## Problem Statement

Calculate the limit of the function:

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

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{3^{5 x}-2^{-7 x}}{2 x-\operatorname{tg} x}=\lim _{x \rightarrow 0} \frac{\left(243^{x}-1\right)-\left(128^{-x}-1\right)}{2 x-\operatorname{tg} x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 243}\right)^{x}-1\right)-\left(\left(e^{\ln 128}\right)^{-x}-1\right)}{2 x-\operatorname{tg} x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)}{2 x-\operatorname{tg} x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)\right)}{\frac{1}{x}(2 x-\operatorname{tg} x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 243}-1\right)-\left(e^{-x \ln 128}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 x-\operatorname{tg} x)}=
\end{aligned}
$$

$$
=\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 243}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 128}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x}\right)=
$$

Using the substitution of equivalent infinitesimals:

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

$e^{-x \ln 128}-1 \sim -x \ln 128$, as $x \rightarrow 0(-x \ln 128 \rightarrow 0)$

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

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 243}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 128}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 243-\lim _{x \rightarrow 0}-\ln 128}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}= \\
& =\frac{\ln 243+\ln 128}{2-1}=\ln (243 \cdot 128)=\ln \left(3^{5} \cdot 2^{7}\right)
\end{aligned}
$$

## Problem Kuznetsov Limits 15-27