## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{2 \sin (\pi(x+1))}{\ln (1+2 x)}$
## Solution

We will use the substitution of equivalent infinitesimals:

$\ln (1+2 x) \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

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

We obtain:

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

## Problem Kuznetsov Limits 12-17