## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0}\left(\frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 x+1}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 x+1}=\left(\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{\lim _{x \rightarrow 0} 2 x+1}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{2 \cdot 0+1}=\left(\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}\right)^{1}= \\
& =\lim _{x \rightarrow 0} \frac{\arcsin ^{2} x}{\arcsin ^{2} 4 x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

$\arcsin 4 x \sim 4 x$, as $x \rightarrow 0(4 x \rightarrow 0)$

We get:

$=\lim _{x \rightarrow 0} \frac{x^{2}}{(4 x)^{2}}=\lim _{x \rightarrow 0} \frac{1}{16}=\frac{1}{16}$

## Problem Kuznetsov Limits 18-30