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?
 A: 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, $1-2\cos(x)=1-\cos(u)\color{blue}{+}\sqrt 3 \sin(u)$

A: 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.
A: Note that $$\frac{\sin u}{u}\to 1$$ and 
$$\frac{1-\cos(u)}{u}=u\frac{1-cos u}{u^2}\to 0$$
A: Note that $\cos (\pi/3) = 1/2$ and hence we can write $$\lim_{x \to \pi/3}\frac{1 - 2\cos x}{\sin (x - \pi/3)} = -2\lim_{x \to \pi/3}\frac{\cos x - \cos (\pi/3)}{x - \pi/3}\cdot\frac{x - \pi/3}{\sin (x - \pi/3)} = -2(-\sin \pi/3)\cdot 1 = \sqrt{3}$$ here we have used the fact that $(\cos x)' = -\sin x$ so that $$\lim_{x \to a}\frac{\cos x - \cos a}{x - a} = -\sin a$$
A: 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
$$\begin{align}
\lim_{u\to0} \frac{1-\cos^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}&=\lim_{u\to0} \frac{\sin^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}\\
&=\lim_{u\to0} \frac{\sin(u)}{1+\cos(u)} + \sqrt{3}\\
&=\sqrt{3}
\end{align}$$
