## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\sin (5(x+\pi))}{e^{3 x}-1}
$$
## Solution

We will use the substitution of equivalent infinitesimals:

$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

$\sin 5 x \sim 5 x$, as $x \rightarrow 0(5 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin (5(x+\pi))}{e^{3 x}-1}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\sin (5 x+5 \pi)}{3 x}= \\
& =\lim _{x \rightarrow 0} \frac{\sin (5 x+\pi)}{3 x}=\lim _{x \rightarrow 0} \frac{-\sin (5 x)}{3 x}= \\
& =\lim _{x \rightarrow 0} \frac{-5}{3}=-\frac{5}{3}=-1 \frac{2}{3}
\end{aligned}
$$