A) \[\frac{\pi }{6}\]
B) \[\frac{2\pi }{3}\]
C) \[1\]
D) \[\frac{\pi }{3}\]
Correct Answer: B
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,n.\,\sin \frac{2\pi }{3n}.\cos \frac{2\pi }{3n}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,n.\left\{ \frac{\left( \sin \frac{2\pi }{3n} \right)}{\left( \frac{2\pi }{3n} \right)} \right\}.\cos \frac{2\pi }{3n}\times \frac{2\pi }{3n}\] \[=(1).\cos ({{0}^{o}})\times \frac{2\pi }{3}\] \[\left\{ \because \,\,\underset{0\to \infty }{\mathop{\lim }}\,\frac{\sin 1/\theta }{1/\theta }=1 \right\}\] \[=1.\frac{2\pi }{3}=\frac{2\pi }{3}\]
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