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Question Number 37817 by prof Abdo imad last updated on 17/Jun/18

calculate lim_(n→+∞) x^n (1−cos((π/x^n ))) with x from R and x≠0

$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:{x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)\:{with}\:{x} \\ $$$${from}\:{R}\:{and}\:{x}\neq\mathrm{0} \\ $$

Commented by prof Abdo imad last updated on 24/Jun/18

if ∣x∣<1 lim_(n→+∞) x^n =0 chang.x^n =(1/t) lim_(n→+∞) x^n (1−cos((π/x^n )))=lim_(t→+∞) (1/t)(1 −cos(πt)) =lim_(t→+∞) ((2sin^2 (((πt)/2)))/t) =0 because0≤ sin^2 (((πt)/2)) ≤1 if x>1 lim_(n→+∞) x^n =+∞ chang.x^n =(1/t) lim_(n→+∞) x^n (1−cos((π/x^n )))=lim_(t→0) (1/t)(1−cos(πt)) but 1−cos(πt) ∼ ((π^2 t^2 )/2) ( t →0)⇒ ((1−cos(πt))/t) ∼ ((π^2 t)/2) →0(t→0) so lim_(n→+∞) x^n (1−cos((π/x^n )))=0.

$${if}\:\:\mid{x}\mid<\mathrm{1}\:\:\:{lim}_{{n}\rightarrow+\infty} {x}^{{n}} =\mathrm{0}\:{chang}.{x}^{{n}} =\frac{\mathrm{1}}{{t}} \\ $$$${lim}_{{n}\rightarrow+\infty} {x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{1}}{{t}}\left(\mathrm{1}\:−{cos}\left(\pi{t}\right)\right) \\ $$$$={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi{t}}{\mathrm{2}}\right)}{{t}}\:=\mathrm{0}\:{because}\mathrm{0}\leqslant\:{sin}^{\mathrm{2}} \left(\frac{\pi{t}}{\mathrm{2}}\right)\:\leqslant\mathrm{1} \\ $$$$\:{if}\:{x}>\mathrm{1}\:{lim}_{{n}\rightarrow+\infty} {x}^{{n}} =+\infty\:{chang}.{x}^{{n}} =\frac{\mathrm{1}}{{t}} \\ $$$${lim}_{{n}\rightarrow+\infty} {x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)={lim}_{{t}\rightarrow\mathrm{0}} \frac{\mathrm{1}}{{t}}\left(\mathrm{1}−{cos}\left(\pi{t}\right)\right) \\ $$$${but}\:\mathrm{1}−{cos}\left(\pi{t}\right)\:\sim\:\frac{\pi^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{2}}\:\left(\:{t}\:\rightarrow\mathrm{0}\right)\Rightarrow \\ $$$$\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:\sim\:\frac{\pi^{\mathrm{2}} {t}}{\mathrm{2}}\:\rightarrow\mathrm{0}\left({t}\rightarrow\mathrm{0}\right)\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{x}^{{n}} \left(\mathrm{1}−{cos}\left(\frac{\pi}{{x}^{{n}} }\right)\right)=\mathrm{0}. \\ $$

Answered by behi83417@gmail.com last updated on 17/Jun/18

(1/x^n )=t,x→+∞⇒t→0 L=lim_(x→0) ((1−cosπt)/t)=lim_(x→0) ((((πt)^2 )/2)/t)=0 . ■

$$\frac{\mathrm{1}}{{x}^{{n}} }={t},{x}\rightarrow+\infty\Rightarrow{t}\rightarrow\mathrm{0} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\pi{t}}{{t}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\left(\pi{t}\right)^{\mathrm{2}} }{\mathrm{2}}}{{t}}=\mathrm{0}\:.\:\blacksquare \\ $$

Commented by math khazana by abdo last updated on 18/Jun/18

sir Behi be carreful n→+∞ not x...

$${sir}\:{Behi}\:{be}\:{carreful}\:{n}\rightarrow+\infty\:{not}\:{x}... \\ $$

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