The value of lim _{n rightarrow infty} n sin | vedclass

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(B) Let $L = \lim _{n ightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.We know that $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,so

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$\frac{2 \pi}{3}$

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(B) Let $L = \lim _{n ightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.We know that $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,so $\sin 	heta \cos 	heta = \frac{1}{2} \sin(2	heta)$.Here,$	heta = \frac{2 \pi}{3 n}$.Thus,$L = \lim _{n ightarrow \infty} n \cdot \frac{1}{2} \sin \left( 2 \cdot \frac{2 \pi}{3 n} ight) = \lim _{n ightarrow \infty} \frac{n}{2} \sin \left( \frac{4 \pi}{3 n} ight)$.Using the standard limit $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$,we rewrite the expression:$L = \lim _{n ightarrow \infty} \frac{n}{2} \cdot \left( \frac{4 \pi}{3 n} ight) \cdot \frac{\sin \left( \frac{4 \pi}{3 n} ight)}{\left( \frac{4 \pi}{3 n} ight)}$.As $n ightarrow \infty$,$\frac{4 \pi}{3 n} ightarrow 0$,so $\frac{\sin \left( \frac{4 \pi}{3 n} ight)}{\left( \frac{4 \pi}{3 n} ight)} ightarrow 1$.Therefore,$L = \lim _{n ightarrow \infty} \frac{n}{2} \cdot \frac{4 \pi}{3 n} \cdot 1 = \frac{4 \pi}{6} = \frac{2 \pi}{3}$.

The value of $\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$ is

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