📝 题目
例 6 求极限 $\mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{1}{{x}^{2}} - {\cot }^{2}x}\right)$ .
💡 答案与解析
解 这是一个 $\displaystyle{\infty - \infty}$ 型未定式,它容易转化成 $\frac{0}{0}$ 型未定式:
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{1}{{x}^{2}} - {\cot }^{2}x}\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{\sin }^{2}x - {x}^{2}{\cos }^{2}x}{{x}^{2}{\sin }^{2}x} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x + x\cos x}{\sin x} \cdot \frac{\sin x - x\cos x}{{x}^{2}\sin x} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {1 + \frac{x}{\sin x}\cos x}\right) \left( \frac{\sin x - x\cos x}{{x}^{2}\sin x}\right) . $$
前一因式的极限为
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {1 + \frac{x}{\sin x}\cos x}\right) = 2. $$
为求得后一因式的极限, 我们用洛必达法则:
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x - x\cos x}{{x}^{2}\sin x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x\sin x}{{2x}\sin x + {x}^{2}\cos x} $$
$$ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{2\sin x + x\cos x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos x}{3\cos x - x\sin x} = \frac{1}{3}. $$
于是, 所求的极限为
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{1}{{x}^{2}} - {\cot }^{2}x}\right) = \frac{2}{3}. $$