## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow u} \frac{2^{x}-1}{\ln (1+2 x)}
$$
## Solution

We will use the substitution of equivalent infinitesimals:

\(\ln (1+2 x) \sim 2 x\), as \(x \rightarrow 0\) (i.e., \(2 x \rightarrow 0\))

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

We obtain:

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

## Problem Kuznetsov Limits 12-9