## Problem Statement

Calculate the limit of the function:

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

We will use the substitution of equivalent infinitesimals:

$\arcsin 3 x \sim 3 x$, as $x \rightarrow 0 (3 x \rightarrow 0)$

We get:

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

## Problem Kuznetsov Limits 12-8