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Question Number 57405 by Abdo msup. last updated on 03/Apr/19

calculate lim_(x→0) ((1−cosx.cos(2x)....cos(nx))/x^2 )  with n integr natural not 0.

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)....{cos}\left({nx}\right)}{{x}^{\mathrm{2}} } \\ $$$${with}\:{n}\:{integr}\:{natural}\:{not}\:\mathrm{0}. \\ $$

Answered by Smail last updated on 05/Apr/19

cosx=1−(x^2 /2)+o(x^2 )  cos2x=1−(((2x)^2 )/2)+o(x^2 )  cos3x=1−(((3x)^2 )/2)+o(x^2 )  cosx.cos(2x).cos(3x)...cos(nx)=(1−(x^2 /2))(1−(((2x)^2 )/2))...(1−(((nx)^2 )/2))+o(x^2 )  =1−((x^2 /2)+(((2x)^2 )/2)+(((3x)^2 )/2)+...(((nx)^2 )/2))+o(x^2 )  =1−(x^2 /2)(1+2^2 +3^2 +...+n^2 )+o(x^2 )  1−cosx.cos2x...cos(nx)∼_0 (x^2 /2)(((n(n+1)(2n+1))/6))  ((1−cosx.cos(2x)...cos(nx))/x^2 )∼_0 ((n(n+1)(2n+1))/(12))  lim_(x→0) ((1−cosx.cos(2x)...cos(nx))/x^2 )=((n(n+1)(2n+1))/(12))

$${cosx}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cos}\mathrm{2}{x}=\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cos}\mathrm{3}{x}=\mathrm{1}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cosx}.{cos}\left(\mathrm{2}{x}\right).{cos}\left(\mathrm{3}{x}\right)...{cos}\left({nx}\right)=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right)...\left(\mathrm{1}−\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+...\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}−{cosx}.{cos}\mathrm{2}{x}...{cos}\left({nx}\right)\underset{\mathrm{0}} {\sim}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right) \\ $$$$\frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)...{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }\underset{\mathrm{0}} {\sim}\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)...{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$

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