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?

This is actually much easier than it looks. By letting $x=u+\frac\pi3$, we get
$$
\lim_{x\to\pi/3} \frac{1-2\cos(x)}{\sin(x-\frac\pi3)} = \lim_{u\to0}\frac{1-\cos(u)+\sqrt{3}\sin(u)}{\sin(u)}
$$
Now, we write the limit on the right as
$$
\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \lim_{u\to0}\frac{\sqrt{3}\sin(u)}{\sin(u)}=\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \sqrt{3}
$$
Multiplying the remaining limit by $\frac{1+\cos(u)}{1+\cos(u)}$ gives
$$\begin{align}
\lim_{u\to0} \frac{1-\cos^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}&=\lim_{u\to0} \frac{\sin^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}\\
&=\lim_{u\to0} \frac{\sin(u)}{1+\cos(u)} + \sqrt{3}\\
&=\sqrt{3}
\end{align}$$
