## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{2^{-3 x}-1} \cdot \ln 2$
## Solution

We will use the substitution of equivalent infinitesimals:

$\arcsin 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

$e^{-3 x \ln 2}-1 \sim -3 x \ln 2$, as $x \rightarrow 0(-3 x \ln 2 \rightarrow 0)$

We get:

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