📝 题目
例 3 求极限 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{\alpha }}\left( {\alpha > 0}\right)$ .
💡 答案与解析
解 $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{a}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\frac{1}{x}}{\alpha {x}^{\alpha - 1}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{\alpha {x}^{a}} = 0}$ .