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Question Number 73334 by mathmax by abdo last updated on 10/Nov/19
calculate lim_(n→+∞)    n^2 ( e^(sin((π/n^2 ))) −cos((π/n)))
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:{n}^{\mathrm{2}} \left(\:{e}^{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} −{cos}\left(\frac{\pi}{{n}}\right)\right) \\ $$
Answered by Smail last updated on 10/Nov/19
sin((π/n^2 ))∼_∞ (π/n^2 )  and e^(1/n) ∼_∞ 1+(1/n)  e^(sin(π/n^2 )) ∼_∞ 1+(π/n^2 )   and  cos((π/n))∼_∞ 1−(π^2 /(2n^2 ))  e^(sin(π/n^2 )) −cos(π/n)∼_∞ 1+(π/n^2 )−1+(π^2 /(2n^2 ))  ∼_∞ ((2π+π^2 )/(2n^2 ))  so  lim_(n→∞) n^2 (e^(sin(π/n^2 )) −cos(π/n))=lim_(n→∞) n^2 ×(((2π+π^2 )/(2n^2 )))  =((2π+π^2 )/2)
$${sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)\underset{\infty} {\sim}\frac{\pi}{{n}^{\mathrm{2}} }\:\:{and}\:{e}^{\mathrm{1}/{n}} \underset{\infty} {\sim}\mathrm{1}+\frac{\mathrm{1}}{{n}} \\ $$$${e}^{{sin}\left(\pi/{n}^{\mathrm{2}} \right)} \underset{\infty} {\sim}\mathrm{1}+\frac{\pi}{{n}^{\mathrm{2}} }\:\:\:{and}\:\:{cos}\left(\frac{\pi}{{n}}\right)\underset{\infty} {\sim}\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$${e}^{{sin}\left(\pi/{n}^{\mathrm{2}} \right)} −{cos}\left(\pi/{n}\right)\underset{\infty} {\sim}\mathrm{1}+\frac{\pi}{{n}^{\mathrm{2}} }−\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\underset{\infty} {\sim}\frac{\mathrm{2}\pi+\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$${so}\:\:\underset{{n}\rightarrow\infty} {{lim}n}^{\mathrm{2}} \left({e}^{{sin}\left(\pi/{n}^{\mathrm{2}} \right)} −{cos}\left(\pi/{n}\right)\right)=\underset{{n}\rightarrow\infty} {{lim}n}^{\mathrm{2}} ×\left(\frac{\mathrm{2}\pi+\pi^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{2}\pi+\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 10/Nov/19
are you now in usa sir smail..
$${are}\:{you}\:{now}\:{in}\:{usa}\:{sir}\:{smail}.. \\ $$
Commented by Smail last updated on 10/Nov/19
Yes, I currently live in the US
$${Yes},\:{I}\:{currently}\:{live}\:{in}\:{the}\:{US} \\ $$

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