## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

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

We get:

$=\lim _{x \rightarrow 0} \frac{2 x \ln 2}{x}=\lim _{x \rightarrow 0} 2 \ln 2=2 \ln 2$

## Problem Kuznetsov Limits $18-17$