## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(2-e^{\arcsin ^{2} \sqrt{x}}\right)^{\frac{3}{x}}
$$
## Solution

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

Using the substitution of equivalent infinitesimals:

$$
\ln \left(1+\left(1-e^{\arcsin ^{2} \sqrt{x}}\right)\right) \sim 1-e^{\arcsin ^{2} \sqrt{x}}, \text { as } x \rightarrow 0\left(1-e^{\arcsin ^{2} \sqrt{x}} \rightarrow 0\right)
$$

We get:

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

Using the substitution of equivalent infinitesimals:

$$
e^{\arcsin ^{2} \sqrt{x}}-1 \sim \arcsin ^{2} \sqrt{x}, \text { as } x \rightarrow 0\left(\arcsin ^{2} \sqrt{x} \rightarrow 0\right)
$$

We get:

$$
=\exp \left\{\lim _{x \rightarrow 0}-\frac{3}{x} \cdot \arcsin ^{2} \sqrt{x}\right\}=
$$

Using the substitution of equivalent infinitesimals:

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

We get:

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

## Problem Kuznetsov Limits 17-8