📝 题目
例 12 求 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos x}{{x}^{2}}}$ 和 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\tan x - \sin x}{x}}$ .
💡 答案与解析
解 利用 $1 - \cos x = 2{\sin }^{2}\frac{x}{2}$ 得
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos x}{{x}^{2}} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{2{\sin }^{2}\frac{x}{2}}{{x}^{2}} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1}{2}{\left( \frac{\sin \frac{x}{2}}{\frac{x}{2}}\right) }^{2} $$
$$ = \mathop{\lim }\limits_{{y \rightarrow 0}}\frac{1}{2}{\left( \frac{\sin y}{y}\right) }^{2} = \frac{1}{2}. $$
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\tan x - \sin x}{{x}^{3}} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1}{\cos x} \cdot \frac{\sin x}{x} \cdot \frac{1 - \cos x}{{x}^{2}} $$
$$ = \frac{1}{2}\text{ . } $$