## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{\tan 4 x}{x}\right)^{2+x}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{\lim _{x \rightarrow 0} 2+x}=$

$=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2+0}=\left(\lim _{x \rightarrow 0} \frac{\tan 4 x}{x}\right)^{2}=$

Using the substitution of equivalent infinitesimals:

$\tan 4 x \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$

We get:

$=\left(\lim _{x \rightarrow 0} \frac{4 x}{x}\right)^{2}=\left(\lim _{x \rightarrow 0} \frac{4}{1}\right)^{2}=4^{2}=16$

## Problem Kuznetsov Limits 18-8