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. $$\lim_{x \to {\pi \over 3}} {\sin (x-{\pi \over 3})\over {1 - 2\cos x}}$$

Proceed as follows:
$$\lim_{x\to\frac{\pi}{3}} \frac{\sin(x-\pi/3)}{1-2\cos x}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos (t+\frac{\pi}{3})}$$
$$=\lim_{t\to0} \frac{\sin t}{1-2\cos t\cos \frac{\pi}{3}+2\sin t \sin\frac{\pi}{3}}$$
$$=\lim_{t\to0} \frac{\sin t}{(1-\cos t)+2\sin t \sin\frac{\pi}{3}}=\frac{1}{2\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$
Maybe a little bit of reasoning: $1-\cos t\approx t^2/2$ for small $t$, which is insignificant compared to $\sin t$. If unsure, maybe use $1-\cos t = 2\sin^2 \frac{t}{2}$ and put everything else into half angles to. In that case, you get
$$=\lim_{t\to0} \frac{2\sin \frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}+4\sin\frac{t}{2}\cos\frac{t}{2} \sin\frac{\pi}{3}}=\lim_{t\to0} \frac{2\cos\frac{t}{2}}{2\sin\frac{t}{2}+4\cos\frac{t}{2} \sin\frac{\pi}{3}}=\frac{2}{4\sin\frac{\pi}{3}}=\frac{1}{\sqrt3}$$
