Example 2. Find $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}$.
Solution. Since

$$
\frac{\ln (1+x)}{x}=\frac{1}{x} \ln (1+x)=\ln (1+x)^{\frac{1}{x}}
$$

then based on formula (1.58) we find

$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=\lim _{x \rightarrow 0}\left[\ln (1+x)^{\frac{1}{x}}\right]=\ln \left[\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}\right]=\ln e=1
$$

Therefore,

$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=1
$$