方企勤 第二章 一元函数微分学 第23题
📝 题目
例 23 由拉格朗日中值定理,对 $\forall x > - 1,\exists \theta \in \left( {0,1}\right)$ ,使得
$$ \ln \left( {1 + x}\right) = \ln \left( {1 + x}\right) - \ln \left( {1 + 0}\right) = \frac{x}{1 + {\theta x}}. $$
求证: $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\theta = \frac{1}{2}}$ .
💡 答案解析
证法 1 用带皮亚诺余项的泰勒公式, 得
$$ \ln \left( {1 + x}\right) = x - \frac{{x}^{2}}{2} + o\left( {x}^{2}\right) ,\;\frac{1}{1 + {\theta x}} = 1 - {\theta x} + o\left( x\right) . $$
于是
$$ \ln \left( {1 + x}\right) = \frac{x}{1 + {\theta x}} $$
$$ \Rightarrow x - \frac{{x}^{2}}{2} + o\left( {x}^{2}\right) = x - \theta {x}^{2} + o\left( {x}^{2}\right) \;\left( {x \rightarrow 0}\right) $$
$$ \Rightarrow \theta {x}^{2} = \frac{{x}^{2}}{2} + o\left( {x}^{2}\right) \;\left( {x \rightarrow 0}\right) , $$
即
$$ \theta = \frac{1}{2} + o\left( 1\right) \;\left( {x \rightarrow 0}\right) , $$
即得 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\theta = \frac{1}{2}}$ .
证法 2 由 $\ln \left( {1 + x}\right) = \frac{x}{1 + {\theta x}}$ 解出 $\theta = \frac{x - \ln \left( {1 + x}\right) }{x\ln \left( {1 + x}\right) }$ . 由洛必达法则及 $\ln \left( {1 + x}\right) \sim x\left( {x \rightarrow 0}\right)$ ,得
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\theta = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x - \ln \left( {1 + x}\right) }{x\ln \left( {1 + x}\right) } = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x - \ln \left( {1 + x}\right) }{{x}^{2}} $$
$$ \overset{\text{ 洛必达法则 }}{ = }\frac{1 - {\left( 1 + x\right) }^{-1}}{2x} = \frac{1}{2}. $$