## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{4^{5 x}-9^{-2 x}}{\sin x-\tan x^{3}}
$$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{4^{5 x}-9^{-2 x}}{\sin x-\operatorname{tg} x^{3}}=\lim _{x \rightarrow 0} \frac{\left(1024^{x}-1\right)-\left(81^{-x}-1\right)}{\sin x-\operatorname{tg} x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 1024}\right)^{x}-1\right)-\left(\left(e^{\ln 81}\right)^{-x}-1\right)}{\sin x-\operatorname{tg} x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 1024}-1\right)-\left(e^{-x \ln 81}-1\right)}{\sin x-\operatorname{tg} x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 1024}-1\right)-\left(e^{-x \ln 81}-1\right)\right)}{\frac{1}{x}\left(\sin x-\operatorname{tg} x^{3}\right)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 1024}-1\right)-\left(e^{-x \ln 81}-1\right)\right)}{\lim \frac{1}{x}\left(\sin x-\operatorname{tg} x^{3}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 1024}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 81}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \frac{\operatorname{tg} x^{3}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 1024}-1 \sim x \ln 1024$, as $x \rightarrow 0(x \ln 1024 \rightarrow 0)$

$e^{-x \ln 81}-1 \sim -x \ln 81$, as $x \rightarrow 0(-x \ln 81 \rightarrow 0)$

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

$\operatorname{tg} x^{3} \sim x^{3}$, as $x \rightarrow 0\left(x^{3} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 1024}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 81}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}-\lim _{x \rightarrow 0} \frac{x^{3}}{x}}=\frac{\lim _{x \rightarrow 0} \ln 1024-\lim _{x \rightarrow 0}-\ln 81}{\lim _{x \rightarrow 0} 1-\lim _{x \rightarrow 0} x^{2}}= \\
& =\frac{\ln 1024+\ln 81}{1-0}=\ln \left(2^{10} \cdot 9^{2}\right)
\end{aligned}
$$