## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\ln (e-x)-1}$
## Solution

We will use the substitution of equivalent infinitesimals:

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

$\ln \left(1+\left(-\frac{x}{e}\right)\right) \sim -\frac{x}{e}$, as $x \rightarrow 0 \left(-\frac{x}{e} \rightarrow 0\right)$

We obtain:

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

## Problem Kuznetsov Limits 12-26