\input /home/pasky/school/fastex/lib.tex

\subchapter{Vlastn� limita posloupnosti}{


\definition{
	Posloupnost $\{a_n\}_{n=1}^\infty$ m� vlastn� limitu (je {\bf konvergentn�}) $A \in \real$, jestli�e:
	$$ \forall \eps > 0\ \existss n_0 \in \nat,\ \forall n \in \nat, n \ge n_0: |a_n - A| < \eps $$

	\shortdef{Zna��me}{\M{\lim_{n\to\infty} a_n = A}}

	\smallskip

	Posloupnost $\{a_n\}_{n=1}^\infty$ nem� vlastn� limitu (je {\bf divergentn�}), jestli�e:
	$$ \forall A \in \real\ \existss \eps > 0,\ \forall n_0 \in \nat\ \existss n \in \nat, n \ge n_0: |a_n - A| > \eps $$

	\shortnote{Divergentn� posloupnost� mysl�me i posloupnost osciluj�c�.}
}


\examples{
\list{
	\listItem{\M{\lim_{n\to\infty} 1/n = 0}}
	\listItem{\M{\lim_{n\to\infty} \broot{n}\of{n} = 1}}
	\listItem{\M{\lim_{n\to\infty} (-1)^n} nen� definovan�.}
	\listItem{\M{\lim_{n\to\infty} (1+1/n)^n = e}}
	\listItem{\M{\lim_{n\to\infty} a = a}}
}
}

\proof{
\list{
	\listItem{$A = 0$, d�no $\eps > 0$. Zvolme $n_0 > 1/\eps$:
		$$ \forall n \in \nat, n \ge n_0: 1/n \le 1/n_0 < \eps $$
		$$ \Longrightarrow \lim_{n\to\infty} 1/n = 0$$}

	\listItem{$A = 1$, m�jme $a_n = \broot{n}\of{n} = 1 + \delta_n$, kde $\delta_n > 0$ pro $\forall n \ge 2$.

		Sporem: nech� $\lim \ne 1$:
		$$ \existss \eps > 0,\ \forall n_0 \in \nat\ \existss n \in \nat, n \ge n_0: \broot{n}\of{n} \ge 1 + \eps $$

		Pak mus� existovat rostouc� posloupnost $\{n_k\}_{k=1}^\infty$, kde $n_k \in \nat$ a $a_{n_k} > 1 + \eps$.
		Tedy:
		$$ \broot{n_k}\of{n_k} > 1 + \eps $$
		$$ 1 + \delta_{n_k} > 1 + \eps $$
		$$ n_k = (1 + \delta_{n_k})^{n_k} > (1 + \eps)^{n_k} = 1 + n_k\eps + \cn{n_k}{2}\eps^2 + \cdots $$
		$$ > n_k\eps + {n_k(n_k-1)\over2}\eps^2 $$
		$$ \Longrightarrow 1 > \eps + {n_k-1\over2}\eps^2\qquad \forall n_k $$
		$$ \Longrightarrow {1-\eps\over\eps^2}\cdot2 + 1 > n_k\qquad \forall n_k $$
		\XXX
	}

	\listItem{$\existss \eps$ takov�, �e pro $\forall n_0\ \existss n: |a_n - A| \ge \eps$
		pro ka�d� p�edem dan� $A \in \real$.

		Sporem: p�edpokl�dejme $\lim_{n\to\infty} (-1)^n = A$, pak k $\eps = 1/3$ existuje
		$n_0$ takov�, �e $|(-1)^n - A| < 1/3$ pro $\forall n > n_0$.

		$$ 2 = |(-1)^n - (-1)^{n+1}| = |(-1)^n - A + A - (-1)^{n+1}| \le $$
		$$ \le |(-1)^n - A| + |A - (-1)^{n+1}| < 1/3 + 1/3 = 2/3 $$
		\XXX
	}

	\listItem{$e$ si pozd�ji nadefinujeme pr�v� jako tuto limitu.}

	\listItem{Trivi�ln�.}
}
}


\theorem{1}{o jednozna�nosti limity posloupnosti}{
	Ka�d� posloupnost m� nejv��e jednou limitu.

	\proof{
		Nech� \M{\lim_{n\to\infty} a_n = A_1 \et \lim_{n\to\infty} a_n = A_2}, bez �jmy na obecnosti (dichotomie)
		nech� $A_1 > A_2$.

		Zvolme $\eps < (A_1 - A_2) / 2$. Pak $\existss n_1, n_2$ takov�, �e:
		$$ \sesac{\eqalign{\forall n \ge n_1: |a_n - A_1| < \eps\cr
		                   \forall n \ge n_2: |a_n - A_2| < \eps}}
			{\hbox{pro $n \ge \max(n_1,n_2)$ plat� z�rove� oboj�.}} $$

		\XXX
	}
}

\theorem{2}{o omezenosti konvergentn� posloupnosti}{
	Ka�d� konvergentn� posloupnost je omezen�.

	\proof{
		Zvol $\eps = 1$. Nech� \M{\lim_{n\to\infty} a_n = A \in \real}.
		Pak dle definice $\existss n_0 \in \nat$ takov�, �e
		$$\forall n \ge n_0, n \in \nat: |a_n - A| < \eps \Rightarrow |a_n| < |A| + 1$$

		Zvol $K := \max \{ |A|+1; |a_1|, |a_2|, \ldots, |a_{n_0}| \}$.
		Pak pro $\forall n \in \nat: |a_n| \le K$.
	}
}

\definition{
	�ekneme, �e posloupnost $\{b_k\}_{k=1}^\infty$ je {\bf vybran�} z $\{a_n\}$,
	existuje-li {\it rostouc�} posloupnost $$\{n_k\}_{k=1}^\infty \subseteq \nat$$
	takov�, �e $b_k = a_{n_k}$ pro $\forall k \in \nat$.
}

\theorem{3}{o limit� vybran� posloupnosti}{
	Nech� \M{\lim_{n\to\infty} a_n = A} a $\{b_k\}$ je vybran� posloupnost z $\{a_n\}$.
	Pak \M{\lim_{k\to\infty} b_k = A}.

	\proof{
		$$\forall \eps > 0\ \existss n_0: |a_n - A| < \eps\quad \forall n > n_0
			\Longrightarrow \existss k_0 \ge n_0: a_{n_{k_0}} = b_{k_0}$$
		$$\forall l \ge l_0: |b_l - A| < \eps \textnote{nebo� $b_l$ je tak� n�jak� $a_n$}$$

		\qed
	}
}

\theorem{4}{o aritmetice limit}{
	Nech� $a_n \opupon{$\scriptstyle{n\to\infty}$}{\longrightarrow} A \in \real$,
	$b_n \opupon{$\scriptstyle{n\to\infty}$}{\longrightarrow} B \in \real$,
	pak plat�:
	\list{
	\listItem{\M{\lim_{n\to\infty} (-a_n) = -A}}
	\listItem{\M{\lim_{n\to\infty} (a_n+b_n) = A+B}}
	\listItem{\M{\lim_{n\to\infty} (1/a_n) = 1/A} (pokud $a_n \ne 0$ pro $\forall n$ a $A \ne 0$)}
	\listItem{\M{\lim_{n\to\infty} (a_nb_n) = AB}

		Tedy tak� \M{\lim_{n\to\infty} (\alpha a_n) = \alpha A}
		a za dobr�ho po�as� i \M{\lim_{n\to\infty} (a_n/b_n) = A/B}.
	}
	}

	Pozor! Obecn� neplat� \M{\lim_{n\to\infty} (c_n+d_n) = \lim_{n\to\infty} c_n + \lim_{n\to\infty} d_n},
	nemus� jedna z nich existovat.

	\proof{
	\list{
	\listItem{Trivi�ln�, sta�� volit $n_0' = n_0$ (jako pro p�vodn� posloupnost).}
	\listItem{Pro dan� $\eps > 0$ plat� limita $\{a_n\}$ od $n_1$, $\{b_n\}$ od $n_2$, tedy ob�
		plat� od $n_3 = \max\{n_1,n_2\}$ d�le. Tedy:
		$$ |(a_n + b_n) - (A+B)| \le |a_n - A| + |b_n - B| < 2\eps\qquad \forall n \ge n_3 $$
	}
	\listItem{Nejd��ve dok�eme lemmu.
		\lemma{}{
			$$ \existss K > 0,\ \forall n: |a_n| \ge K $$

			\proof{
				M�jme $\eps < |A|/2$:
				$$\existss n_0, \forall n \ge n_0: |a_n| \ge 2\eps \Rightarrow K \ge 2\eps$$
				Pro $n \in \{1,\ldots,n_0-1\}: K < \min\{|a_n|\} \Rightarrow \existss K$.
				\qed
			}
		}

		\medskip
		Zvol�me $\eps$-p�s kolem nuly o velikosti $K = \eps_0$ takov�, �e $|a_n \ge \eps_0|$,
		tedy $|1/a_n| \le 1/\eps_0$.

		$$ \left|{1 \over a_n} - {1 \over A}\right| = \left| A - a_n \over a_n A \right|
		\le {\eps \over |a_n A|} \le \eps \cdot {1\over KA} $$
	}
	\listItem{M�jme $n \ge n_3 = \max\{n_1,n_2\}$:
		$$ |(a_nb_n) - (AB)| = |a_nb_n - a_nB + a_nB - AB| \le |a_nb_n - a_nB| + |a_nB - AB| = $$
		$$ = |a_n|\undernote{|b_n - B|}{<\eps} + |B|\undernote{|a_n - A|}{<\eps} < |a_n|\eps + |B|\eps $$
		P�itom $|a_n|$ je omezeno n�jakou konstantou $K$ a $|B|$ je konstantn� rovnou, tedy:
		$$ |a_n|\eps + |B|\eps \le \undernote{(K+|B|)}{\hbox{konst.}}\eps $$
	}
	}

	\qed
	}
}

\penalty-100

\theorem{5}{o uspo��d�n� limit}{
	M�jme \M{\lim_{n\to\infty} a_n = A}, \M{\lim_{n\to\infty} b_n = B}. Pak:
	\list{
	\listItem{Pokud $A>B$, pak $\existss n_0$, od kter�ho $a_n>b_n$ pro $\forall \ge n_0$.}
	\listItem{Pokud $\existss n_0$ takov�, �e $a_n \ge b_n$ pro $\forall \ge n_0$, pak $A \ge B$. (Pozor, neplat�
		$a_n > b_n \Rightarrow A > B$! Nap�. $a_n = 1/n$, $b_n = -1/n$.)}
	}

	\proof{
		Zvol $\eps > 0$, $\eps < {1\over4}(A-B)$, pak pro $n_1$: $|a_n - A| < \eps$, pro $n_2$: $|b_n - B| < \eps$,
		pro $n_3 = \max \{n_1,n_2\}$ jsou tyto p�sy ji� disjunktn�:

		$$a_n > A - \eps > B + \eps > b_n\qquad\forall n \ge n_3$$

		\qed
	}
}


\theorem{6}{o dvou policajtech}{
	Nech�:
	$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = A \in \real $$
	$$ \existss n_0 \in \nat,\ \forall n \ge n_0: a_n \le c_n \le b_n $$

	Pak \M{\lim_{n\to\infty} c_n = A}.

	\proof{
		Zvolme $\eps > 0$, pak:
		$$ \eqalign{\existss n_1 \in \nat,\ \forall n \ge n_1: |a_n - A| < \eps\cr
		            \existss n_2 \in \nat,\ \forall n \ge n_2: |b_n - A| < \eps} $$
		Definujme $n_3 := \max \{n_0, n_1, n_2\}$. Pak jist�
		$$ A - \eps \le a_n\qquad \forall n > n_3 $$
		$$ A - \eps < a_n \le c_n \le b_n < A + \eps $$
		$$ \lim_{n\to\infty} c_n = A $$
		\qed
	}

	\example{
		Nech� $a > 0$. Pak \M{\lim_{n\to\infty} \broot{n}\of{a} = 1}.

		\proof{
		\list{
			\namedListItem{$a > 1$}{$\existss n_0 \in \nat, n_0 \ge a,\ \forall n \ge n_0: a \le n$.
				\smallskip
			
				Tedy \M{1 \le \broot{n}\of{a} \le \broot{n}\of{n}} pro $\forall n \ge n_0$.

				$$ \lim_{n\to\infty} 1 = 1 \et \lim_{n\to\infty} \broot{n}\of{n} = 1
					\Longrightarrow \lim_{n\to\infty} \broot{n}\of{a} = 1 $$
			}
			\namedListItem{$a = 1$}{Trivi�ln�.}
			\namedListItem{$a < 1$}{\M{\lim_{n\to\infty} \broot{n}\of{a}
					= \lim_{n\to\infty} {1\over\broot{n}\of{1/a}}}

				$$ 1/a > 1 \Longrightarrow \lim_{n\to\infty} \broot{n}\of{1/a} = 1 $$
				(Jak bylo dok�z�no.)

				$$ 1 \ne 0 \et \broot{n}\of{1/n} \ne 0\quad \forall n
					\becauseof{\ias{}{4}{}}{\Longrightarrow}
					\lim_{n\to\infty} \broot{n}\of{a} - 1/n = 1 $$
			}
		}

			\qed
		}
	}
}


\theorem{7}{o sou�inu omezen� a ``nulov�'' posloupnosti}{
	Nech� \M{\lim_{n\to\infty} a_n = 0} a $\{b_n\}$ je omezen�.
	Pak \M{\lim_{n\to\infty} a_nb_n = 0}.

	\shortnote{Plat�, �e \M{\lim_{n\to\infty} a_n = 0 \Longleftrightarrow \lim_{n\to\infty} |a_n| = 0}.
		(D�kaz: \DCV.)}

	\proof{
		V�me, �e pro $\forall n \in \nat$ plat� $|b_n| \le K$. Pak dle p�edpokladu a dvou policajt�:
		$$ \undernote{0}{\to 0} \le
			\undernote{|a_nb_n| = |a_n||b_n|}{\to 0} \le
			\undernote{K|a_n|}{\to 0} $$
		$$ \Longrightarrow \lim_{n\to\infty} |a_nb_n| = 0 \Longrightarrow \lim_{n\to\infty} a_nb_n = 0 $$
		\qed
	}

	\shortdef{P��klad}{\M{\lim_{n\to\infty} {1\over n}\sin n = 0}}
}


}

\bend
