## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{12^{x}-5^{-3 x}}{2 \arcsin x-x}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{12^{x}-5^{-3 x}}{2 \arcsin x-x}=\lim _{x \rightarrow 0} \frac{\left(12^{x}-1\right)-\left(125^{-x}-1\right)}{2 \arcsin x-x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 12}\right)^{x}-1\right)-\left(\left(e^{\ln 125}\right)^{-x}-1\right)}{2 \arcsin x-x}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 12}-1\right)-\left(e^{-x \ln 125}-1\right)}{2 \arcsin x-x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 12}-1\right)-\left(e^{-x \ln 125}-1\right)\right)}{\frac{1}{x}(2 \arcsin x-x)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 12}-1\right)-\left(e^{-x \ln 125}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}(2 \arcsin x-x)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 12}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{-x \ln 125}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{2 \arcsin x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 12}-1 \sim x \ln 12$, as $x \rightarrow 0(x \ln 12 \rightarrow 0)$

$e^{-x \ln 125}-1 \sim -x \ln 125$, as $x \rightarrow 0(-x \ln 125 \rightarrow 0)$

$\arcsin x \sim x$, as $x \rightarrow 0$

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 12}{x}-\lim _{x \rightarrow 0} \frac{-x \ln 125}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} \ln 12-\lim _{x \rightarrow 0}-\ln 125}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}= \\
& =\frac{\ln 12^{x \rightarrow \ln 125^{x \rightarrow 0}}}{2-1}=\ln \left(12 \cdot 5^{3}\right)
\end{aligned}
$$

## Problem Kuznetsov Limits 15-9