KCET · Maths · Limits
\(\lim _{\mathrm{n} \rightarrow \infty} \mathrm{n} \sin \frac{2 \pi}{3 \mathrm{n}} \cdot \cos \frac{2 \pi}{3 \mathrm{n}}\) is
- A \(\frac{\pi}{6}\)
- B \(\frac{2 \pi}{3}\)
- C 1
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \lim _{n \rightarrow \infty} n \cdot \sin \frac{2 \pi}{3 n} \cdot \cos \frac{2 \pi}{3 n} \\ &=\lim _{n \rightarrow \infty} n \cdot\left\{\frac{\left(\sin \frac{2 \pi}{3 n}\right)}{\left(\frac{2 \pi}{3 n}\right)}\right\} \cdot \cos \frac{2 \pi}{3 n} \times \frac{2 \pi}{3 n} \\ &=(1) \cdot \cos \left(0^{\circ}\right) \times \frac{2 \pi}{3}\left\{\because \lim _{\theta \rightarrow \infty} \frac{\sin 1 / \theta}{1 / \theta}=1\right\} \\ &=1 \cdot \frac{2 \pi}{3}=\frac{2 \pi}{3} \end{aligned}\)
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