Question

Find the following limit:
$\underset{x\to \pi /6}{lim}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x+1}.$

Answer 1

$\underset{x\to \frac{\pi }{6}}{lim}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x+1}$

$=\underset{x\to \frac{\pi }{6}}{lim}\frac{2{\sin }^{2}x+2\sin x-\sin x-1}{2\sin x-2\sin x-\sin x+1}$

$=\underset{x\to \frac{\pi }{6}}{lim}\frac{2\sin x(\sin x+1)-1(\sin x+1)}{2\sin x(\sin x-1)-1(\sin x-1)}$

$=\underset{x\to \frac{\pi }{6}}{lim}\frac{(2\sin x-1)(\sin x+1)}{(2\sin x-1)(\sin x-1)}$

$=\frac{\sin \frac{\pi }{6}+1}{\sin \frac{\pi }{6}-1}=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{\frac{-1}{2}}=-3$.