📝 题目
例 15 若 $q > 0$ ,证明:
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}q{x}_{n} = q\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n},\;\mathop{\lim }\limits_{{n \rightarrow \infty }}q{x}_{n} = q\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}. $$
💡 答案与解析
证 当 $k \geq n$ 时,由
$$ {x}_{k} \leq \mathop{\sup }\limits_{{k \geq n}}{x}_{k} \Rightarrow q{x}_{k} \leq q\mathop{\sup }\limits_{{k \geq n}}{x}_{k} \Rightarrow \mathop{\sup }\limits_{{k \geq n}}q{x}_{k} \leq q\mathop{\sup }\limits_{{k \geq n}}{x}_{k}; $$
又当 $k \geq n$ 时,由
$$ q{x}_{k} \leq \mathop{\sup }\limits_{{k \geq n}}q{x}_{k} \Rightarrow {x}_{k} \leq \frac{1}{q}\mathop{\sup }\limits_{{k \geq n}}q{x}_{k} $$
$$ \Rightarrow \mathop{\sup }\limits_{{k \geq n}}{x}_{k} \leq \frac{1}{q}\mathop{\sup }\limits_{{k \geq n}}q{x}_{k} $$
$$ \Rightarrow q\mathop{\sup }\limits_{{k \geq n}}q{x}_{k} \leq \mathop{\sup }\limits_{{k \geq n}}q{x}_{k}. $$
合起来即得出
$$ q\mathop{\sup }\limits_{{k \geq n}}{x}_{k} = \mathop{\sup }\limits_{{k \geq n}}q{x}_{k} $$
令 $\displaystyle{n \rightarrow \infty}$ ,即得
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}q{x}_{n} = q\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} $$
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}q{x}_{n} = - \left\lbrack {\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {-q{x}_{n}}\right) }\right\rbrack = - \left\lbrack {q\mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {-{x}_{n}}\right) }\right\rbrack $$
$$ = - \left\lbrack {-q\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}}\right\rbrack = q\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}. $$