## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin 3(x+\pi)}$
Solution

$\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin 3(x+\pi)}=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin (3 x+3 \pi)}=$

$=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{\sin (3 x+\pi)}=\lim _{x \rightarrow 0} \frac{\arcsin 2 x}{-\sin 3 x}=$

Using the substitution of equivalent infinitesimals:

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

$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

We get:

$=\lim _{x \rightarrow 0} \frac{2 x}{-3 x}=\lim _{x \rightarrow 0} \frac{2}{-3}=-\frac{2}{3}$

Problem Kuznetsov Limits 12-22