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} $
 A: Method $\#1:$
Express  as $$F=2\lim_{x\to\frac\pi3}\dfrac{\cos\dfrac\pi3-\cos x}{\sin3x}$$
Set $\frac\pi3-x=y$ to find $$F=\lim_{y\to0}\dfrac{\cos\dfrac\pi3-\cos\left(\dfrac\pi3-y\right)}{\sin3y}$$
Now apply $\cos C-\cos D$ in the numerator
for the denominator, use $\sin3A=\sin A(3-4\sin^2A)$
Method $\#2:$
Express  as $$F=\sec A\cdot\lim_{x\to A}\dfrac{\cos A-\cos x}{\sin3x-\sin3A}$$
Method $\#2A:$  $$F=-\sec A\cdot\dfrac{\lim_{x\to A}\dfrac{\cos x-\cos A}{x-A}}{\lim_{x\to A}\dfrac{\sin3x-\sin3A}{x-A}}=-\sec A\cdot\dfrac{\dfrac{d(\cos x)}{dx}_{(\text{ at }x=A)}}{\dfrac{d(\sin3x)}{dx}_{(\text{ at }x=A)}}$$
Method $\#2B:$ Using Prosthaphaeresis Formulas $$F=\sec A\cdot\lim_{x\to A}\dfrac{2\sin\dfrac{x-A}2\sin\dfrac{x+A}2}{2\sin\dfrac{3(x-a)}2\cos\dfrac{3(x+A)}2}$$
Now use $\lim_{u\to0}\dfrac{\sin u}u=1$
A: $$\frac{1 - 2 \cos x}{\sin (3x)}=\frac{1- 2 \cos (x)}{-4\sin^3 x+3\sin x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4\sin^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{3-4+4\cos^2 x}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{4\cos^2 x-1}=\frac{1}{\sin x}\frac{1- 2 \cos (x)}{(2\cos x-1)(2\cos x+1)}=\frac{1}{\sin x}\frac{-1}{(2\cos x+1)}\to\frac{2}{\sqrt3}\frac{-1}{2}=\frac{-1}{\sqrt3}=-\frac{\sqrt3}{3}$$
A: 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.
A: If we use the translation as suggested in the OP, then we begin by letting $x=t+\pi/3$.  Proceeding, we find 
$$1-2\cos(x)=1-\cos(t)+\sqrt3 \sin(t)$$
and 
$$\sin(3x)=\sin(3t+\pi)=-\sin(3t)$$
Hence, we can write $\frac{1-2\cos(x)}{\sin(3x)}$ in terms of $t$ as 
$$\begin{align}
\frac{1-2\cos(x)}{\sin(3x)}&=-\frac{1-\cos(t)+\sqrt3 \sin(t)}{\sin(3t)}\\\\
&=-\frac13 \left(\frac{1-\cos(t)}{t}+\sqrt3 \frac{\sin(t)}{t}\right)\,\left(\frac{3t}{\sin(3t)}\right)
\end{align}$$
Finally, applying the "well-known" limits, $\lim_{t\to 0}\frac{\sin(t)}{t}=1$ and $\lim_{t\to 0}\frac{1-\cos(t)}{t}=0$ yields the coveted limit 
$$\begin{align}
\lim_{x\to \pi/3}\frac{1-2\cos(x)}{\sin(3x)}&=\lim_{t\to 0}\left(-\frac13 \left(\underbrace{\frac{1-\cos(t)}{t}}_{\to 0}+\sqrt3 \underbrace{\frac{\sin(t)}{t}}_{\to 1}\right)\,\underbrace{\left(\frac{3t}{\sin(3t)}\right)}_{\to 1}\right)\\\\
&=-\frac{\sqrt 3}3
\end{align}$$
A: Since we have a case of $$0/0$$ we use L'Hospital rule. $$\lim_{x\to \frac{\pi }{3} }  \frac{1 - 2 \cos (x)}{\sin (3x)}=\lim_{x\to \pi/3}\frac{2\sin(x)}{3\cos(3x)}=-\frac{\sqrt 3}3$$
A: by L'Hospital we get $$\lim_{x\to \pi/3}\frac{2\sin(x)}{3\cos(3x)}=$$
