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

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

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.

Note that $$\frac{\sin u}{u}\to 1$$ and $$\frac{1-\cos(u)}{u}=u\frac{1-cos u}{u^2}\to 0$$