## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\sin (\pi-3 x)}$
## Solution

Substitution:

$x=y+\frac{\pi}{3} \Rightarrow y=x-\frac{\pi}{3}$

$x \rightarrow \frac{\pi}{3} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{3}} \frac{1-2 \cos x}{\sin (\pi-3 x)}=\lim _{y \rightarrow 0} \frac{1-2 \cos \left(y+\frac{\pi}{3}\right)}{\sin \left(\pi-3\left(y+\frac{\pi}{3}\right)\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2\left(\cos y \cos \frac{\pi}{3}-\sin y \sin \frac{\pi}{3}\right)}{\sin (\pi-3 y-\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2\left(\cos y \cdot \frac{1}{2}-\sin y \cdot \frac{\sqrt{3}}{2}\right)}{\sin (\pi-3 y-\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{1-\cos y+\sqrt{3} \cdot \sin y}{\sin (-3 y)}= \\
& =\lim _{y \rightarrow 0} \frac{1-\cos y}{\sin (-3 y)}+\lim _{y \rightarrow 0} \frac{\sqrt{3} \cdot \sin y}{\sin (-3 y)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$1-\cos y \sim \frac{y^{2}}{2}$, as $y \rightarrow 0$

$\sin (-3 y) \sim-3 y$, as $y \rightarrow 0(-3 y \rightarrow 0)$

$\sin y \sim y$, as $y \rightarrow 0$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\frac{y^{2}}{2}}{-3 y}+\lim _{y \rightarrow 0} \frac{\sqrt{3} y}{-3 y}=\lim _{y \rightarrow 0} \frac{y}{-6}+\lim _{y \rightarrow 0} \frac{\sqrt{3}}{-3}= \\
& =\frac{0}{-6}+\frac{\sqrt{3}}{-3}=-\frac{\sqrt{3}}{3}
\end{aligned}
$$

Problem Kuznetsov Limits 16-9