求极限 $\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\tan (\tan x)}{\sin x-\tan x}$.
【答案】 因 $\sin x-\tan x=(\cos x-1) \tan x \sim-\frac{1}{2} x^3$, 则
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\tan (\tan x)}{\sin x-\tan x} &=\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x+\sin x-\tan x+\tan x-\tan (\tan x)}{\sin x-\tan x} \\
&=\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{\sin x-\tan x}-\lim _{x \rightarrow 0} \frac{\tan (\tan x)-\tan x}{\sin x-\tan x}+1 \\
&=-2 \lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{x^3}+2 \lim _{x \rightarrow 0} \frac{\tan (\tan x)-\tan x}{x^3}+1 \\
&=-2 \lim _{x \rightarrow 0} \frac{[\cos (\sin x)-1] \cos x}{3 x^2}+2 \lim _{x \rightarrow 0} \frac{\left[\sec ^2(\tan x)-1\right] \cdot \sec ^2 x}{3 x^2}+1 \\
&=-2 \lim _{x \rightarrow 0} \frac{-\frac{1}{2}(\sin x)^2 \cos x}{3 x^2}+2 \lim _{x \rightarrow 0} \frac{\tan ^2(\tan x) \cdot \sec ^2 x}{3 x^2}+1 \\
&=-2 \lim _{x \rightarrow 0} \frac{-\frac{1}{2} x^2 \cos x}{3 x^2}+2 \lim _{x \rightarrow 0} \frac{x^2 \sec ^2 x}{3 x^2}+1=\frac{1}{3}+\frac{2}{3}+1=2
\end{aligned}
$$


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