Question

$ \underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{\sin 2x}{\sin x} $ is equal to

• A. $ \sqrt{3} $
• B. $ \frac{1}{\sqrt{3}} $
• C. $ 2 $
• D. $ \frac{1}{2} $

Answer 1

Answer: (A)

$\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin 2 x}{\sin x}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin x \cos x}{\sin x}=2 \lim \cos x=2 \frac{\sqrt{3}}{2}=\sqrt{3}$
Alternate Method: $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin 2 x}{\sin x}=\frac{\sin 2 \frac{\pi}{6}}{\sin \frac{\pi}{6}}=\frac{\sqrt{3} / 2}{1 / 2}=\sqrt{3}$