I need to evaluate following limit:
$$\lim_{x\to \frac{\pi}{3}} \frac{\sin(x-\frac{\pi}{3})}{1-2 \cos{x}}$$
Tried multiplying with the argument inside the sinus function but finished with this limit:
$$\lim_{x\to \frac{\pi}{3}} \frac{x - \frac{\pi}{3}}{1-2\cos{x}}$$
I know this limit is $\frac{1}{\sqrt{3}}$, but I have no steps which would show how I obtained this value
PS: l'Hospital is forbidden
Set $x-\tfrac{\pi}{3} = t$
$$\begin{align} \lim_{t\to 0}\frac{\sin t}{1-2\cos(t+\tfrac{\pi}{3})} &= \lim_{t\to 0}\frac{\sin t}{1-\cos t+ \sqrt 3\sin t} \\ &= \lim_{t\to 0}\frac{2\sin (\tfrac{t}{2}) \cos(\tfrac{t}{2})}{2 \sin^2 (\tfrac{t}{2})+ 2\sqrt 3\sin (\tfrac{t}{2}) \cos(\tfrac{t}{2})}\\ &= \lim_{t\to 0}\frac{ \cos(\tfrac{t}{2})}{2 \sin (\tfrac{t}{2})+ \sqrt 3 \cos(\tfrac{t}{2})} \end{align}$$
Now it should be clear :) The limit according to me is $\dfrac{1}{\sqrt{3}}$