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Question Number 226815 by CrispyXYZ last updated on 16/Dec/25
lim_(x→0) (lim_(n→∞) (cos (x/2) cos (x/2^2 ) ... cos (x/2^n ))) = ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{{n}} }\right)\right)\:=\:? \\ $$
Answered by AgniMath last updated on 16/Dec/25
sin2θ = 2sinθcosθ sin x = 2sin(x/2)cos(x/2) = 2.2sin(x/4)cos(x/4)cos(x/2) = 2^2 .2sin(x/8)cos(x/8)cos(x/4)cos(x/2) = 2^3 .sin(x/2^3 )cos(x/2^3 )cos(x/2^2 )cos(x/2) ............ = 2^n .sin(x/2^n )cos(x/2^n )cos(x/2^(n−1) )....cos(x/2) ∴ cos(x/2)cos(x/2^2 )...cos(x/2^n )=((sin x)/(2^n .sin(x/2^n ))) lim_(n → ∞) cos(x/2)cos(x/2^2 )...cos(x/2^n ) = lim_(n → ∞) ((sin x)/(2^n .sin(x/2^n ))) = lim_(n → ∞) ((sin x)/(2^n .((sin (x/2^n ))/(x/2^n ))×(x/2^n ))) = ((sin x)/x) lim_(x → 0) ((sin x)/x) = 1 (Ans)
$${sin}\mathrm{2}\theta\:=\:\mathrm{2}{sin}\theta{cos}\theta \\ $$$${sin}\:{x}\:=\:\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}} \\ $$$$=\:\mathrm{2}.\mathrm{2}{sin}\frac{{x}}{\mathrm{4}}{cos}\frac{{x}}{\mathrm{4}}{cos}\frac{{x}}{\mathrm{2}} \\ $$$$=\:\mathrm{2}^{\mathrm{2}} .\mathrm{2}{sin}\frac{{x}}{\mathrm{8}}{cos}\frac{{x}}{\mathrm{8}}{cos}\frac{{x}}{\mathrm{4}}{cos}\frac{{x}}{\mathrm{2}} \\ $$$$=\:\mathrm{2}^{\mathrm{3}} .{sin}\frac{{x}}{\mathrm{2}^{\mathrm{3}} }{cos}\frac{{x}}{\mathrm{2}^{\mathrm{3}} }{cos}\frac{{x}}{\mathrm{2}^{\mathrm{2}} }{cos}\frac{{x}}{\mathrm{2}} \\ $$$$………… \\ $$$$=\:\mathrm{2}^{{n}} .{sin}\frac{{x}}{\mathrm{2}^{{n}} }{cos}\frac{{x}}{\mathrm{2}^{{n}} }{cos}\frac{{x}}{\mathrm{2}^{{n}−\mathrm{1}} }….{cos}\frac{{x}}{\mathrm{2}} \\ $$$$\therefore\:{cos}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}^{\mathrm{2}} }…{cos}\frac{{x}}{\mathrm{2}^{{n}} }=\frac{{sin}\:{x}}{\mathrm{2}^{{n}} .{sin}\frac{{x}}{\mathrm{2}^{{n}} }} \\ $$$$\:\underset{{n}\:\rightarrow\:\infty} {\mathrm{lim}}\:{cos}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}^{\mathrm{2}} }…{cos}\frac{{x}}{\mathrm{2}^{{n}} }\: \\ $$$$=\:\underset{{n}\:\rightarrow\:\infty} {\mathrm{lim}}\:\frac{{sin}\:{x}}{\mathrm{2}^{{n}} .{sin}\frac{{x}}{\mathrm{2}^{{n}} }} \\ $$$$=\:\underset{{n}\:\rightarrow\:\infty} {\mathrm{lim}}\:\frac{{sin}\:{x}}{\cancel{\mathrm{2}^{{n}} }.\frac{{sin}\:\frac{{x}}{\mathrm{2}^{{n}} }}{\frac{{x}}{\mathrm{2}^{{n}} }}×\frac{{x}}{\cancel{\mathrm{2}^{{n}} }}} \\ $$$$=\:\frac{{sin}\:{x}}{{x}} \\ $$$$ \\ $$$$\underset{{x}\:\rightarrow\:\mathrm{0}} {\mathrm{lim}}\frac{{sin}\:{x}}{{x}}\:=\:\mathrm{1}\:\left({Ans}\right) \\ $$$$\: \\ $$

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