遇到关于对数函数的式子，先将乘除变加减

一、题目

设可导函数 $f(x)>0$, 则：

$$
\lim \limits_{n \rightarrow \infty} n \ln \frac{f\left(\frac{1}{n}\right)}{f(0)} = ?
$$

难度评级：

二、解析

首先：

$$
\lim \limits_{n \rightarrow \infty} n \textcolor{orangered}{\ln \frac{f\left(\frac{1}{n}\right)}{f(0)} } =
$$

$$
\lim \limits_{n \rightarrow \infty} n \textcolor{orangered}{\left[ \ln f\left(\frac{1}{n}\right) – \ln f(0) \right] } =
$$

$$
\lim \limits_{n \rightarrow \infty} \frac{\ln f\left(\frac{1}{n}\right) – \ln f(0)}{\frac{1}{n} – 0} =
$$

$$
\lim \limits_{x \rightarrow 0} \frac{\ln f\left( x \right) – \ln f(0)}{x – 0} =
$$

$$
\lim \limits_{x \rightarrow 0} \left[ \ln f(x) \right]^{\prime} =
$$

$$
\lim \limits_{x \rightarrow 0} \frac{f^{\prime}(x)}{f(x)} = \textcolor{springgreen}{ \frac{f^{\prime}(0)}{f(0)} }
$$

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