lim-n-nsin-2pi-1-n-2-n-N-

Question Number 36953 by rahul 19 last updated on 07/Jun/18

lim_(n→∞) nsin (2π(√(1+n^2 )) ) ,( n∈N).

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{nsin}\:\left(\mathrm{2}\pi\sqrt{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\right)\:,\left(\:\mathrm{n}\in\mathbb{N}\right). \\ $$

Commented by rahul 19 last updated on 07/Jun/18

Thank you Prof ��

Commented by prof Abdo imad last updated on 07/Jun/18

we have (√(1+n^2 ))=(√(n^2 (1+(1/n^2 ))))=n(1+(1/n^2 ))^(1/2) but we have (1+u)^α ∼ 1+α (u→0) so (1+(1/n^2 ))^(1/2) ∼ 1 +(1/(2n^2 )) (n→+∞) and 2π(√(1+n^2 ))∼ 2πn(1+(1/(2n^2 ))) =2πn +(π/n) ⇒ sin(2π(√(1+n^2 ))) ∼ sin((π/n))⇒ nsin(2π(√(1+n^2 ))) ∼ n sin((π/n)) but lim_(n→+∞) n sin((π/n)) =lim_(n→+∞) ((sin((π/n)))/(π/n)) .π =π×1 =π ⇒ lim_(n→+∞) n sin(2π(√(1+n^2 )))=π .

$${we}\:{have}\:\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }=\sqrt{{n}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}={n}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${but}\:{we}\:{have}\:\left(\mathrm{1}+{u}\right)^{\alpha} \:\sim\:\mathrm{1}+\alpha\:\left({u}\rightarrow\mathrm{0}\right)\:{so} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \sim\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\:\left({n}\rightarrow+\infty\right)\:{and} \\ $$$$\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\sim\:\mathrm{2}\pi{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)\:=\mathrm{2}\pi{n}\:+\frac{\pi}{{n}}\:\Rightarrow \\ $$$${sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)\:\:\sim\:{sin}\left(\frac{\pi}{{n}}\right)\Rightarrow \\ $$$${nsin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)\:\sim\:{n}\:{sin}\left(\frac{\pi}{{n}}\right)\:{but} \\ $$$${lim}_{{n}\rightarrow+\infty} {n}\:{sin}\left(\frac{\pi}{{n}}\right)\:={lim}_{{n}\rightarrow+\infty} \frac{{sin}\left(\frac{\pi}{{n}}\right)}{\frac{\pi}{{n}}}\:.\pi \\ $$$$=\pi×\mathrm{1}\:=\pi\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} {n}\:{sin}\left(\mathrm{2}\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)=\pi\:. \\ $$

Commented by math khazana by abdo last updated on 09/Jun/18

$${nevermind}\:{sir} \\ $$

Answered by ajfour last updated on 07/Jun/18

L=lim_(n→∞) nsin [2nπ(1+(1/(2n^2 ))+...)] =πlim_(n→∞) ((sin ((π/n)))/(((π/n)))) = 𝛑 .

$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\mathrm{sin}\:\left[\mathrm{2}{n}\pi\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }+…\right)\right] \\ $$$$\:\:\:\:=\pi\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\pi}{{n}}\right)}{\left(\frac{\pi}{{n}}\right)}\:=\:\boldsymbol{\pi}\:. \\ $$

Commented by Joel579 last updated on 07/Jun/18

Sir, pls explain how to get sin (2nπ(1 + (1/(2n^2 )) + ...))?

$$\mathrm{Sir},\:\mathrm{pls}\:\mathrm{explain}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{sin}\:\left(\mathrm{2}{n}\pi\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\:+\:…\right)\right)? \\ $$

Commented by rahul 19 last updated on 07/Jun/18

sin 2πn (√((1/n^2 )+1))= sin(2πn(1+(1/(2n^2 ))+...))

$$\mathrm{sin}\:\mathrm{2}\pi\mathrm{n}\:\sqrt{\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\mathrm{1}}=\:\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }+…\right)\right) \\ $$

Commented by rahul 19 last updated on 07/Jun/18

thank you sir��

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