## Problem Statement

Calculate the limit of the function:

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

We will use the substitution of equivalent infinitesimals:

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

We get:

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

## Problem Kuznetsov Limits 12-30