A) \[\sqrt{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[2\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
\[\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{\sin 2x}{\sin x}=\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{2\sin x\cos x}{\sin x}\] \[=2\lim \cos x\] \[=2\frac{\sqrt{3}}{2}=\sqrt{3}\] Alternate Method: \[\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{\sin 2x}{\sin x}=\frac{\sin 2\frac{\pi }{6}}{\sin \frac{\pi }{6}}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}\]
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