Calculus - limit of a function: $\lim\limits_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$ How do you compute the following limit without using the l'Hopital rule? 
If you were allowed to use it, it becomes easy and the result is $\sqrt{3}\over 3$ but without it, I am not sure how to proceed. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$

$\frac {\sin x\cos \frac {\pi}{3}- \cos x\sin\frac {\pi}{3}}{1+2\cos x}\\
\frac {\frac 12\sin x- \frac {\sqrt 3}{2} \cos x}{1-2\cos x}\\
\frac {\sin x- \sqrt 3\cos x}{2-4\cos x}\\
\frac {\sin^2 x- 3\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {\sin^2 x + \cos^2 x- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {1- 4\cos^2 x}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {(1- 2\cos x)(1+2\cos x)}{(2-4\cos x)(\sin x + \sqrt 3 \cos x)}\\
\frac {(1+2\cos x)}{2(\sin x + \sqrt 3 \cos x)}$
And now we can plug $x = \frac {\pi}{3}$
$\frac {2}{2(\frac {\sqrt 3}{2} + \frac {\sqrt 3}{2})}$
$\frac {1}{\sqrt 3}$
