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Question Number 129731 by Adel last updated on 18/Jan/21

lim_(x→∞) cos (x/2)×cos (x/4).......cos (x/2^n )=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}cos}\:\frac{\mathrm{x}}{\mathrm{2}}×\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{4}}.......\mathrm{cos}\:\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{n}} }=? \\ $$

Answered by Dwaipayan Shikari last updated on 18/Jan/21

cos(x/2)=((sinx)/(2sin(x/2)))  Π_(n=1) ^∞ cos((x/2^n ))=lim_(n→∞) ((sin(x))/(2sin((x/2)))).((sin((x/2)))/(2sin((x/4)))).((sin((x/4)))/(2sin((x/8))))...((sin((x/2^(n−1) )))/(2sin((x/2^n ))))  =((sin(x))/(2^n sin((x/2^n ))))=((sinx)/(2^n .(x/2^n )))=((sinx)/x)           lim_(n→∞) sin((x/2^n ))∼(x/2^n )

$${cos}\frac{{x}}{\mathrm{2}}=\frac{{sinx}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}{cos}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{sin}\left({x}\right)}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}.\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{4}}\right)}.\frac{{sin}\left(\frac{{x}}{\mathrm{4}}\right)}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{8}}\right)}...\frac{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)} \\ $$$$=\frac{{sin}\left({x}\right)}{\mathrm{2}^{{n}} {sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}=\frac{{sinx}}{\mathrm{2}^{{n}} .\frac{{x}}{\mathrm{2}^{{n}} }}=\frac{{sinx}}{{x}}\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)\sim\frac{{x}}{\mathrm{2}^{{n}} } \\ $$

Commented by Adel last updated on 18/Jan/21

tanks bro

$$\mathrm{tanks}\:\mathrm{bro} \\ $$

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