Question

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

• A. $3$
• B. $-3$
• C. $6$
• D. $0$

Answer 1

The correct option is D $0$

$L={lim}_{x\to \pi /6}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x-1}L=\frac{2{\sin }^2\left(\frac{\pi }{6}\right)+\sin \left(\frac{\pi }{6}\right)-1}{2{\sin }^2\left(\frac{\pi }{6}\right)-3\sin \left(\frac{\pi }{6}\right)-1}L=\frac{2.\frac{1}{4}+\frac{1}{2}-1}{2.\frac{1}{4}-3.\frac{1}{2}-1}=\frac{0}{-2}=0$

So option $D$ is correct