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Mathematics Question on limits and derivatives

lim⁡n→∞nsin⁡2π3n⋅cos⁡2π3n \displaystyle\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cdot \cos \frac{2 \pi}{3 n} is

A

11

B

π3\frac {\pi} {3}

C

π6\frac {\pi}{6}

D

2π3\frac {2\pi}{3}

Answer

2π3\frac {2\pi}{3}

Explanation

Solution

lim⁡n→∞n⋅sin⁡2π3n⋅cos⁡2π3n\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)⋅cos⁡(0∘)×2π3=(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⋅2π3=2π3=1 \cdot \frac{2 \pi}{3}=\frac{2 \pi}{3}