Evaluate trig limit without l'hopital's rule

Question

I need help evaluating the following limit without using l'Hopital's rule:

$$\lim_{x\to \pi/3} \frac{\sin{x}-\sqrt{3}\cos{x}}{\sin{(x-\pi/3)}}$$

I have tried converting $\sin{(x-\pi/3)}$ to $-\cos{(\pi/6+x)}$ and seeing if I can cancel out terms. I have also tried rationalizing the expression by multiplying both numerator and denominator with $(\sin{x} + \sqrt{3}\cos{x})$. Neither seemed to lead me where I needed and I am out of ideas

Answer 1

$$\lim_{x\to \pi/3} \frac{\sin{x}-\sqrt{3}\cos{x}}{\sin{(x-\pi/3)}}$$

note that $$\sin \left(x-\frac{\pi}{3}\right) = \sin x \cos \frac{\pi}{3}-\cos x \sin \frac{\pi}{3}$$

$$\sin \left(x-\frac{\pi}{3}\right)= \frac{1}{2}\sin x-\frac{\sqrt{3}}{2} \cos x$$

substitute this in denominator,to get the answer