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

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 $\dfrac{1}{\sqrt{3}}$

$$\lim_{x\to A}\dfrac{\sin(x-A)}{\sec A(\cos A-\cos x)}$$

$$=\cos A\lim_{x\to A}\dfrac{2\sin\dfrac{x-A}2\cos\dfrac{x-A}2}{2\sin\dfrac{x-A}2\sin\dfrac{x+A}2}=?$$

Use $\sin\dfrac{x-A}2\ne0$ as $x\to A,x-A\ne0$

\begin{align}\lim_{x\to\frac\pi3}\frac{\sin\left(x-\frac\pi3\right)}{1-2\cos x}&=\lim_{x\to\frac\pi3}\frac{\sin\left(x-\frac\pi3\right)}{1-2\cos\left( \left(x-\frac\pi3\right)+\frac\pi3\right)}\\&=\lim_{x\to\frac\pi3}\frac{\sin\left(x-\frac\pi3\right)}{1-\cos\left(x-\frac\pi3\right)+\sqrt3\sin\left(x-\frac\pi3\right)}\\&=\lim_{x\to0}\frac{\sin x}{1-\cos(x)+\sqrt3\sin(x)}\\&=\lim_{x\to0}\frac{\frac{\sin x}x}{-\frac{\cos(x)-1}x+\sqrt3\frac{\sin x}x}\\&=\frac{\lim_{x\to0}\frac{\sin x}x}{-\lim_{x\to0}\frac{\cos(x)-1}x+\sqrt3\lim_{x\to0}\frac{\sin x}x}\\&=\frac1{-0+\sqrt3}\\&=\frac1{\sqrt3}.\end{align}

$$ \frac {\sin(x-\frac{\pi}{3})} {1-2 \cos{x}} = \frac {\sin(x-\frac{\pi}{3})} {x-\frac{\pi}{3}} \frac {x-\frac{\pi}{3}} {1-2 \cos{x}}=\frac {\sin y} {y} \frac {{y}} {1-2 \cos{\left(y+\frac{\pi}{3}\right)}}=\frac {\sin y} {y} \frac {{y}} {1-\cos y+\sqrt{3}\sin y}\to \frac{1}{\sqrt{3}} $$