# 極限 (數列)

(已重新導向自 收敛)

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

## 定義

$\{ x_n \}, x_n \in \mathrm R,n=1,2,\ldots,x_0 \in \mathrm R$

## 數列極限的性質

   證：設數列$\left\{ {{x_n}} \right\}$有兩個不相符的極限值a、b，則對應於$d = \left| {a - b} \right| > 0$，可找到正數$N$，使$n > N$時，恆有
$\left| {{x_n} - a} \right| < \frac{d}{2}\begin{array}{*{20}{c}} ,&{\left| {{x_n} - b} \right|} \end{array} < \frac{d}{2}$，
從而$\left| {a - b} \right| = \left| {\left( {a - {x_n}} \right) - \left( {b - {x_n}} \right)} \right| \le \left| {a - {x_n}} \right| + \left| {b - {x_n}} \right| < d.$
這與假設$d = \left| {a - b} \right|$不符.
故$\left\{ {{x_n}} \right\}$不可能以兩個不相等的數為極限.


   證：應$\lim _{n \to \infty} x_n=x_0$，所以對$\varepsilon = 1$，$\exists N \in {\rm N}$，當$n > N$時有
$\left| {{x_n} - {x_0}} \right| < \varepsilon = 1$，
從而
$\left| {{x_n}} \right| = \left| {{x_n} - {x_0} + {x_0}} \right| \le \left| {{x_n} - {x_0}} \right| + \left| {{x_0}} \right| < 1 + \left| {{x_0}} \right|$.
令$M = \max \left( {\left| {{x_1}} \right|,\left| {{x_2}} \right|, \cdots ,\left| {{x_N}} \right|,1 + \left| {{x_0}} \right|} \right)$，於是，$\forall n \in {\rm N}$，有$\left| {{x_n}} \right| \le M$，即$\left\{ {{x_n}} \right\}$有界.


$1,0,1,0, \cdots \frac{{1 - {{\left( { - 1} \right)}^n}}}{2}, \cdots$

   證：已知$\lim _{n \to \infty} x_n=a$，$\lim _{n \to \infty} y_n=b$，且$a>b$.取$\varepsilon = \frac{{a - b}}{2} > 0$，由極限定義知：
$\exists {N_1} \in {\rm N},\forall n > {N_1}$，有
$\left| {{x_n} - a} \right| < \frac{{a - b}}{2}$，
從而
${x_n} > a - \frac{{a - b}}{2} = \frac{{a + b}}{2}$.
$\exists {N_2} \in {\rm N},\forall n > {N_2}$，有
$\left| {{y_n} - b} \right| < \frac{{a - b}}{2}$，
從而
${y_n} < b + \frac{{a - b}}{2} = \frac{{a + b}}{2}$.
所以當$n > N = \max \left( {{N_1},{N_2}} \right)$時，有
${y_n} < \frac{{a + b}}{2} < {x_n}$，
即                                            ${x_n} > {y_n}$.


## 數列的四則運算

$\lim _{n \to \infty} x_n=a$$\lim _{n \to \infty} y_n=b$，則

（1）$\lim _{n \to \infty} \left( {{x_n} \pm {y_n}} \right)= \lim _{n \to \infty } {x_n} \pm \lim _{n \to \infty } {y_n}$

（2）$\lim _{n \to \infty } {x_n} \cdot {y_n} = \lim _{n \to \infty } {x_n} \cdot \lim _{n \to \infty } {y_n}$

（3）若$b \ne 0,{y_n} \ne 0$,則$\lim _{n \to \infty } \frac{{{x_n}}}{{{y_n}}} = \frac{{\lim _{n \to \infty } {x_n}}}{{\lim _{n \to \infty } {y_n}}}$.