## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{\sin 3 x}$
## Solution

$\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{\sin 3 x}=$
$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{2}}{3 x}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{x+2}-\sqrt{2})(\sqrt{x+2}+\sqrt{2})}{3 x(\sqrt{x+2}+\sqrt{2})}= \\
& =\lim _{x \rightarrow 0} \frac{x+2-2}{3 x(\sqrt{x+2}+\sqrt{2})}=\lim _{x \rightarrow 0} \frac{x}{3 x(\sqrt{x+2}+\sqrt{2})}= \\
& =\lim _{x \rightarrow 0} \frac{1}{3(\sqrt{x+2}+\sqrt{2})}=\frac{1}{3(\sqrt{0+2}+\sqrt{2})}= \\
& =\frac{1}{3 \cdot 2 \sqrt{2}}=\frac{1}{6 \sqrt{2}}
\end{aligned}
$$

## Problem Kuznetsov Limits 16-20