无穷大转无穷小的一个常用策略：提取公因式构造分式

一、题目

难度评级：

二、解析

$$\begin{array}{r}I\\ & =\underset{x\to +\mathrm{\infty }}{lim}(\sqrt[6]{x^6+x^5}-\sqrt[6]{x^6-x^5})\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}{(x^6+x^5)}^{\frac{1}{6}}-\underset{x\to +\mathrm{\infty }}{lim}{(x^6-x^5)}^{\frac{1}{6}}\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}{\left(x^6\right)}^{\frac{1}{6}}{(1+\frac{1}{x})}^{\frac{1}{6}}-\underset{x\to +\mathrm{\infty }}{lim}{\left(x^6\right)}^{\frac{1}{6}}{\left(1–\frac{1}{x}\right)}^{\frac{1}{6}}\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}x\left[{(1+\frac{1}{x})}^{\frac{1}{6}}–{\left(1–\frac{1}{x}\right)}^{\frac{1}{6}}\right]\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}x\left[\left({(1+\frac{1}{x})}^{\frac{1}{6}}–1\right)–\left({\left(1\overline{x}\frac{1}{x}\right)}^{\frac{1}{6}}–1\right)\right]\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}x(\frac{1}{6}x+\frac{1}{6}x)\\ \\ & =\underset{x\to +\mathrm{\infty }}{lim}x\cdot \frac{2}{6}x\\ \\ & =\frac{2}{6}\\ \\ & =\frac{\mathbf{1}}{\mathbf{3}}\end{array}$$

Note

本题中用到的等价无穷小公式可以在这里找到。

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