看到无穷大，就要想转为无穷小：因为无穷小能用的解题工具更多

一、题目

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(2^{\frac{1}{x}}–2^{\frac{1}{x+1}})=?$$

难度评级：

二、解析

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(2^{\frac{1}{x}}–2^{\frac{1}{x+1}})=$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2[(2^{\frac{1}{x}}–1)–(2^{\frac{1}{x+1}}–1)]=$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2(\frac{1}{x}\ln 2–\frac{1}{x+1}\ln 2)=$$

$$\underset{x\to +\mathrm{\infty }}{lim}{x}^2\ln 2(\frac{1}{x}–\frac{1}{x+1})=$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2\ln 2}{x(x+1)}=$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2\ln 2}{x^2+x}=$$

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{x^2\ln 2}{x^2}=\ln 2.$$

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