方企勤 第二章 一元函数微分学 第5题
📝 题目
例 5 设 $f\left( x\right)$ 在(a, b)内可导,对 $\forall {x}_{0} \in \left( {a,b}\right)$ ,求证:
$$ \exists {x}_{n} \in \left( {a,b}\right) \;\left( {n = 1,2,\cdots }\right) , $$
使得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = {x}_{0}}$ ,且 $\mathop{\lim }\limits_{{n \rightarrow \infty }}{f}^{\prime }\left( {x}_{n}\right) = {f}^{\prime }\left( {x}_{0}\right)$ .
💡 答案解析
证 取 $y > 0$ 足够大,使得
$$ {y}_{n}\overset{\text{ 定义 }}{ = }{x}_{0} + \frac{1}{n + y} \in \left( {a,b}\right) \;\left( {n = 1,2,\cdots }\right) . $$
则有
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{y}_{n} = {x}_{0} \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{f\left( {y}_{n}\right) - f\left( {x}_{0}\right) }{{y}_{n} - {x}_{0}} = {f}^{\prime }\left( {x}_{0}\right) . \tag{2.4} $$
再由拉格朗日定理,对 $\forall n \in N,\exists {x}_{n} \in \left( {{x}_{0},{y}_{n}}\right)$ ,使得
$$ \frac{f\left( {y}_{n}\right) - f\left( {x}_{0}\right) }{{y}_{n} - {x}_{0}} = {f}^{\prime }\left( {x}_{n}\right) \;\left( {n = 1,2,\cdots }\right) . \tag{2.5} $$
联合 (2.4) 与 (2.5) 式,即得 $\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}\overset{\text{ 由夹挤准则 }}{ = }{x}_{0}}$ ,且
$$ \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}^{\prime }\left( {x}_{n}\right) = {f}^{\prime }\left( {x}_{0}\right) . $$