Q74 / 180 · KCET 2010 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'B'.

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