## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{\sin \left(\pi\left(\frac{x}{2}+1\right)\right)}$
## Solution

Let's use the substitution of equivalent infinitesimals:

$e^{4 x}-1 \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$

$\sin \left(\frac{\pi x}{2}\right) \sim \sin \frac{\pi x}{2}$, as $x \rightarrow 0\left(\frac{\pi x}{2} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{4 x}-1}{\sin \left(\pi\left(\frac{x}{2}+1\right)\right)}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{4 x}{\sin \left(\frac{\pi x}{2}+\pi\right)}= \\
& =\lim _{x \rightarrow 0} \frac{4 x}{-\sin \left(\frac{\pi x}{2}\right)}=\lim _{x \rightarrow 0} \frac{4 x}{-\frac{\pi x}{2}}=\lim _{x \rightarrow 0} \frac{4}{-\frac{\pi}{2}}=-\frac{8}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-23