## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{2^{3 x}-3^{2 x}}{x+\arcsin x^{3}}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2^{3 x}-3^{2 x}}{x+\arcsin x^{3}}=\lim _{x \rightarrow 0} \frac{\left(8^{x}-1\right)-\left(9^{x}-1\right)}{x+\arcsin x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 8}\right)^{x}-1\right)-\left(\left(e^{\ln 9}\right)^{x}-1\right)}{x+\arcsin x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 9}-1\right)}{x+\arcsin x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 9}-1\right)\right)}{\frac{1}{x}\left(x+\arcsin x^{3}\right)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 8}-1\right)-\left(e^{x \ln 9}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(x+\arcsin x^{3}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 8-1}}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 9}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{\arcsin x^{3}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

$e^{x \ln 9}-1 \sim x \ln 9$, as $x \rightarrow 0(x \ln 9 \rightarrow 0)$

$\arcsin x^{3} \sim x^{3}$, as $x \rightarrow 0(x \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 8}{x}-\lim _{x \rightarrow 0} \frac{x \ln 9}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{x^{3}}{x}}=\frac{\lim _{x \rightarrow 0} \ln 8-\lim _{x \rightarrow 0} \ln 9}{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0} x^{2}}= \\
& =\frac{\ln 8-\ln 9}{1+0}=\ln \frac{8}{9}=\ln \frac{2^{3}}{3^{2}}
\end{aligned}
$$

## Problem Kuznetsov Limits 15-30