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Question Number 30282 by Tinkutara last updated on 19/Feb/18

$$Find{\underset{n\to \mathrm{\infty }}{\lim cos}}^n\left(\frac{2\pi }{n}\right)$$

Commented by prof Abdo imad last updated on 20/Feb/18

$$wehavecosx\sim 1-\frac{x^2}{2}forx\in v\left(0\right)\Rightarrow$$$$cos\left(\frac{2\pi }{n}\right)\sim 1-\frac{4{\pi }^2}{2n^2}forn\to \mathrm{\infty }and$$$${cos}^n\left(\frac{2\pi }{n}\right)\sim {(1-2\frac{{\pi }^2}{n^2})}^n\sim 1-\frac{2{\pi }^2}{n}because$$$${(1-u)}^n\sim 1-nuatv\left(0\right)forthat$$$${lim}_{n\to \mathrm{\infty }}{cos}^n\left(\frac{2\pi }{n}\right)=1.$$

Commented by Tinkutara last updated on 20/Feb/18

$$Whatisv\left(0\right)andnuatv\left(0\right)andhow$$$$1-\frac{4{\pi }^2}{2n^2}came?$$

Commented by abdo imad last updated on 20/Feb/18

$$x\in V\left(0\right)meansherexismorenearfrom0orx\to 0and$$$$1-\frac{4{\pi }^2}{2n^2}comefromlimiteddeveloppementbych.x=\frac{2\pi }{n}.$$

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