📝 题目
例 24 设 $y = f\left( x\right)$ 在(-1,1)内具有二阶连续导数,且 ${f}^{\prime \prime }\left( x\right) \neq$ 0. 试证:
(1)对于 $\forall x \in \left( {-1,1}\right) ,x \neq 0$ ,存在惟一的 $\theta \left( x\right) \in \left( {0,1}\right)$ ,使 $f\left( x\right) = f\left( 0\right) + x{f}^{\prime }\left( {\theta \left( x\right) x}\right)$ 成立;
(2) $\mathop{\lim }\limits_{{x \rightarrow 0}}\theta \left( x\right) = \frac{1}{2}$ .
💡 答案与解析
证 (1) $\forall x \in \left( {-1,1}\right) ,x \neq 0$ ,由拉格朗日中值定理, $\exists \theta \left( x\right) \in$ (0,1),使得
$$ f\left( x\right) = f\left( 0\right) + x{f}^{\prime }\left( {\theta \left( x\right) x}\right) . \tag{6.30} $$
又因为 ${f}^{\prime \prime }\left( x\right)$ 连续,且在(-1,1)内 ${f}^{\prime \prime }\left( x\right) \neq 0$ ,所以 ${f}^{\prime \prime }\left( x\right)$ 在(-1,1) 内不变号,从而 ${f}^{\prime }\left( x\right)$ 在(-1,1)内严格单调增加或严格单调下降, 故 $\theta \left( x\right)$ 惟一.
(2)证法 1 为了使方程 (6.30) 中的 $\theta \left( x\right)$ 从 ${f}^{\prime }\left( {\theta \left( x\right) x}\right)$ 的括号内“解放”出来,对 $\forall x \neq 0$ ,我们首先将 (6.30) 式改写为
$$ \frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( 0\right) = {f}^{\prime }\left( {\theta \left( x\right) x}\right) - {f}^{\prime }\left( 0\right) . \tag{6.31} $$
接着在 (6.31) 式的右端对 ${f}^{\prime }\left( x\right)$ 用拉格朗日中值定理,存在 $\eta$ 于 $\theta \left( x\right) x$ 与 0 之间,使得
$$ {f}^{\prime }\left( {\theta \left( x\right) x}\right) - {f}^{\prime }\left( 0\right) = {f}^{\prime \prime }\left( \eta \right) {x\theta }\left( x\right) . \tag{6.32} $$
注意到 ${f}^{\prime \prime }\left( \eta \right) \neq 0$ ,联立 (6.31) 和 (6.32) 式我们得到 $\theta \left( x\right)$ 的表达式:
$$ \theta \left( x\right) = \frac{\frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( 0\right) }{x{f}^{\prime \prime }\left( \eta \right) }. \tag{6.33} $$
到此,为了证明结论只需证明 (6.33) 式的右端当 $x \rightarrow 0$ 时极限是 $\frac{1}{2}$ 即可. 因为 $\eta$ 在 $\theta \left( x\right) x$ 与 0 之间,所以
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\eta = 0 \Rightarrow \mathop{\lim }\limits_{{x \rightarrow 0}}{f}^{\prime \prime }\left( \eta \right) = \mathop{\lim }\limits_{{\eta \rightarrow 0}}{f}^{\prime \prime }\left( \eta \right) = {f}^{\prime \prime }\left( 0\right) \neq 0. \tag{6.34} $$
又用洛必达法则及导数定义, 我们有
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( 0\right) }{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{f\left( x\right) - f\left( 0\right) - x{f}^{\prime }\left( 0\right) }{{x}^{2}} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{f}^{\prime }\left( x\right) - {f}^{\prime }\left( 0\right) }{2x} = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{f}^{\prime }\left( x\right) - {f}^{\prime }\left( 0\right) }{x - 0} $$
$$ = \frac{1}{2}{f}^{\prime \prime }\left( 0\right) , \tag{6.35} $$
联立 (6.33),(6.34) 和 (6.35) 式,即得 $\mathop{\lim }\limits_{{x \rightarrow 0}}\theta \left( x\right) = \frac{1}{2}$ .
证法 2 为了从方程 (6.30) 中分离出 $\left( {\theta \left( x\right) - \frac{1}{2}}\right)$ 的因子,对 $\forall x \neq 0$ ,我们首先将 (6.30) 式改写为
$$ \frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( {\frac{1}{2}x}\right) = {f}^{\prime }\left( {\theta \left( x\right) x}\right) - {f}^{\prime }\left( {\frac{1}{2}x}\right) . \tag{6.36} $$
接着在 (6.36) 式的右端对 ${f}^{\prime }\left( x\right)$ 用拉格朗日中值定理,存在 $\eta$ 于 $\theta \left( x\right) x$ 与 $\frac{1}{2}x$ 之间,使得
$$ {f}^{\prime }\left( {\theta \left( x\right) x}\right) - {f}^{\prime }\left( {\frac{1}{2}x}\right) = {f}^{\prime \prime }\left( \eta \right) x\left( {\theta \left( x\right) - \frac{1}{2}}\right) . \tag{6.37} $$
注意到 ${f}^{\prime \prime }\left( \eta \right) \neq 0$ ,联立 (6.36) 和 (6.37) 式,我们得到 $\theta \left( x\right) - \frac{1}{2}$ 的表达式:
$$ \theta \left( x\right) - \frac{1}{2} = \frac{\frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( {\frac{1}{2}x}\right) }{x{f}^{\prime \prime }\left( \eta \right) }. \tag{6.38} $$
到此,为了证明结论只需证明 (6.38) 式的右端当 $x \rightarrow 0$ 时极限是零即可. 因为 $\eta$ 在 $\theta \left( x\right) x$ 与 $\frac{1}{2}x$ 之间,所以
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\eta = 0 \Rightarrow \mathop{\lim }\limits_{{x \rightarrow 0}}{f}^{\prime \prime }\left( \eta \right) = \mathop{\lim }\limits_{{\eta \rightarrow 0}}{f}^{\prime \prime }\left( \eta \right) = {f}^{\prime \prime }\left( 0\right) \neq 0. \tag{6.39} $$
又用洛必达法则两次, 我们有
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\frac{f\left( x\right) - f\left( 0\right) }{x} - {f}^{\prime }\left( {\frac{1}{2}x}\right) }{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{f\left( x\right) - f\left( 0\right) - x{f}^{\prime }\left( {\frac{1}{2}x}\right) }{{x}^{2}} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{f}^{\prime }\left( x\right) - {f}^{\prime }\left( {\frac{1}{2}x}\right) - \frac{x}{2}{f}^{\prime \prime }\left( {\frac{1}{2}x}\right) }{2x} $$
$$ = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{f}^{\prime }\left( x\right) - {f}^{\prime }\left( \frac{x}{2}\right) }{x} - \frac{1}{4}\mathop{\lim }\limits_{{x \rightarrow 0}}{f}^{\prime \prime }\left( \frac{x}{2}\right) $$
$$ = \frac{1}{2}{f}^{\prime \prime }\left( 0\right) - \frac{1}{4}{f}^{\prime \prime }\left( 0\right) - \frac{1}{4}{f}^{\prime \prime }\left( 0\right) = 0, \tag{6.40} $$
联立 (6.38),(6.39) 和 (6.40) 式,即得 $\mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\theta \left( x\right) - \frac{1}{2}}\right) = 0$ .
证法 3 为了从 (6.30) 式获取 $\theta \left( x\right)$ 的信息,需要从 (6.30) 式右端着手. 将 (6.30) 式看做 $f\left( x\right)$ 在点 $x = 0$ 处展开的零阶泰勒公式,其中 $x{f}^{\prime }\left( {\theta \left( x\right) x}\right)$ 表示的是用 $f\left( 0\right)$ 去近似 $f\left( x\right)$ 的误差,而关于这个误差的更多信息蕴藏在更高一阶的泰勒公式中. 这启发我们用泰勒公式,则对 $\forall x \neq 0$ ,存在 $\xi$ 于 0 与 $x$ 之间,使得
$$ f\left( x\right) = f\left( 0\right) + {f}^{\prime }\left( 0\right) x + \frac{1}{2}{f}^{\prime \prime }\left( \xi \right) {x}^{2}\;\left( {0 < \left| \xi \right| < \left| x\right| }\right) . $$
(6.41)
联立 (6.30) 和 (6.41) 式, 得到
$$ x{f}^{\prime }\left( {\theta \left( x\right) x}\right) = {f}^{\prime }\left( 0\right) x + \frac{1}{2}{f}^{\prime \prime }\left( \xi \right) {x}^{2} $$
$$ \xrightarrow[{\text{ 等式两边消去 }x}]{\because x \neq 0}{f}^{\prime }\left( {\theta \left( x\right) x}\right) = {f}^{\prime }\left( 0\right) + \frac{1}{2}{f}^{\prime \prime }\left( \xi \right) x. $$
为了在上面最后一个方程中,使 $\theta \left( x\right)$ 从 ${f}^{\prime }\left( {\theta \left( x\right) x}\right)$ 的括号内 “解放” 出来,令 $t = {x\theta }\left( x\right)$ ,我们得到 $\theta \left( x\right)$ 的表达式,再用导数定义即得结论. 即
$$ \theta \left( x\right) = \frac{1}{2}\frac{{f}^{\prime \prime }\left( \xi \right) }{\frac{{f}^{\prime }\left( t\right) - {f}^{\prime }\left( 0\right) }{t}} \Rightarrow \mathop{\lim }\limits_{{x \rightarrow 0}}\theta \left( x\right) $$
$$ = \frac{1}{2}\frac{\mathop{\lim }\limits_{{x \rightarrow 0}}{f}^{\prime \prime }\left( \xi \right) }{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{f}^{\prime }\left( t\right) - {f}^{\prime }\left( 0\right) }{t}} = \frac{1}{2}\frac{{f}^{\prime \prime }\left( 0\right) }{{f}^{\prime \prime }\left( 0\right) } = \frac{1}{2}. $$