## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{(\sqrt{1+\tan x}-\sqrt{1+\sin x})(\sqrt{1+\tan x}+\sqrt{1+\sin x})}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{1+\tan x-(1+\sin x)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{\tan x-\frac{\sin x \cdot \cos x}{\cos x}}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{\tan x-\tan x \cdot \cos x}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{\tan x(1-\cos x)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=
\end{aligned}
$$

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

$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{x\left(\frac{x^{2}}{2}\right)}{x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\lim _{x \rightarrow 0} \frac{x^{3}}{2 x^{3}(\sqrt{1+\tan x}+\sqrt{1+\sin x})}= \\
& =\lim _{x \rightarrow 0} \frac{1}{2(\sqrt{1+\tan x}+\sqrt{1+\sin x})}=\frac{1}{2(\sqrt{1+\tan 0}+\sqrt{1+\sin 0})}= \\
& =\frac{1}{2(\sqrt{1+0}+\sqrt{1+0})}=\frac{1}{4}
\end{aligned}
$$

## Problem Kuznetsov Limits 16-5