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

$$
\begin{aligned}
I \\
& = \lim_{x \rightarrow + \infty} \left(\sqrt[6]{x^{6} + x^{5}}−\sqrt[6]{x^{6}−x^{5}}\right) \\
& = ?
\end{aligned}
$$

难度评级：

二、解析

$$
\begin{aligned}
I \\
& = \lim_{x \rightarrow + \infty} \left(\sqrt[6]{x^{6} + x^{5}}−\sqrt[6]{x^{6}−x^{5}}\right) \\ \\
& = \lim_{x \rightarrow + \infty} \left(x^{6} + x^{5}\right)^{\frac{1}{6}} − \lim_{x \rightarrow + \infty} \left(x^{6}−x^{5}\right)^{\frac{1}{6}} \\ \\
& = \lim_{x \rightarrow + \infty} \left(x^{6}\right)^{\frac{1}{6}}\left(1 + \frac{1}{x}\right)^{\frac{1}{6}} − \lim_{x \rightarrow + \infty} \left(x^{6}\right)^{\frac{1}{6}} \left( 1 – \frac{1}{x} \right)^{\frac{1}{6}} \\ \\
& = \lim_{x \rightarrow + \infty} x \left[ \left( 1 + \frac { 1 } { x } \right) ^ { \frac { 1 } { 6 } } – \left( 1 – \frac { 1 } { x } \right) ^ { \frac { 1 } { 6 } } \right] \\ \\
& = \lim_{x \rightarrow + \infty} x \left[ \left( \left( 1 + \frac { 1 } { x } \right) ^ { \frac { 1 } { 6 } } – 1 \right) – \left( \left( 1 \bar { x } \frac { 1 } { x } \right) ^ { \frac { 1 } { 6 } } – 1 \right) \right] \\ \\
& = \lim_{x \rightarrow + \infty} x \left( \frac { 1 } { 6 } x + \frac { 1 } { 6 } x \right) \\ \\
& = \lim_{x \rightarrow + \infty} x \cdot \frac { 2 } { 6 } x \\ \\
& = \frac { 2 } { 6 } \\ \\
& = \textcolor{springgreen}{\boldsymbol{\frac { 1 } { 3 }}}
\end{aligned}
$$