<s>[INST] 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?
 [/INST] You cannot partially pass to the limit. See this example:
$$
1 = \lim_{x \to 1} 1 = \lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1} = \lim_{x \to 1} 0 = 0.
$$
The error lies here:
$$
\lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1}
$$
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