## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{x}\right)^{\cos ^{2}\left(\frac{\pi}{4}+x\right)}
$$
## Solution

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

Using the substitution of equivalent infinitesimals:

$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$$
=\left(\lim _{x \rightarrow 0} \frac{3 x}{x}\right)^{\frac{1}{2}}=\left(\lim _{x \rightarrow 0} \frac{3}{1}\right)^{\frac{1}{2}}=(3)^{\frac{1}{2}}=\sqrt{3}
$$

## Problem Kuznetsov Limits 18-4