## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{7^{2 x}-5^{3 x}}{2 x-\operatorname{arctg} 3 x}$
## Solution

$\lim _{x \rightarrow 0} \frac{7^{2 x}-5^{3 x}}{2 x-\operatorname{arctg} 3 x}=\lim _{x \rightarrow 0} \frac{\left(49^{x}-1\right)-\left(125^{x}-1\right)}{2 x-\operatorname{arctg} 3 x}=$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 49}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\right)^{x}-1\right)}{2 x-\operatorname{arctg} 3 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 49}-1\right)-\left(e^{x \ln 125}-1\right)}{2 x-\operatorname{arctg} 3 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 49}-1\right)-\left(e^{x \ln 125}-1\right)\right)}{\frac{1}{x}(2 x-\operatorname{arctg} 3 x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 49}-1\right)-\left(e^{x \ln 125}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 x-\operatorname{arctg} 3 x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 49}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 125}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 49}-1 \sim x \ln 49$, as $x \rightarrow 0(x \ln 49 \rightarrow 0)$

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

$\operatorname{arctg} 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 49}{x}-\lim _{x \rightarrow 0} \frac{x \ln 125}{x}}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} \frac{3 x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 49-\lim _{x \rightarrow 0} \ln 125}{2-\lim _{x \rightarrow 0} 3}= \\
& =\frac{\ln 49-\ln 125}{2-3}=-\ln \frac{49}{125}=\ln \frac{125}{49}
\end{aligned}
$$

## Problem Kuznetsov Limits 15-1