Checking the argument for calculation of the limit $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$ 
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$

In computing this limit, i have used following steps:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}(\sin(x)-\frac{1}{2})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}\sin(x)-\frac{\sqrt3}{2}\frac{1}{2}} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(\frac{\pi}{6})\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$            , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x-\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
from here by the use of $\lim_{\theta\to 0} \frac{\sin(\theta)}{\theta}$ and theorem for limits of composite functions, it can be proved that:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $2 \over \sqrt3$, which is incorrect.
Can anyone tell me, what i did wrong in this computation?
It is likely that the assumption in this step is wrong:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$            , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
If so, then can anyone tell me, what is wrong with this assumption?

We can not replace $\cos\dfrac\pi6$ with $\cos x$  just because $x\to\dfrac\pi6$
Better use Prosthaphaeresis Formula $$\sin x-\sin\dfrac\pi6=2\sin\dfrac{x-\dfrac\pi6}2\cos\dfrac{x+\dfrac\pi6}2$$
Then apply $\lim_{h\to0}\dfrac{\sin h}h=1$
