分子是两式相减等于零的极限怎么算？先做分子有理化

一、题目

已知：

$$
\begin{aligned}
I & = \\
& \lim \limits_{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\arctan x}}{x-\sin x}
\end{aligned}
$$

则：

$$
I \ = \ ?
$$

难度评级：

二、解析

分析可知，式子 $\lim \limits_{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\arctan x}}{x-\sin x}$ 是一个 $\frac{1 – 1}{0 – 0} = \frac{0}{0}$ 型的极限，因此，首先尝试对分子进行有理化：

$$
\begin{aligned}
I & \\
& = \lim \limits_{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\arctan x}}{x-\sin x} \\ \\
& = \lim \limits_{x \rightarrow 0} \frac{(\sqrt{1+\tan x} – \sqrt{1+\arctan x}) (\sqrt{1+\tan x} + \sqrt{1+\arctan x})}{x-\sin x} \\ \\
& = \lim \limits_{x \rightarrow 0} \frac{1+\tan x – 1+\arctan x }{(x-\sin x) (\sqrt{1+\tan x} + \sqrt{1+\arctan x})} \\ \\
& = \lim \limits_{x \rightarrow 0} \frac{\tan x + \arctan x }{(x-\sin x) (1 + 1)} \\ \\
& = \textcolor{orangered}{ \frac{1}{2} \lim \limits_{x \rightarrow 0} \frac{\tan x + \arctan x }{(x-\sin x)} }
\end{aligned}
$$

又根据等价无穷小公式可知，当 $x \rightarrow 0$ 时, 有：

$$
\textcolor{yellow}{
\begin{aligned}
x-\sin x & \sim \frac{1}{6} x^{3} \\ \\
\tan x – x & \sim \frac{1}{3} x^{3} \\ \\
x – \arctan x & \sim \frac{1}{3} x^{3}
\end{aligned}
}
$$

于是：

$$
\begin{aligned}
I \\
& = \textcolor{orangered}{ \frac{1}{2} \lim \limits_{x \rightarrow 0} \frac{\tan x + \arctan x }{(x-\sin x)} } \\ \\
& = \frac{1}{2} \lim \limits_{x \rightarrow 0} \frac{\tan x-\arctan x}{\textcolor{yellow}{ \frac{x^{3}}{6} }} \\ \\
& = 3 \lim \limits_{x \rightarrow 0} \frac{\tan x-\arctan x}{x^{3}} \\ \\
& = 3 \lim \limits_{x \rightarrow 0} \frac{\tan x – x + x – \arctan x}{x^{3}} \\ \\
& = 3 \left(\lim \limits_{x \rightarrow 0} \frac{\tan x-x}{x^{3}}+\lim \limits_{x \rightarrow 0} \frac{x-\arctan x}{x^{3}}\right) \\ \\
& = 3 \left(\lim \limits_{x \rightarrow 0} \frac{\textcolor{yellow}{\frac{1}{2} x^{3} } }{x^{3}} + \lim \limits_{x \rightarrow 0} \frac{ \textcolor{yellow}{\frac{1}{3} x^{3} }}{x^{3}} \right) \\ \\
& = 3\left(\frac{1}{3}+\frac{1}{3}\right) \\ \\
& = \textcolor{springgreen}{ 2 }
\end{aligned}
$$

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