📝 题目
例 9 求 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x - \arctan x}{\tan x - \arcsin x}}$ .
💡 答案与解析
解 我们有
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x - \arctan x}{\tan x - \arcsin x} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\left( {x - \frac{{x}^{3}}{3!} + o\left( {x}^{3}\right) }\right) - \left( {x - \frac{{x}^{3}}{3} + o\left( {x}^{3}\right) }\right) }{\left( {x + \frac{{x}^{3}}{3} + o\left( {x}^{3}\right) }\right) - \left( {x + \frac{{x}^{3}}{3!} + o\left( {x}^{3}\right) }\right) } $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\frac{{x}^{3}}{6} + o\left( {x}^{3}\right) }{\frac{{x}^{3}}{6} + o\left( {x}^{3}\right) } = 1. $$