## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{3^{5 x}-2^{x}}{x-\sin 9 x}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{x}}{x-\sin 9 x}=\lim _{x \rightarrow 0} \frac{\left(243^{x}-1\right)-\left(2^{x}-1\right)}{x-\sin 9 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 243}\right)^{x}-1\right)-\left(\left(e^{\ln 2}\right)^{x}-1\right)}{x-\sin 9 x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 243}-1\right)-\left(e^{x \ln 2}-1\right)}{x-\sin 9 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{x \ln 2}-1\right)\right)}{\frac{1}{x}(x-\sin 9 x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{x \ln 2}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(x-\sin 9 x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 243}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 2}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{\sin 9 x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

$e^{x \ln 2}-1 \sim x \ln 2$, as $x \rightarrow 0(x \ln 2 \rightarrow 0)$

$\sin 9 x \sim 9 x$, as $x \rightarrow 0(9 x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 243}{x}-\lim _{x \rightarrow 0} \frac{x \ln 2}{x}}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} \frac{9 x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 243-\lim _{x \rightarrow 0} \ln 2}{1-\lim _{x \rightarrow 0} 9}= \\
& =\frac{\ln 243-\ln 2}{1-9}=-\frac{1}{8} \ln \frac{243}{2}=\frac{1}{8} \ln \frac{2}{243}
\end{aligned}
$$

## Problem Kuznetsov Limits 15-7