## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{2 x}{\tan\left(2 \pi\left(x+\frac{1}{2}\right)\right)}$
## Solution

We will use the substitution of equivalent infinitesimals:

$\operatorname{tg} 2 \pi x \sim 2 \pi x$, as $x \rightarrow 0 (2 \pi x \rightarrow 0)$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2 x}{\operatorname{tg}\left(2 \pi\left(x+\frac{1}{2}\right)\right)}=\lim _{x \rightarrow 0} \frac{2 x}{\operatorname{tg}(2 \pi x+\pi)}=\lim _{x \rightarrow 0} \frac{2 x}{\operatorname{tg}(2 \pi x)}= \\
& =\lim _{x \rightarrow 0} \frac{2 x}{2 \pi x}=\lim _{x \rightarrow 0} \frac{1}{\pi}=\frac{1}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-6