## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{4 x}{\tan(\pi(2+x))}$
## Solution

We will use the substitution of equivalent infinitesimals:

$\operatorname{tg} \pi x \sim \pi x$, as $x \rightarrow 0 (\pi x \rightarrow 0)$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{4 x}{\operatorname{tg}(\pi(2+x))}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{4 x}{\operatorname{tg}(2 \pi+\pi x)}= \\
& =\lim _{x \rightarrow 0} \frac{4 x}{\operatorname{tg}(\pi x)}=\lim _{x \rightarrow 0} \frac{4 x}{\pi x}=\lim _{x \rightarrow 0} \frac{4}{\pi}=\frac{4}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits $12-5$