## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{6^{2 x}-7^{-2 x}}{\sin 3 x-2 x}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{6^{2 x}-7^{-2 x}}{\sin 3 x-2 x}=\lim _{x \rightarrow 0} \frac{\left(36^{x}-1\right)-\left(49^{-x}-1\right)}{\sin 3 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 36}\right)^{x}-1\right)-\left(\left(e^{\ln 49}\right)^{-x}-1\right)}{\sin 3 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 36}-1\right)-\left(e^{-x \ln 49}-1\right)}{\sin 3 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 36}-1\right)-\left(e^{-x \ln 49}-1\right)\right)}{\frac{1}{x}(\sin 3 x-2 x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 36}-1\right)-\left(e^{-x \ln 49}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 3 x-2 x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 36}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 49}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 36}-1 \sim x \ln 36$, as $x \rightarrow 0(x \ln 36 \rightarrow 0)$

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

$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 36}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 49}{x}}{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} 2}=\frac{\lim _{x \rightarrow 0} \ln 36-\lim _{x \rightarrow 0}-\ln 49}{\lim _{x \rightarrow 0} 3-2}= \\
& =\frac{\ln 36+\ln 49}{3-2}=\ln \left(6^{2} \cdot 7^{2}\right)
\end{aligned}
$$

## Problem Kuznetsov Limits $15-3$