## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& e^{x \ln 3}-1 \sim x \ln 3, \text { as } x \rightarrow 0(x \ln 3 \rightarrow 0) \\
& \ln \left(1+x \sqrt{1+x e^{x}}\right) \sim x \sqrt{1+x e^{x}}, \text { as } \\
& \ln \rightarrow 0\left(x \sqrt{1+x e^{x}} \rightarrow 0\right)
\end{aligned}
$$

We get:

$=\lim _{x \rightarrow 0} \frac{3 x \ln 3}{x \sqrt{1+x e^{x}}}=\lim _{x \rightarrow 0} \frac{3 \ln 3}{\sqrt{1+x e^{x}}}=$

$$
=\frac{3 \ln 3}{\sqrt{1+0 \cdot e^{0}}}=\frac{3 \ln 3}{1}=3 \ln 3
$$

## Problem Kuznetsov Limits 16-26