## Problem Statement

Calculate the limit of the function:

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

Let's use the substitution of equivalent infinitesimals:

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

We get:

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