Question Number 37817 by prof Abdo imad last updated on 17/Jun/18
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:{x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)\:{with}\:{x} \\ $$$${from}\:{R}\:{and}\:{x}\neq\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 24/Jun/18
$${if}\:\:\mid{x}\mid<\mathrm{1}\:\:\:{lim}_{{n}\rightarrow+\infty} {x}^{{n}} =\mathrm{0}\:{chang}.{x}^{{n}} =\frac{\mathrm{1}}{{t}} \\ $$$${lim}_{{n}\rightarrow+\infty} {x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{1}}{{t}}\left(\mathrm{1}\:−{cos}\left(\pi{t}\right)\right) \\ $$$$={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi{t}}{\mathrm{2}}\right)}{{t}}\:=\mathrm{0}\:{because}\mathrm{0}\leqslant\:{sin}^{\mathrm{2}} \left(\frac{\pi{t}}{\mathrm{2}}\right)\:\leqslant\mathrm{1} \\ $$$$\:{if}\:{x}>\mathrm{1}\:{lim}_{{n}\rightarrow+\infty} {x}^{{n}} =+\infty\:{chang}.{x}^{{n}} =\frac{\mathrm{1}}{{t}} \\ $$$${lim}_{{n}\rightarrow+\infty} {x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)={lim}_{{t}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{{t}}\left(\mathrm{1}−{cos}\left(\pi{t}\right)\right) \\ $$$${but}\:\mathrm{1}−{cos}\left(\pi{t}\right)\:\sim\:\frac{\pi^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{2}}\:\left(\:{t}\:\rightarrow\mathrm{0}\right)\Rightarrow \\ $$$$\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:\sim\:\frac{\pi^{\mathrm{2}} {t}}{\mathrm{2}}\:\rightarrow\mathrm{0}\left({t}\rightarrow\mathrm{0}\right)\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)=\mathrm{0}. \\ $$
Answered by behi83417@gmail.com last updated on 17/Jun/18
$$\frac{\mathrm{1}}{{x}^{{n}} }={t},{x}\rightarrow+\infty\Rightarrow{t}\rightarrow\mathrm{0} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\pi{t}}{{t}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\left(\pi{t}\right)^{\mathrm{2}} }{\mathrm{2}}}{{t}}=\mathrm{0}\:.\:\blacksquare \\ $$
Commented by math khazana by abdo last updated on 18/Jun/18
$${sir}\:{Behi}\:{be}\:{carreful}\:{n}\rightarrow+\infty\:{not}\:{x}… \\ $$