Question

$ \displaystyle\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cdot \cos \frac{2 \pi}{3 n}$ is

• A. $1$
• B. $\frac {\pi} {3}$
• C. $\frac {\pi}{6}$
• D. $\frac {2\pi}{3}$

Answer 1

The Correct Option is D

$\displaystyle \lim _{n \rightarrow \infty} n \cdot \sin \frac{2 \pi}{3 n} \cdot \cos \frac{2 \pi}{3 n}$
$=\displaystyle\lim _{n \rightarrow \infty} n\left\{\frac{\left(\sin \frac{2 \pi}{3 n}\right)}{\left(\frac{2 \pi}{3 n}\right)}\right\} \cdot \cos \frac{2 \pi}{3 n} \times \frac{2 \pi}{3 n}$
$=(1) \cdot \cos \left(0^{\circ}\right) \times \frac{2 \pi}{3}$
$\left\{\because \displaystyle\lim _{\theta \rightarrow \infty} \frac{\sin 1 / \theta}{1 / \theta}=1\right\}$
$=1 \cdot \frac{2 \pi}{3}=\frac{2 \pi}{3}$