## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^{2}}-1}
$$
## Solution

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

Using the substitution of equivalent infinitesimals:

$e^{x^{2}}-1 \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

$\sin x \sim x$, as $x \rightarrow 0$

We get:

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

## Problem Kuznetsov Limits 16-7