📝 题目
例 9 设当 $x \rightarrow a$ 时, ${f}_{1}\left( x\right) ,{f}_{2}\left( x\right)$ 为不为零的等价无穷小量. 若广义极限 $\mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{1}\left( x\right) }$ 存在,求证: $\mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{2}\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{1}\left( x\right) }$ ,并利用此结论求极限 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos x}{{\sin }^{2}x}}$ .
💡 答案与解析
解 由广义极限的四则运算法则, 有
$$ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{2}\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{1}\left( x\right) } \cdot \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{1}\left( x\right) } \cdot \mathop{\lim }\limits_{{x \rightarrow a}}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } $$
$$ = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( x\right) }{{f}_{1}\left( x\right) }. $$
利用此结果,因为 ${\sin }^{2}x \sim {x}^{2}\left( {x \rightarrow 0}\right)$ ,立即推出
$$ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos x}{{\sin }^{2}x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1 - \cos x}{{x}^{2}} = \frac{1}{2}. $$
提问 如下两式是否成立:
$$ \frac{o\left( {x}^{2}\right) }{x} = o\left( x\right) ;\;\frac{o\left( {x}^{2}\right) }{o\left( x\right) } = o\left( x\right) . $$
\subsubsection{四、否定命题的肯定叙述}