lim _{n rightarrow infty} n sin frac{2 pi}{3 | vedclass

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Question

(B) ધારો કે $L = \lim _{n ightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.આપણે જાણીએ છીએ કે $\sin(2	heta) = 2 \sin 	heta \cos \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $L = \lim _{n ightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.આપણે જાણીએ છીએ કે $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,તેથી $\sin 	heta \cos 	heta = \frac{1}{2} \sin(2	heta)$.અહીં,$	heta = \frac{2 \pi}{3 n}$.તેથી,$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)$.પ્રમાણિત લક્ષ $\lim _{x ightarrow 0} \frac{\sin x}{x} = 1$ નો ઉપયોગ કરીને,પદાવલિને ફરીથી લખતા:$L = \lim _{n ightarrow \infty} \frac{n}{2} \cdot \left( \frac{4 \pi}{3 n} ight) \cdot \frac{\sin \left( \frac{4 \pi}{3 n} ight)}{\left( \frac{4 \pi}{3 n} ight)}$.જેમ $n ightarrow \infty$,તેમ $\frac{4 \pi}{3 n} ightarrow 0$,તેથી $\frac{\sin \left( \frac{4 \pi}{3 n} ight)}{\left( \frac{4 \pi}{3 n} ight)} ightarrow 1$.તેથી,$L = \lim _{n ightarrow \infty} \frac{n}{2} \cdot \frac{4 \pi}{3 n} \cdot 1 = \frac{4 \pi}{6} = \frac{2 \pi}{3}$.

$\lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$ ની કિંમત શોધો.

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