Given a limit like: $$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$$
How did I solve it:
$$\begin{align}\lim_{x\rightarrow \frac{\pi }{3}}\frac {-2 (-\frac {1}{2}\sin x+\frac {\sqrt {3}}{2}\cos x)}{\sin 3x} &= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2 (-\sin\frac {\pi}{6}\sin x+\cos \frac {\pi}{6}\cos x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos (\frac{\pi }{6}+x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos (\frac{\pi }{6}+x)}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\cos \left[\frac{\pi }{2}-(\frac{\pi}{3}-x)\right]}{\sin 3x}\\&= \lim_{x\rightarrow \frac{\pi }{3}}\frac {-2\sin (\frac{\pi}{3}-x)}{\sin 3x}\\&= \lim_{t\rightarrow 0}\frac {-2\sin t}{\sin \left[3 (\frac{\pi}{3}-t)\right]}\\&= \lim_{t\rightarrow 0}\frac {-2\sin t}{-\sin 3t}\\&= \frac {2}{3}\end{align}$$
I don't know if this is correct but Wolfram Alpha points out it's $-\frac {2}{3}$ instead (L'hopital Rules). Can anyone show me if there's any error above? Or the limit really has two answers? Thanks in advance.
Your mistake: $$ \sin 3 (\frac{\pi}{3}-t)=\color{red}{-}\sin 3t $$ instead of $$ \sin\left[ 3 \left(\frac{\pi}{3}-t\right)\right]=\sin 3t. $$ (one may recall that $\cos (3\pi)=-1$)