%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Sunday, February 13, 2005 17:53:10} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - Limita funkcie - Cvi\U{10d}enia\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Limita funkcie} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \section{Cvi\v{c}enia} \textbf{Pr\'{\i}klad 1. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 2}% \frac{x^{2}+5}{x^{2}-3}.% \CustomNote{Answer}{$9.$}$ \textbf{Pr\'{\i}klad 2. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 1}% \frac{x^{2}-2x+1}{x^{3}-x}.% \CustomNote{Answer}{$0.$}$ \textbf{Pr\'{\i}klad 3. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \frac{1}{2}}\frac{8x^{3}-1}{6x^{2}-5x+1}.% \CustomNote{Answer}{$6.$}$ \textbf{Pr\'{\i}klad 4. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 1}% \frac{x^{m}-1}{x^{n}-1},\,m,\,n\in \mathbf{N}.% \CustomNote{Prob_Solv_Hint}{$\frac{m}{n}.$ N\'{a}vod: rozlo\v{z}te v\'{y}% razy $x^{k}-1=\left( x-1\right) \left( x^{k-1}+x^{k-2}+...+x+1\right) $ a funkciu zjednodu\v{s}te.}$ \textbf{Pr\'{\i}klad 5. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 1^{+}}\frac{x^{3}+x-2}{x^{3}-x^{2}-x+1}.% \CustomNote{Answer}{$+\infty $}$ \textbf{Pr\'{\i}klad 6.} Vypo\v{c}\'{\i}tajte limity v bodoch, v ktor\'{y}ch funkcia $f\left( x\right) =\frac{1}{\left| x^{2}-16\right| }$\ nie je definovan\'{a}. Zistite, \v{c}i je ohrani\v{c}en\'{a} a na\v{c}rtnite pribli% \v{z}ne jej graf$.% \CustomNote{Answer}{% Funkcia nie je definovan\'{a} v bodoch, kde je $\left| x^{2}-16\right| =0$, t.j. $x_{1,2}=\pm 4.$ Tak m\'{a}me $D\left( f\right) =\left( -\infty ,-4\right) \cup \left( -4,4\right) \cup \left( 4,\infty \right) =\mathbf{% R\setminus }\left\{ -4,4\right\} .$ Vypo\v{c}\'{\i}tame limity iba v bode $% a=-4.$ Limity v bode $a=4$ vypo\v{c}\'{\i}tajte sami. $\lim_{x% \longrightarrow -4^{-}}\frac{1}{\left| x^{2}-16\right| }=\infty ,\,\lim_{x\longrightarrow -4^{+}}\frac{1}{\left| x^{2}-16\right| }=\infty ,$ odkia\v{l} plynie, \v{z}e funkcia nie je zhora ohrani\v{c}en\'{a}. Preto\v{z}% e plat\'{\i} $\frac{1}{\left| x^{2}-16\right| }>0$, funkcia $f$ je zdola ohrani\v{c}en\'{a}. Na\v{c}rtnite graf:.}$ \textbf{Pr\'{\i}klad 7.} Vypo\v{c}\'{\i}tajte limity v bodoch, v ktor\'{y}ch funkcia $f\left( x\right) =\frac{x}{x^{2}-4}$ nie je definovan\'{a}. Zistite, \v{c}i je ohrani\v{c}en\'{a} a na\v{c}rtnite pribli\v{z}ne jej graf$% .% \CustomNote{Answer}{% Funkcia nie je definovan\'{a} v bodoch, kde je $x^{2}-4=0$, t.j. $% x_{1,2}=\pm 2.$ Tak m\'{a}me $D\left( f\right) =\left( -\infty ,-2\right) \cup \left( -2,2\right) \cup \left( 2,\infty \right) =\mathbf{R\setminus }% \left\{ -2,2\right\} .$ Vypo\v{c}\'{\i}tame limity iba v bode $a=-2.$ Limity v bode $a=2$ vypo\v{c}\'{\i}tajte sami. $\lim_{x\longrightarrow -2^{-}}\frac{% x}{x^{2}-4}=-\infty ,\,\lim_{x\longrightarrow -2^{+}}\frac{x}{x^{2}-4}% =\infty ,$ odkia\v{l} plynie, \v{z}e funkcia nie je zdola ani zhora ohrani% \v{c}en\'{a}. Na\v{c}rtnite graf:.}$ \textbf{Pr\'{\i}klad 8.} Vypo\v{c}\'{\i}tajte limity funkcie $f$\ v bodoch, v ktor\'{y}ch nie je definovan\'{a}. Zistite, \v{c}i je ohrani\v{c}en\'{a} a na\v{c}rtnite pribli\v{z}ne jej graf, ak $f\left( x\right) =\left\{ \begin{tabular}{cc} $\frac{x^{2}}{x^{2}-9},$ & $x\in \left( -3,3\right) \,$ \\ $\frac{x^{3}-16x}{x^{2}-9},$ & $x\notin \left\langle -3,3\right\rangle $% \end{tabular}% .\right. \CustomNote{Answer}{% Funkcia nie je definovan\'{a} v bodoch $x_{1,2}=\pm 3.$ Vypo\v{c}\'{\i}tame limity iba v bode $a=3.$ Limity v bode $a=-$ vypo\v{c}\'{\i}tajte sami. $% \lim_{x\longrightarrow 3^{-}}\frac{x^{2}}{x^{2}-9}=-\infty ,\,\lim_{x\longrightarrow 3^{+}}\frac{x^{3}-16x}{x^{2}-9}=-\infty ,$ odkia% \v{l} plynie, \v{z}e funkcia nie je zdola ohrani\v{c}en\'{a}. Preto\v{z}e plat\'{\i} $\lim_{x\longrightarrow \infty }\frac{x^{3}-16x}{x^{2}-9}=\infty , $ funkcia $f$ nie je zhora ohrani\v{c}en\'{a}. Na\v{c}rtnite graf:.}$ \textbf{Pr\'{\i}klad 9. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( \frac{x^{3}}{2x^{2}-1}-\frac{x^{2}}{2x+1}\right) .% \CustomNote{Answer}{$\frac{1}{4}$}$ \textbf{Pr\'{\i}klad 10. }Dan\'{a} je funkcia $f\left( x\right) =\frac{x}{% \left\vert \limfunc{tg}x\right\vert },\;x\in \left( -\frac{\pi }{2},0\right) \cup \left( 0,\frac{\pi }{2}\right) .% \CustomNote{Answer}{$\lim_{x\rightarrow -\frac{\pi }{2}^{+}}\frac{x}{% \left\vert \func{tg}x\right\vert }=0,\,\lim_{x\rightarrow \frac{\pi }{2}^{-}}% \frac{x}{\left\vert \limfunc{tg}x\right\vert }=0,\,$% \par $\lim_{x\rightarrow 0^{+}}\frac{x}{\left\vert \tan x\right\vert }% =\allowbreak 1,\,\lim_{x\rightarrow 0^{-}}\frac{x}{\left\vert \limfunc{tg}% x\right\vert }=-1.$}$ a) Vypo\v{c}\'{\i}tajte limity v krajn\'{y}ch bodoch defini\v{c}n\'{e}ho oboru a v bode $0$. b) Zistite, \v{c}i je dan\'{a} funkcia p\'{a}rna, alebo nep\'{a}rna.% \CustomNote{Answer}{% nep\'{a}rna} \textbf{Pr\'{\i}klad 11.} Vypo\v{c}\'{\i}tajte limity: $\lim_{h\rightarrow 0}% \frac{\sqrt{3+h}-\sqrt{3}}{h},\;\lim_{x\rightarrow 3}\frac{\sqrt{x}-\sqrt{3}% }{x^{2}-3x}.% \CustomNote{Answer}{$\frac{1}{6}\sqrt{3},\,\frac{1}{18}\sqrt{3}.$}$ \bigskip \textbf{Pr\'{\i}klad 12. }Vypo\v{c}\'{\i}tajte limitu $% \lim_{x\rightarrow \infty }\frac{\sqrt{x^{2}+1}+\sqrt{x}}{\sqrt[4]{x^{3}+x}-x% }.% \CustomNote{Answer}{$-1$}$ \textbf{Pr\'{\i}klad 13. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\frac{\sqrt[5]{x^{7}+3}+\sqrt[4]{2x^{3}-1}}{\sqrt[6]{x^{8}+x}-x}.% \CustomNote{Prob_Solv_Hint}{% Vyberte z \v{c}itate\v{l}a aj menovate\v{l}a v\'{y}razu maxim\'{a}lne mocniny $x$: \par {}% \[ \frac{\sqrt[5]{x^{7}+3}+\sqrt[4]{2x^{3}-1}}{\sqrt[6]{x^{8}+x}-x}= \]% \[ =\frac{x^{\frac{7}{5}}\left( \sqrt[5]{1+\frac{3}{x^{7}}}+\sqrt[4]{\frac{2}{% x^{\frac{13}{5}}}-\frac{1}{x^{\frac{28}{5}}}}\right) }{x^{\frac{4}{3}}\left( \sqrt[6]{1+\frac{1}{x^{7}}}-\frac{1}{x^{\frac{1}{3}}}\right) }= \]% \[ =\frac{\sqrt[15]{x}\left( \sqrt[5]{1+\frac{3}{x^{7}}}+\sqrt[4]{\frac{2}{x^{% \frac{13}{5}}}-\frac{1}{x^{\frac{28}{5}}}}\right) }{\left( \sqrt[6]{1+\frac{1% }{x^{7}}}-\frac{1}{x^{\frac{1}{3}}}\right) } \]% \par potom \[ \lim_{x\longrightarrow \infty }\frac{\sqrt[5]{x^{7}+3}+\sqrt[4]{2x^{3}-1}}{% \sqrt[6]{x^{8}+x}-x}= \]% \[ =\lim_{x\longrightarrow \infty }\frac{\sqrt[12]{x}\left( \sqrt[5]{1+\frac{3}{% x^{7}}}+\sqrt[4]{\frac{2}{x^{\frac{13}{5}}}-\frac{1}{x^{\frac{28}{5}}}}% \right) }{\left( \sqrt[6]{1+\frac{1}{x^{7}}}-\frac{1}{x^{\frac{1}{3}}}% \right) }=\infty . \]% }$ \textbf{Pr\'{\i}klad 14. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( \sqrt{x+3}-\sqrt{x}\right) .% \CustomNote{Answer}{$0$}$ \textbf{Pr\'{\i}klad 15. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }x\left( \sqrt{x^{2}+1}-x\right) .% \CustomNote{Answer}{$\lim_{x\rightarrow \infty }x\left( \sqrt{x^{2}+1}% -x\right) =\frac{1}{2}.$}$ \textbf{Pr\'{\i}klad 16. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow -\infty }x\left( \sqrt{x^{2}+1}-x\right) .% \CustomNote{Answer}{$\lim_{x\rightarrow -\infty }x\left( \sqrt{x^{2}+1}% -x\right) =\allowbreak -\infty .$}$ \textbf{Pr\'{\i}klad 17. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( \sqrt{x^{2}+1}-\sqrt{x}\right) .% \CustomNote{Answer}{$% \begin{tabular}{ccc} $\infty $ & pre & $x\longrightarrow \infty $% \end{tabular}% $}$ \textbf{Pr\'{\i}klad 18. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 0}% \frac{\limfunc{tg}5x}{\limfunc{tg}6x}.% \CustomNote{Answer}{$\frac{5}{6}$}$ \textbf{Pr\'{\i}klad 19. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 0}% \frac{1-\cos x}{x^{2}}.% \CustomNote{Prob_Solv_Hint}{% Pou\v{z}ite rovnos\v{t}: $\sin ^{2}\frac{x}{2}=\frac{1}{2}\left( 1-\cos x\right) ,$ v\'{y}sledok bude $\frac{1}{2}$}$ \textbf{Pr\'{\i}klad 20. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( 1-\frac{1}{x}\right) ^{x}.% \CustomNote{Answer}{$e^{-1}.$}$ \textbf{Pr\'{\i}klad 21. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( \frac{3x-4}{3x+2}\right) ^{\frac{x+1}{2}}.% \CustomNote{Prob_Solv_Hint}{% \[ \lim_{x\rightarrow \infty }\left( \frac{3x-4}{3x+2}\right) ^{\frac{x+1}{2}}= \]% \[ =\lim_{x\rightarrow \infty }\left( 1+\frac{1}{\frac{3x+2}{-6}}\right) ^{% \frac{3x+2}{-6}\frac{-6}{3x+2}\frac{x+1}{2}}= \]% \[ =\lim_{x\rightarrow \infty }\left[ \left( 1+\frac{1}{\frac{3x+2}{-6}}\right) ^{\frac{3x+2}{-6}}\right] ^{\frac{-3x-3}{3x+2}}=e^{-1}. \]% }$ \textbf{Pr\'{\i}klad 22. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 0}% \frac{\ln \left( a+x\right) -\ln a}{x}.% \CustomNote{Answer}{$\frac{1}{a}$}$ \textbf{Pr\'{\i}klad 23. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\frac{\sin x}{x}.% \CustomNote{Answer}{% Plat\'{\i} nerovnica: \par $-1\leq \sin x\leq 1\Longrightarrow -\frac{1}{x}\leq \frac{\sin x}{x}\leq \frac{1}{x},$ pre $x>0,$ potom z vety o nerovnostiach medzi limitami plat% \'{\i}: $\lim_{x\longrightarrow \infty }\frac{\sin x}{x}=0.$}$ \textbf{Pr\'{\i}klad 24. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\frac{\limfunc{arctg}x}{x}.% \CustomNote{Answer}{$0.$}$ \textbf{Pr\'{\i}klad 25. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\frac{x+\sin x}{x+\cos x}.% \CustomNote{Answer}{$1$}$ \textbf{Pr\'{\i}klad 26. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow \infty }\left( \cos \sqrt{x+1}-\cos \sqrt{x}\right) .% \CustomNote{Prob_Solv_Hint}{% Pou\v{z}ite s\'{u}\v{c}tov\'{y} vzorec: \par $\cos x-\cos y=-2\sin \frac{x-y}{2}\sin \frac{x+y}{2}.$ V\'{y}sledok je $0.$} $ \textbf{Pr\'{\i}klad 27. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x\rightarrow 0}% \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2}x}.% \CustomNote{Answer}{$\frac{\sqrt{2}}{8}$}$ \textbf{Pr\'{\i}klad 28. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x% \longrightarrow 0}x\func{cotg}x.% \CustomNote{Answer}{$1$}$ \textbf{Pr\'{\i}klad 29. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x% \longrightarrow -1^{-}}\limfunc{arctg}\frac{1}{1+x}.% \CustomNote{Answer}{$-\frac{\pi }{2}.$}$ \textbf{Pr\'{\i}klad 30. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x% \longrightarrow -2}\frac{\sqrt{6+x}-2}{x+2}.% \CustomNote{Prob_Solv_Hint}{$\lim_{x\longrightarrow -2}\frac{\sqrt{6+x}-2}{% x+2}=\lim_{x\longrightarrow -2}\frac{\sqrt{6+x}-2}{x+2}\frac{\sqrt{6+x}+2}{% \sqrt{6+x}+2}=$% \par $=\lim_{x\longrightarrow -2}\frac{6+x-4}{x+2}\frac{1}{\sqrt{6+x}+2}% =\lim_{x\longrightarrow -2}\frac{1}{\sqrt{6+x}+2}=\frac{1}{4}$}$ \textbf{Pr\'{\i}klad 31. }Vypo\v{c}\'{\i}tajte limitu $\lim_{x% \longrightarrow 0}\frac{\cos x-\cos ^{3}x}{x^{2}}.% \CustomNote{Answer}{$1$}$ \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za I} \section{Limita funkcie} \end{document}