Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - \displaystyle \frac{\pi}{3}$)
$\displaystyle \lim_{u\to 0} \frac{1 - 2\cos(u + \frac{\pi}{3})}{\sin(u)} = \lim_{u\to 0} \frac{1 - \cos(u) + \sqrt{3}\sin(u)}{u} \approx \lim_{u\to 0} \frac{1 - 1 + \sqrt{3}u}{u} = \sqrt{3}$
But it bothers me using that sort of linear approximation, I want to get the result in a more formal way. I have tried using double angle properties 
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
$$\sin(2x) = 2\sin(x)\cos(x)$$
But I reach to a point of nowhere, I cannot come up with a way of simplifying expressions to get the results:
$\displaystyle\lim_{x\to\pi/3} 2\frac{3\sin^2(\frac{x}{2}) - \cos^2(\frac{x}{2})}{\sqrt{3}\sin(x) - \cos(x)}$
Is there a way of computing this limit without approximations and without L'Hopital?

HINT:
Using Prosthaphaeresis & Double angle Formula  as $\cos\dfrac\pi3=?$
$$2\cdot\dfrac{\cos\dfrac\pi3-\cos x}{\sin\left(x-\dfrac\pi3\right)}=2\cdot\dfrac{2\sin\dfrac{\pi+3x}6\sin\dfrac{3x-\pi}6}{2\sin\dfrac{3x-\pi}6\cos\dfrac{3x-\pi}6}$$
Now as $x\to\dfrac\pi3\iff3x\to\pi,3x\ne\pi\implies\sin\dfrac{3x-\pi}6\ne0$, so it can be cancelled safely.
