Question

Find the value of limit $\underset{x\to \frac{\pi }{6}}{lim}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x+1}=$.

• A. $0$
• B. $3$
• C. $-3$
• D. $1$

Answer 1

Answer: (C)

$-3$

${lim}_{x\to \frac{\pi }{6}}\frac{2{\sin }^{2}x+\sin x-1}{2{\sin }^{2}x-3\sin x+1}={lim}_{x\to \frac{\pi }{6}}\frac{2{\sin }^{2}x+2\sin x-\sin x-1}{2{\sin }^{2}x-2\sin x-\sin x+1}={lim}_{x\to \frac{\pi }{6}}\frac{2\sin x(\sin x+1)-1(\sin x+1)}{2\sin x(\sin x-1)-1(\sin x-1)}={lim}_{x\to \frac{\pi }{6}}\frac{(2\sin x-1)(\sin x+1)}{(2\sin x-1)(\sin x-1)}={lim}_{x\to \frac{\pi }{6}}\frac{\sin x+1}{\sin x-1}=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3$