方企勤 第二章 一元函数微分学 第5题
📝 题目
例 5 设 $f\left( x\right)$ 一阶可导,且 ${f}^{\prime \prime }\left( {x}_{0}\right)$ 存在. 求证:
$$ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {{x}_{0} + {2h}}\right) - {2f}\left( {{x}_{0} + h}\right) + f\left( {x}_{0}\right) }{{h}^{2}} = {f}^{\prime \prime }\left( {x}_{0}\right) . $$
💡 答案解析
证法 1 用洛必达法则及 ${f}^{\prime \prime }\left( {x}_{0}\right)$ 的定义,得
$$ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {{x}_{0} + {2h}}\right) - {2f}\left( {{x}_{0} + h}\right) + f\left( {x}_{0}\right) }{{h}^{2}} $$
$$ \underline{\text{ 洛必达法则 }}\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{2{f}^{\prime }\left( {{x}_{0} + {2h}}\right) - 2{f}^{\prime }\left( {{x}_{0} + h}\right) }{2h} $$
$$ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{2\left\lbrack {{f}^{\prime }\left( {{x}_{0} + {2h}}\right) - {f}^{\prime }\left( {x}_{0}\right) }\right\rbrack - 2\left\lbrack {{f}^{\prime }\left( {{x}_{0} + h}\right) - {f}^{\prime }\left( {x}_{0}\right) }\right\rbrack }{2h} $$
$$ = 2{f}^{\prime \prime }\left( {x}_{0}\right) - {f}^{\prime \prime }\left( {x}_{0}\right) = {f}^{\prime \prime }\left( {x}_{0}\right) . $$
证法 2 用带皮亚诺余项的泰勒公式, 得
$$ f\left( {{x}_{0} + {2h}}\right) - {2f}\left( {{x}_{0} + h}\right) + f\left( {x}_{0}\right) $$
$$ = \left\lbrack {f\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \cdot {2h} + \frac{{f}^{\prime \prime }\left( {x}_{0}\right) }{2!}{\left( 2h\right) }^{2} + o\left( {h}^{2}\right) }\right\rbrack $$
$$ - 2\left\lbrack {f\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \cdot h + \frac{{f}^{\prime \prime }\left( {x}_{0}\right) }{2!}{h}^{2} + o\left( {h}^{2}\right) }\right\rbrack + f\left( {x}_{0}\right) $$
$$ = {f}^{\prime \prime }\left( {x}_{0}\right) {h}^{2} + o\left( {h}^{2}\right) . $$
由此得 $\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {{x}_{0} + {2h}}\right) - {2f}\left( {{x}_{0} + h}\right) + f\left( {x}_{0}\right) }{{h}^{2}} = {f}^{\prime \prime }\left( {x}_{0}\right)$ .