## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{3^{2 x}-5^{3 x}}{\operatorname{arctg} x+x^{3}}$
## Solution

$\lim _{x \rightarrow 0} \frac{3^{2 x}-5^{3 x}}{\operatorname{arctg} x+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(9^{x}-1\right)-\left(125^{x}-1\right)}{\operatorname{arctg} x+x^{3}}=$

$=\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 9}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\right)^{x}-1\right)}{\operatorname{arctg} x+x^{3}}=$

$=\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 125}-1\right)}{\operatorname{arctg} x+x^{3}}=$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 125}-1\right)\right)}{\frac{1}{x}\left(\operatorname{arctg} x+x^{3}\right)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 9}-1\right)-\left(e^{x \ln 125}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\operatorname{arctg} x+x^{3}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 9}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 125}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} x}{x}+\lim _{x \rightarrow 0} \frac{x^{3}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 9}-1 \sim x \ln 9$, as $x \rightarrow 0(x \ln 9 \rightarrow 0)$

$e^{x \ln 125}-1 \sim x \ln 125$, as $x \rightarrow 0(x \ln 125 \rightarrow 0)$

$\operatorname{arctg} x \sim x$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 9}{x}-\lim _{x \rightarrow 0} \frac{x \ln 125}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} x^{2}}=\frac{\lim _{x \rightarrow 0} \ln 9-\lim _{x \rightarrow 0} \ln 125}{\lim _{x \rightarrow 0} 1+0}= \\
& =\frac{\ln 9-\ln 125}{1+0}=\ln \frac{9}{125}
\end{aligned}
$$

## Problem Kuznetsov Limits $15-5$