# Evaluate trig limit without l'hopital's rule

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

• what have you tried?
– IITM
May 31, 2021 at 0:59
• 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 (sinx + v3cosx). Neither seemed to lead me where I needed and I am out of ideas. May 31, 2021 at 1:03
• It would be great if you can add this in the question,otherwise the question might attract downvotes,close votes
– IITM
May 31, 2021 at 1:08
• Thanks for the tip! I'll add it May 31, 2021 at 1:09
• Even faster. Let $x=y+\frac \pi 3$ and expand the numerator May 31, 2021 at 5:37

$$\lim_{x\to \pi/3} \frac{\sin{x}-\sqrt{3}\cos{x}}{\sin{(x-\pi/3)}}$$
note that $$\sin \left(x-\frac{\pi}{3}\right) = \sin x \cos \frac{\pi}{3}-\cos x \sin \frac{\pi}{3}$$
$$\sin \left(x-\frac{\pi}{3}\right)= \frac{1}{2}\sin x-\frac{\sqrt{3}}{2} \cos x$$