<s>[INST] 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} $
 [/INST] Taking into account "I would like to not do a translation with the change of variable ...", let us use the Taylor series around $x=a$.
$$\cos(x)=\cos (a)-(x-a) \sin (a)-\frac{1}{2} (x-a)^2 \cos (a)+O\left((x-a)^3\right)$$
$$\sin(x)=sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)+O\left((x-a)^3\right)$$ So, using $a=\frac \pi 3$
$$\cos(x)=\frac{1}{2}-\frac{1}{2} \sqrt{3} \left(x-\frac{\pi }{3}\right)-\frac{1}{4}
   \left(x-\frac{\pi }{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$1-2\cos(x)=\sqrt{3} \left(x-\frac{\pi }{3}\right)+\frac{1}{2} \left(x-\frac{\pi
   }{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$\sin(3x)=-3 \left(x-\frac{\pi }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
$$\dfrac{1 - 2\cos (x)}{\sin (3x)}=\frac{\sqrt{3} \left(x-\frac{\pi }{3}\right)+\frac{1}{2} \left(x-\frac{\pi
   }{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right) } {-3 \left(x-\frac{\pi }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^3\right) }$$ $$\dfrac{1 - 2\cos (x)}{\sin (3x)}=-\frac{1}{\sqrt{3}}-\frac{1}{6} \left(x-\frac{\pi
   }{3}\right)+O\left(\left(x-\frac{\pi }{3}\right)^2\right)$$ which shows the limit and how it is approached.
</s>