每日一题/012/数学分析/求极限/拉格拉日中值定理/幂指函数求导

题目：
求：
$$\underset{x\to 0}{\lim }\frac{{\left(e^{\sin x}+\sin x\right)}^{\frac{1}{\sin x}}-{\left(e^{\tan x}+\tan x\right)}^{\frac{1}{\tan x}}}{x^3}$$

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参考答案：
设函数

$$f(x)=(e^x+x{)}^{\frac{1}{x}}$$

取对数

$$\ln f(x)=\frac{\ln (e^x+x)}{x}$$

求导

$$\frac{f^{\prime }(x)}{f(x)}=\frac{\frac{e^x+1}{e^x+x}\cdot x-\ln (e^x+x)}{x^2}$$

于是

$$f^{\prime }(x)=\frac{\frac{e^x+1}{e^x+x}\cdot x-\ln (e^x+x)}{x^2}\cdot (e^x+x{)}^{\frac{1}{x}}$$

下面来分别求出

$$\underset{x\to 0}{\lim }\frac{\frac{e^x+1}{e^x+x}\cdot x-\ln (e^x+x)}{x^2}\text{和}\underset{x\to 0}{\lim }(e^x+x{)}^{\frac{1}{x}}$$

$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\frac{e^x+1}{e^x+x}\cdot x-\ln (e^x+x)}{x^2}& =\underset{x\to 0}{\lim }\frac{\frac{e^x(e^x+x)-(e^x+1)(e^x+1)}{(e^x+x{)}^2}\cdot x}{2x}\\ & =\underset{x\to 0}{\lim }\frac{e^x(e^x+x)-(e^x+1{)}^2}{2(e^x+x{)}^2}\\ & =-\frac{3}{2}\end{array}$$

$$\begin{array}{rl}\underset{x\to 0}{\lim }(e^x+x{)}^{\frac{1}{x}}& =e^{\underset{x\to 0}{\lim }\frac{\ln (e^x+x)}{x}}\\ & =e^{\underset{x\to 0}{\lim }\frac{e^x+x-1}{x}}\\ & =e^{\underset{x\to 0}{\lim }{e}^x+1}\\ & =e^2\end{array}$$
所以
$$\underset{x\to 0}{\lim }f(x)=-\frac{3}{2}{e}^2$$

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根据拉格拉日中值定理
$$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{{\left(e^{\sin x}+\sin x\right)}^{\frac{1}{\sin x}}-{\left(e^{\tan x}+\tan x\right)}^{\frac{1}{\tan x}}}{x^3}& =\underset{x\to 0}{\lim }\frac{f(\sin x)-f(\tan x)}{x^3}\\ & =\underset{x\to 0}{\lim }\frac{f^{\prime }(\xi )(\sin x-\tan x)}{x^3}\xi \in (\sin x,\tan x)\\ & =\underset{x\to 0}{\lim }{f}^{\prime }(\xi )\underset{x\to 0}{\lim }\frac{\sin x-\tan x}{x^3}\\ & =-\frac{3}{2}{e}^2\cdot -\frac{1}{2}\\ & =\frac{3}{4}{e}^2\end{array}$$

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2021年1月7日23:24:58