Question Number 30282 by Tinkutara last updated on 19/Feb/18
$${Find}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}cos}^{{n}} \:\left(\frac{\mathrm{2}\pi}{{n}}\right) \\ $$
Commented by prof Abdo imad last updated on 20/Feb/18
$${we}\:{have}\:\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:{for}\:{x}\in{v}\left(\mathrm{0}\right)\Rightarrow \\ $$$${cos}\left(\frac{\mathrm{2}\pi}{{n}}\right)\sim\:\mathrm{1}\:−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:{for}\:{n}\rightarrow\infty\:{and} \\ $$$${cos}^{{n}} \left(\frac{\mathrm{2}\pi}{{n}}\right)\sim\:\left(\mathrm{1}−\mathrm{2}\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)^{{n}} \:\sim\mathrm{1}−\:\frac{\mathrm{2}\pi^{\mathrm{2}} }{{n}}\:\:{because} \\ $$$$\left(\mathrm{1}−{u}\right)^{{n}} \sim\mathrm{1}−{nu}\:{atv}\left(\mathrm{0}\right)\:{for}\:{that} \\ $$$${lim}_{{n}\rightarrow\infty} {cos}^{{n}} \left(\frac{\mathrm{2}\pi}{{n}}\right)=\:\mathrm{1}\:\:\:. \\ $$
Commented by Tinkutara last updated on 20/Feb/18
$${What}\:{is}\:{v}\left(\mathrm{0}\right)\:{and}\:{nu}\:{atv}\left(\mathrm{0}\right)\:{and}\:{how} \\ $$$$\mathrm{1}−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:{came}? \\ $$
Commented by abdo imad last updated on 20/Feb/18
$${x}\in{V}\left(\mathrm{0}\right)\:{means}\:{here}\:{x}\:{is}\:{more}\:{near}\:{from}\mathrm{0}\:{or}\:{x}\rightarrow\mathrm{0}\:{and} \\ $$$$\mathrm{1}−\:\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\:{come}\:{from}\:{limited}\:{developpement}\:{by}\:{ch}.{x}=\frac{\mathrm{2}\pi}{{n}}. \\ $$