📝 题目
例 4 求下列极限:
(1) $\mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{a}}\left( {a > 0}\right)$ ; (2) $\displaystyle{\mathop{\lim }\limits_{{x \rightarrow + \infty }}{x}^{\frac{1}{x}}}$ .
💡 答案与解析
解 (1) 令 $x = {\mathrm{e}}^{t},b = {\mathrm{e}}^{a}$ ,则有 $b > 1$ ,且
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{a}} = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{t}{{\mathrm{e}}^{at}} = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{t}{{b}^{t}}\overset{\text{ 用例 }3}{ = }0. $$
(2)用第(1)小题结果,当 $a = 1$ 时,有
$$ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{x} = 0 \Rightarrow \mathop{\lim }\limits_{{x \rightarrow + \infty }}{x}^{\frac{1}{x}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\mathrm{e}}^{\frac{\ln x}{x}} = {\mathrm{e}}^{0} = 1. $$