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\Nadpis{\bf 3. cvi�en� z MA --- 20. a 21.10.2008}
\nadpis{Suprema a infima}

Nap�ed si zopakujte definici suprema a infima, rozd�l oproti maximu
a minimu, uve�te jednoduch� p��klad mno�iny, kter� m� supremum,
ale nem� maximum. Naleznete mno�inu, kter� nem� supremum?

Najd�te suprema a infima n�sleduj�c�ch mno�in (pokud existuj�).
Existuj� maxima a minima?
(Dohodn�me se, �e $\en$ zna�� mno�inu $\{1, 2, 3, \dots\}$, 
zat�mco $\en_0 = \en \cup \{0\}$.)

\pr \cast $A_1 =  \{(n-1)/n;\ n \in \Bbb N\}$, 
    \cast $A_2 = \{p/(p+q);\ p \in \Bbb N, \ q \in \Bbb N\}$,

\pr \cast $B_1 = \{\sin x;\ x \in \langle 0, 2\pi \rangle\}$,
    \cast $B_2 = \{\sin x;\ x \in ( 0, 2\pi )\}$,
    \cast $B_3 = \{\sin x;\ x \in ( 0, \pi )\}$,

\pr \cast $C_1 = \{n^2 - m^2;\ n \in \Bbb N, \ m \in \Bbb N\}$,
    \cast $C_2 = \{n^2 - m^2;\ n \in \Bbb N, \ m \in \Bbb N, n>m \}$,
    \cast $C_3 = \{n^2 - m^2;\ n \in \Bbb N, \ m \in \Bbb N, n\le m \}$,

\pr \cast $D_1 = \{2^{-n} + 3^{-n};\ n \in \Bbb N\}$,
    \cast $D_2 = \{2^{-n} + 3^{-n};\ n \in \Bbb Z\}$,

\pr $E = \{5^{(-1)^j3^k};\ j \in \Bbb Z, \ k \in \Bbb Z\}$.

\pr \cast $F_1=\{\cos(n+1/n)\pi;\ n\in \Bbb N\}$,
    \cast $F_2=\{\cos(n+1/n)\pi;\ n\in \Bbb N \text{ sud�}\}$,
    \cast $F_3=\{\cos(n+1/n)\pi;\ n\in \Bbb N \text{ lich�}\}$.


\nadpis{Limity posloupnost�}

Co ��k� definice limity? Spo�t�te p��mo podle definice limitu posloupnost� 
$(\frac 1{1+n^2})_{n=1}^\infty$ a $(\frac {n+1}{n+2})_{n=1}^\infty$.
Spo�t�te n�sleduj�c� limity (nebo doka�te, �e neexistuj�).
Budou se v�m k tomu hodit (mimo snad prvn� p��klad) v�ty 
o aritmetice limit.

\pr
\cast $\l (-1)^n$, 
\cast $\l \cos (-1)^n$, 
\cast $\l (-1)^{n!}$, 
\cast $\l p^n$.

\pr 
\cast $\l\frac {n+1}{n+2}$,
\cast $\l\frac{2n^2+n-3}{n^3-1}$,
\cast $\l\frac{2n^3+6n}{n^3-7n+7}$,
\cast $\l\frac{2n^5+3n-2}{n^5-3n^3+1}$.

\newpage
\pr
\cast $\l \left(\sqrt{n+1}-\sqrt n\right)$,
\cast $\l \left(\root 3\of{n+11}-\root 3\of n\right)$,
\cast $\l \sqrt{n}(\sqrt{n+1}-\sqrt{n})$.


\pr
\cast $\l{1+2+\dots+n\over n^2}$,
\cast $\l\left(\frac{1+2+\cdots +n}{n+2} - \frac n2\right)$,
\cast $\l\frac{1^2+2^2+\cdots +n^2}{n^3}$.

\oddel

Mali�ko p�itvrd�me \dots\ 
Budeme pou��vat t� v�tu o policajtech.

\pr
\cast $\l\root n\of a,\,(a\ge 0)$,
\cast $\l\root n\of n$.

\pr
\cast $\l\frac{n^a+n^{a-1}+\cdots + n + 1}{n^b+n^{b-1}+\cdots + n + 1}$ ($a, b \in \en$ parametry)
\cast $\l\frac{a^n+a^{n-1}+\cdots + a + 1}{b^n+b^{n-1}+\cdots + b + 1}$ ($a, b \in \er$ parametry, $|a|, |b| < 1$)


\pr $\l\frac {\lfloor{\sqrt n\rfloor}}{\sqrt n}$,

\pr 
\cast $\l n q^n$ ($q > 0$ je parametr),
\cast $\l\frac{q^n}{n!}$,
\cast $\l\frac {n!}{n^n}$,
\cast $\l\frac {n!}{n^3}$,
\cast $\l \root n\of {n!}$,
\cast $\l \root n\of{A^n+B^n+C^n}$ (kde $A,B,C>0$),
\cast $\l{3^n+n^5\over n^6+n!}$.

\pr 
\cast $\l\sum\limits_{k=1}^n{1\over k(k+1)}$,
\cast $\lp{\sum\limits_{k=1}^n \floor{kx}\over n^2}$ (parametr $x\in\Bbb R$),
\cast $\lp\sum\limits_{k=1}^n{1\over\sqrt{n^2+k}}$.

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