Součásti dokumentu Matematika2Priklady
Zdrojový kód
%\wikiskriptum{Matematika2Priklady}
\section{Posloupnosti}
\subsection{Omezenost a monotonost}
\begin{enumerate}
\begin{priklad}
a_n = \frac{n+(-1)^n}{n}
\end{priklad}
;~[omezená zdola 0, shora 3/2, není monotonní]
\begin{priklad}
a_n = \frac{n^2}{n+1}
\end{priklad}
;~[omezená zdola 1/2, neomezená shora, rostoucí]
\begin{priklad}
a_n = \frac{4n}{\sqrt{4n^2 + 1}}
\end{priklad}
;~[shora 2, zdola 4/5$\sqrt5$, rostoucí]
\begin{priklad}
a_n = \frac{4^n}{2^n + 100}
\end{priklad}
;~[zdola 2/51, shora neomezená, rostoucí]
\begin{priklad}
a_n = \frac{10^{10} \sqrt{n}}{n+1}
\end{priklad}
;~[zdola 0, shora 1/2 $10^{10}$, klesající]
\begin{priklad}
a_n = \ln \frac{2n}{n+1}
\end{priklad}
;~[zdola 0, shora $\ln 2$, rostoucí]
\begin{priklad}
a_n = \frac{(n+1)^2}{n^2}
\end{priklad}
;~[shora 4, zdola 1, klesající]
\begin{priklad}
a_n = \sqrt{4 - \frac{1}{n}}
\end{priklad}
;~[zdola $\sqrt 3$, shora 2, rostoucí]
\begin{priklad}
a_n = (-1)^{2n+1}\sqrt n
\end{priklad}
;~[shora -1, není zdola, klesající]
\begin{priklad}
a_n = \frac{2^n - 1}{2^n}
\end{priklad}
;~[zdola 1/2, shora 1, rostoucí]
\begin{priklad}
a_n = \sin \left ( \frac{\pi}{n+1} \right )
\end{priklad}
;~[zdola 0, shora 1, klesající]
\begin{priklad}
a_n = \frac{1}{n} - \frac{1}{n+1}
\end{priklad}
;~[zdola 0, shora 1/2, klesající]
\begin{priklad}
a_n = \frac{\ln (n+2)}{n+2}
\end{priklad}
;~[ zdola 0, shora $1/3 \ln 3$, klesající]
\begin{priklad}
a_n = \frac{3^n}{(n+1)^2}
\end{priklad}
;~[zdola 3/4, shora není, rostoucí]
\end{enumerate}
%\newpage
%\twocolumn
\subsection{Limity posloupností}
\begin{multicols}{2}
\begin{enumerate}
\begin{priklad}
\lim_{n \to +\infty} \frac{2^n}{4^n + 1} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} (-1)^n \sqrt{n} ~ neex.
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( -\frac{1}{2}\right )^n = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \tg \frac{n\pi}{4n+1} = 1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{(2n+1)^2}{(3n-1)^2} = \frac{4}{9}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{n^2}{\sqrt{2n^4+1}} =
\frac{1}{2}\sqrt2
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \cos \pi n ~ neex.
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} e ^ {1/\sqrt n} = 1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \ln(n) - \ln(n+1) = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\sqrt{n+1}}{2 \sqrt{n}} = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( 1 + \frac{1}{n} \right )^{2n} =
e^2
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{2^n}{n^2} = + \infty
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{(n+1) \cos \sqrt n}{n(1+\sqrt n)} =
0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\sqrt n \sin e^n \pi}{n+1} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} 2 \ln 3n - \ln (n^2+1) = \ln 9
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \frac{2}{n} \right )^n = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\ln (n+1)}{n+1} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{x^{100n}}{n!} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} n ^ {\alpha / n} = 1, \alpha > 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{3^{n+1}}{4^{n-1}}
= 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} (n+2)^{1/(n+2)} = 1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} (n+2) ^ {1/n} = 1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_0 ^ n e^{-x} \udx = 1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_{-n}^{n} \frac{\udx}{1+x^2} = \pi
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\ln n^2}{n} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_{-1 + 1/n}^{1-1/n} \frac{\udx}{\sqrt{1-x^2}}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{5^{n+1}}{4^{2n} - 1} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \frac{n+1}{n+2} \right ) ^n =
\frac{1}{e}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_n^{n+1} e ^ {-x^2} \udx = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{n^n}{2^{n^2}} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( 1 + \frac{x}{2n}\right) ^ {2n} =
e^x
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_{-1/n}^{1/n} \sin x^2 \udx = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( t + \frac{x}{n} \right ) ^ n, x >
0, t > 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{n+1} - \sqrt{n} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{n^2+n} - n = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt[3]{n^3+n} - n
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1 + 2 + \cdots + n}{n^2} = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1^3 + 2^3 + \cdots + n^3}{2n^4 + n
-1}= \frac{1}{8}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1^2 + 2 ^2 + \cdots + n^2}{(1+n)(2+n)}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \cos (n\pi) \sin (n\pi) = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \frac{n}{1+n} \right ) ^ {1/n} =
1
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \cos \frac{\pi}{n} \sin \frac{\pi}{n} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( 2 + \frac{1}{n} \right ) ^ n = +
\infty
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\ln (n(n+1))}{n} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \ln \left ( 1 + \frac{1}{n}\right )\right )^n
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\pi}{n} \ln \frac{n}{\pi} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \int_1^n \frac{\udx}{\sqrt x}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1}{n^2} + \frac{2}{n^2} + \cdots +
\frac{n-1}{n^2} = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{2n+1} - \sqrt{n} = + \infty
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{n+1} - \sqrt{n} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} 3 \sqrt{n^2+1} - 2n
= +\infty
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} n(\sqrt{n^2+1} - n) = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{n^2+1} - \sqrt{n^2-1}=0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{3n^2+n+1} - \sqrt{n^2-n+1} =
+\infty
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \sqrt{3n^2+n+1} - \sqrt{3n^2-n+1} =
\frac{1}{\sqrt 3}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\sqrt 1 + \sqrt 2 + \cdots + \sqrt n}{(\sqrt
n)^3}= \frac{2}{3}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1}{n} + \frac{1}{2n} + \cdots +
\frac{1}{n^2} = 0
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1^2+ 2^2 + \cdots + n^2}{n^3} =
\frac{1}{3}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{1^2+2^2 + \cdots + n^2}{n^2} -
\frac{n}{3} = \frac{1}{2}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( 1 + \frac{1}{n+1}\right ) ^ n =
e
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \frac{3n+4}{3n+5}\right ) ^n =
\frac{1}{\sqrt[3] e }
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \left ( \frac{2n+5}{2n+3}\right )^{n+1}
= e
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} (1+\sqrt{n+1} - \sqrt{n}) ^ {\sqrt{n}} =
\sqrt{e}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\ln(n^2 + 3n -2)}{\ln(n^5+n+1)}=
\frac{2}{5}
\end{priklad}
\begin{priklad}
\lim_{n \to +\infty} \frac{\ln(2+e^{3n})}{\ln(3+e^{2n})} =
\frac{3}{2}
\end{priklad}
\end{enumerate}
\end{multicols}
\separator
