Question

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

• A. $-3$
• B. $0$
• C. $\frac{1}{3}$
• D. $\frac{-1}{3}$

Answer 1

The correct option is A $-3$

$L=\underset{x\to \frac{\pi }{6}}{lim}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x+1}$
This is $\frac{0}{0}$ form
$\Rightarrow L=\underset{x\to \frac{\pi }{6}}{lim}\frac{(2\sin x-1)(\sin x+1)}{(2\sin x-1)(\sin x-1)}\Rightarrow L=\underset{x\to \frac{\pi }{6}}{lim}\frac{\sin x+1}{\sin x-1}\Rightarrow L=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3$