式子复杂不要怕，先分析其“型”，再确定求解之“法”

一、题目

已知：

$$f(x)=\underset{t\to x}{lim}\sin x\cdot {\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}$$

则：

$$\underset{x\to 0}{lim}\frac{f(x)–x}{x^3}=?$$

难度评级：

二、解析

首先：

$$\begin{array}{rl}f(x)& =\underset{t\to x}{lim}\sin x\cdot {\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}\\ \\ & =\sin x\cdot \underset{t\to x}{lim}{\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}\end{array}$$

观察可知，当 $t\to x$ 的时候，${\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}$ 是一个 $1^{\mathrm{\infty }}$ 型极限，因此：

$$\begin{array}{rl}f(x)& =\underset{t\to x}{lim}\sin x\cdot {\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}\\ \\ & =\sin x\cdot \underset{t\to x}{lim}{\left(\frac{t}{x}\right)}^{\frac{t^3}{t–x}}\\ \\ & =\sin x\cdot \underset{t\to x}{lim}{(1+\frac{t}{x}–1)}^{\frac{t^3}{t–x}}\\ \\ & =\sin x\cdot \underset{t\to x}{lim}{(1+\frac{t–x}{x})}^{\frac{t^3}{t–x}}\\ \\ & =\sin x\cdot \underset{t\to x}{lim}{(1+\frac{t–x}{x})}^{\frac{x}{t–x}\cdot \frac{t^3}{t–x}\cdot \frac{t–x}{x}}\\ \\ & =\sin x\cdot \underset{x\to x}{lim}{e}^{\frac{t^3}{x}}\\ \\ & =\sin x\cdot e^{x^2}\end{array}$$

于是：

$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{f(x)–x}{x^3}& =\underset{x\to 0}{lim}\frac{\sin x\cdot e^{x^2}–x}{x^3}\\ \\ & =\underset{x\to 0}{lim}\frac{\sin x\cdot e^{x^2}–\sin x+\sin x–x}{x^3}\\ \\ & =\underset{x\to 0}{lim}\frac{\sin x\cdot e^{x^2}–\sin x}{x^3}+\underset{x\to 0}{lim}\frac{\sin x–x}{x^3}\\ \\ & =\underset{x\to 0}{lim}\frac{\sin x\cdot (e^{x^2}–1)}{x^3}+\underset{x\to 0}{lim}\frac{–\frac{1}{6}{x}^3}{x^3}\\ \\ & =1–\frac{1}{6}=\frac{5}{6}\end{array}$$

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