📝 题目
例 4 设 $f\left( x\right)$ 在 $a$ 点有二阶导数 ${f}^{\prime \prime }\left( a\right)$ ,试证
$$ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) + f\left( {a - h}\right) - {2f}\left( a\right) }{{h}^{2}} = {f}^{\prime \prime }\left( a\right) . $$
💡 答案与解析
证明 根据假定, $f$ 在 $a$ 点有二阶导数,因而在 $a$ 点邻近应该具有一阶导数. 按照洛必达法则有
$$ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) + f\left( {a - h}\right) - {2f}\left( a\right) }{{h}^{2}} $$
$$ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{f}^{\prime }\left( {a + h}\right) - {f}^{\prime }\left( {a - h}\right) }{2h} $$
$$ = \frac{1}{2}\mathop{\lim }\limits_{{h \rightarrow 0}}\left\lbrack {\frac{{f}^{\prime }\left( {a + h}\right) - {f}^{\prime }\left( a\right) }{h} + \frac{{f}^{\prime }\left( {a - h}\right) - {f}^{\prime }\left( a\right) }{-h}}\right\rbrack $$
$$ = \frac{1}{2}\left\lbrack {{f}^{\prime \prime }\left( a\right) + {f}^{\prime \prime }\left( a\right) }\right\rbrack = {f}^{\prime \prime }\left( a\right) . $$
\customfootnote{
① 在本书中,全体自然数的集合 $\mathbb{N}$ 不包含 0 .
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