Limit of trigonometric function $lim_{xtopi/3} frac{1 - 2cos(x)}{sin(x - frac{pi}{3})}$

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?

Answer

HINT:

Use $1-\cos(u)=2\sin^2(u/2)$ along with

$$\lim_{u\to 0}\frac{\sin(u)}{u}=1$$

and

$$\lim_{u\to 0}\frac{\sin^2(u/2)}{u}=0$$

Note that in the OP, $1-2\cos(x)=1-\cos(u)\color{blue}{+}\sqrt 3 \sin(u)$