$$
\lim_{x \to \pi/3} \frac{\left(1-2\cos x\right)}{\sin\left(x-\frac{\pi}{3}\right)}
$$
Getting the answer $\sqrt{3}$ is easy with l'Hopital but I want to try to do this in a more formal way.
I got to
$$
\lim_{x \to \pi/3} \frac{\left(2\left(1-2\cos\left(x\right)\right)\right)}{\left(\sin\left(x\right)-\sqrt{3}\cos\left(x\right)\right)}
$$
but im not sure what to do with $1-2\cos x$. The problem is from N. Piskunov Differential and Integral Calculus.
Best Answer
Instead of working on the denominator, it is much more effective to first let $y = x - \frac{\pi}{3}$ so that $$L = \lim_{x \to \pi/3} \frac{1 - 2 \cos x}{\sin (x - \frac{\pi}{3})} = \lim_{y \to 0} \frac{1 - 2 \cos (y + \frac{\pi}{3})}{\sin y}.$$
Now use trigonometric identities on the numerator: $$1 - 2 \cos (y + \tfrac{\pi}{3}) = 1 - \cos y + \sqrt{3} \sin y$$ and now we see $$L = \sqrt{3} + \lim_{y \to 0} \frac{1 - \cos y}{\sin y}.$$ Finally, apply one more trigonometric identity: $$\frac{1 - \cos y}{\sin y} = \frac{1 - (1 - 2 \sin^2 \frac{y}{2})}{2 \sin \frac{y}{2} \cos \frac{y}{2}} = \tan \frac{y}{2},$$ so $$L = \sqrt{3}.$$
Alternatively, we could have also written $$\frac{1 - \cos y}{\sin y} = \frac{(1 - \cos y)(1 + \cos y)}{\sin y (1 + \cos y)} = \frac{1 - \cos^2 y}{\sin y (1 + \cos y)} = \frac{\sin y}{1 + \cos y}$$ and let $y \to 0$.