Q. $\lim_{x \rightarrow \frac{\pi}{6}}\frac{\sin{2x}}{\sin x}$ is equal to
A) $\sqrt{3}$
B) $\frac{1}{\sqrt{3}}$
C) 2
D) $\frac{1}{2}$
Explanation
(a)
$\lim_{x \rightarrow \frac{\pi}{6}}\frac{\sin{2x}}{\sin x} = \lim_{x \rightarrow \frac{\pi}{6}}\frac{2\sin x\cos x}{\sin x}$
$= 2\lim_{x \rightarrow \frac{\pi}{6}}{\cos x} = \sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.