## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{6 x}\right)^{\frac{x}{x+2}}=\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\lim _{x \rightarrow 0} \frac{x}{x+2}}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{\frac{0}{0+2}}=\left(\lim _{x \rightarrow 0} \frac{\ln (1+x)}{6 x}\right)^{0}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln (1+x) \sim x$, as $x \rightarrow 0$

We get:

$$
=\left(\lim _{x \rightarrow 0} \frac{x}{6 x}\right)^{0}=\left(\lim _{x \rightarrow 0} \frac{1}{6}\right)^{0}=\left(\frac{1}{6}\right)^{0}=1
$$

## Problem Kuznetsov Limits 18-7