}$ \pr $\frac{n^2}{3^n}$ \pr $\frac{1}{n^{2/3}}$ \pr $\frac{1}{n^{35/2}}$ \pr $\frac{1}{n^{5/36}}$ \pr $n$ \pr $\frac{1}{n}$\\
\textbf{2.}\resetpr \pr Diverguje (D) \pr D \pr D \pr Konverguje (K) \pr K \pr K \pr K \pr K \pr K \pr D \pr K \pr K \pr K \pr K \pr K\\
\textbf{3.}\resetpr \pr K \pr K \pr K pro $x < 0$, D pro $x \geq 0$ \pr K pro $x \leq 0$, D pro $x > 0$ \pr K \pr D \pr K \pr K \pr D \pr K \pr K \pr D \pr D \pr K\\
\textbf{5.}\resetpr \pr ne AK \pr AK \pr ne AK\\
\textbf{6.}\resetpr \pr K, ale ne absolutn\v{e} \pr D \pr K, ale ne absolutn\v{e} \pr K, ale ne absolutn\v{e} \pr D \pr K, ale ne absolutn\v{e} \pr K, ale ne absolutn\v{e}\\
\textbf{7.}\resetpr \pr AK \pr D \pr K, ale ne absolutn\v{e} \pr K, ale ne absolutn\v{e}\\
%\textbf{5.}\resetpr 
 \pr K, ale ne absolutn\v{e} \pr K, ale ne absolutn\v{e} \pr AK\\
\textbf{8.}\resetpr \pr AK \pr D \pr K, ale ne absolutn\v{e} \pr AK \pr K\\
%\newpage
%\centerline{\MakeUppercase{vi. taylor\r{u}v polynom}}\ \\
%\textbf{1. Spo\v{c}t\v{e}te limity}\\[5pt]
%a) $\lx \frac{\cos x - e^{-\frac{x^2}{2}}}{x^4}$\quad
%b) $\lx \frac{e^x\sin x - x(1+x)}{x^3}$\quad
%c) $\lx[\infty] (\sqrt[6]{x^6+x^5} - \sqrt[6]{x^6-x^5})$\\
%d) $\lx[\infty] x^{3/2}(\sqrt{x+1} + \sqrt{x-1} - 2\sqrt{x})$\quad
%e) $\lx \frac{a^x+a^{-x}-2}{x^2}\;(a>0)$\quad
%f) $\lx\left(\tfrac{1}{x} - \tfrac{1}{\sin x}\right)$\\
%g) $\lx[\infty] (x-x^2\log(1+\tfrac{1}{x}))$\quad
%h) $\lx\tfrac{1}{x}\left(\tfrac{1}{x} - \tfrac{\cos x}{\sin x}\right)$\quad
%i) $\lx[\infty] ((x^3-x^2+\tfrac{x}{2})e^{\frac{1}{x}} - \sqrt{x^6+1})$\\
%j) $\lx \frac{1-(\cos x)^{\sin x}}{x^3}$
%k) $\lx \frac{\sqrt{1-2x+x^3} - \sqrt[3]{1-3x+x^2}-\tfrac{x^2}{6}}{\sin x -x}$\\[10pt]
%\textbf{2. Spo\v{c}t\v{e}te limity (p\v{r}\'iklady ze zkou\v{s}kov\'ych p\'isemek na FSV):}\\[5pt]\resetpr
%\pr $\lx \frac{\sin(e^{(x^2)} - 1) + 2\sin(\cos(x) - 1)}{\log(1 + x^4)}$
%\pr $\lx \frac{\log(1 + x\sin x) - \exp(x\sin x) + \cos(x\sin x)}{\sin^4 x}$\\[10pt]
%\newpage
%\centerline{\MakeUppercase{vi. taylor\r{u}v polynom - v\'ysledky}}\ \\
%\textbf{1.}\resetpr \pr $-\tfrac{1}{12}$ \pr $\tfrac{1}{3}$ \pr $\tfrac{1}{3}$ \pr -$\tfrac{1}{4}$ \pr $\log^2 a$ \pr 0 \pr $\tfrac{1}{2}$
	%\pr $\tfrac{1}{3}$ \pr $\tfrac{1}{6}$ \pr $\tfrac{1}{2}$ \pr -6\\[3pt]
%\textbf{2.}\resetpr \pr $\frac{7}{12}$ \pr $-\tfrac{3}{2}$\\[3pt]
 
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