limit of a trigonometric function $ \lim\limits_{x\to\pi/3} \frac{1 - 2\cos (x)}{\sin (3x)} $ compute the limit of
$$ \lim_{x\to \frac{\pi }{3} }  \frac{1 - 2 \cos (x)}{\sin (3x)} $$ 
I would like to not do a translation with the change  of variable $ t = x - \frac{\pi }{3} $

$$\frac{1 - 2 \cos x}{\sin (3x)}=\frac{1- 2 \cos (x)}{-4\sin^3 x+3\sin x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4\sin^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4+4\cos^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{4\cos^2 x-1}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{(2\cos x-1)(2\cos x+1)}=\frac{1}{\sin x}\frac{-1}{(2\cos x+1)}\to\frac{2}{\sqrt3}\frac{-1}{2}=\frac{-1}{\sqrt3}=-\frac{\sqrt3}{3}$$
