📝 题目
例 6 求证:
(1) $\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{a}^{x} - 1}{x} = \ln a\left( {a > 0}\right)$ ； (2) $\mathop{\lim }\limits_{{n \rightarrow \infty }}n\left( {\sqrt[n]{a} - 1}\right) = \ln a$ .
💡 答案与解析
证 (1) 令 $y = {a}^{x} - 1$ ,则有 $x \rightarrow 0 \Leftrightarrow y \rightarrow 0$ ,且
$$ x = \frac{\ln \left( {1 + y}\right) }{\ln a} \Rightarrow \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{a}^{x} - 1}{x} = \mathop{\lim }\limits_{{y \rightarrow 0}}\frac{y\ln a}{\ln \left( {1 + y}\right) } = \ln a. $$
(2) $\mathop{\lim }\limits_{{n \rightarrow \infty }}n\left( {\sqrt[n]{a} - 1}\right) \overset{x = \frac{1}{n}}{ = }\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{a}^{x} - 1}{x}$ . 由第 (1) 小题 In $a$ .