Explanation

(b)
$\lim_{x \rightarrow \frac{\pi}{6}}\frac{2\sin^{2}{x + \sin{x - 1}}}{2\sin^{2}{x - 3\sin{x + 1}}}$
$= \lim_{x \rightarrow \frac{\pi}{6}}\frac{4\sin x\cos x + \cos x}{4\sin x\cos x - 3\cos x}$
[by L’Hospital’s rule]
$= \lim_{x \rightarrow \frac{\pi}{6}}\frac{\cos x(4\sin{x + 1})}{\cos x(4\sin{x - 3})}$
$= \frac{4\sin{\frac{\pi}{6} + 1}}{4\sin{\frac{\pi}{6} - 3}} = - 3$