$\lim_{x\rightarrow \frac{\pi }{3}}\frac {\sin x-\sqrt {3}\cos x}{\sin 3x}$ has two different values? 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.


Can anyone show me if there's any error above? 

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 $\cos (3\pi)=-1$)
