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Question Number 103749 by bobhans last updated on 17/Jul/20

$$\underset{n\to \mathrm{\infty }}{\lim }{2}^n\sin \left(\frac{\pi }{2^n}\right)=?$$

Answered by bramlex last updated on 17/Jul/20

$$setx=\frac{\pi }{2^n}\Rightarrow 2^n=\frac{\pi }{x};x\to 0$$$$\underset{x\to 0}{\lim }\frac{\pi }{x}.\sin x=\pi \underset{x\to 0}{\lim }\frac{\sin x}{x}=\pi .$$$$◻$$

Answered by Sobir last updated on 17/Jul/20

$$theansveris\pi$$

Answered by Dwaipayan Shikari last updated on 17/Jul/20

$$\underset{n\to \mathrm{\infty }}{\lim }\pi \frac{sin\frac{\pi }{2^n}}{\frac{\pi }{2^n}}=\pi$$

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