## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2^{3 x}-3^{5 x}}{\sin 7 x-2 x}=\lim _{x \rightarrow 0} \frac{\left(8^{x}-1\right)-\left(243^{x}-1\right)}{\sin 7 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 8}\right)^{x}-1\right)-\left(\left(e^{\ln 243}\right)^{x}-1\right)}{\sin 7 x-2 x}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 243}-1\right)}{\sin 7 x-2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 243}-1\right)\right)}{\frac{1}{x}(\sin 7 x-2 x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 243}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(\sin 7 x-2 x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 8}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 243}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin 7 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 8}-1 \sim x \ln 8$, as $x \rightarrow 0(x \ln 8 \rightarrow 0)$

$e^{x \ln 243}-1 \sim x \ln 243$, as $x \rightarrow 0(x \ln 243 \rightarrow 0)$

$\sin 7 x \sim 7 x$, as $x \rightarrow 0(x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 8}{x}-\lim _{x \rightarrow 0} \frac{x \ln 243}{x}}{\lim _{x \rightarrow 0} \frac{7 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 8-\lim _{x \rightarrow 0}-\ln 243}{\lim _{x \rightarrow 0} 7-\lim _{x \rightarrow 0} 2}= \\
& =\frac{\ln 8-\ln 243}{7-2}=\frac{1}{5} \ln \frac{8}{243}=\frac{1}{5} \ln \frac{2^{3}}{3^{5}}
\end{aligned}
$$

Problem Kuznetsov Limits 15-31