## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\ln (1+\sin x)}{\sin 4 x}
$$
## Solution

Let's use the substitution of equivalent infinitesimals:

\(\sin x \sim x\), as \(x \rightarrow 0\)

\(\sin 4x \sim 4x\), as \(x \rightarrow 0\) (or \(4x \rightarrow 0\))

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

We get:

\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\ln (1+\sin x)}{\sin 4 x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{4 x}=\left\{\frac{0}{0}\right\}= \\
& =\lim _{x \rightarrow 0} \frac{x}{4 x}=\lim _{x \rightarrow 0} \frac{1}{4}=\frac{1}{4}
\end{aligned}
\]

## Problem Kuznetsov Limits 12-1