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Question-191887

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Question Number 191887 by Rupesh123 last updated on 03/May/23
Answered by mehdee42 last updated on 03/May/23
hop⇒lim_(x→0) ((sinxcos2xco3x...cosnx+2sin2xcosxcos(3x)...cos(nx)+...+nsin(nx)cosxcos(2x)...cos(n−1)x)/(2x))= lim_(x→0) ((x(1+4+9+...+n^2 ))/x)=((n(n+1)(2n+1))/6)✓
$${hop}\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinxcos}\mathrm{2}{xco}\mathrm{3}{x}…{cosnx}+\mathrm{2}{sin}\mathrm{2}{xcosxcos}\left(\mathrm{3}{x}\right)…{cos}\left({nx}\right)+…+{nsin}\left({nx}\right){cosxcos}\left(\mathrm{2}{x}\right)…{cos}\left({n}−\mathrm{1}\right){x}}{\mathrm{2}{x}}= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\left(\mathrm{1}+\mathrm{4}+\mathrm{9}+…+{n}^{\mathrm{2}} \right)}{{x}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\checkmark \\ $$
Commented by Subhi last updated on 03/May/23
it must be 12 in the denominator
$${it}\:{must}\:{be}\:\mathrm{12}\:{in}\:{the}\:{denominator} \\ $$
Commented by mehdee42 last updated on 03/May/23
yes.i should have put 2x at the beginning of the denominator. thank you
$${yes}.{i}\:{should}\:{have}\:{put}\:\mathrm{2}{x}\:{at}\:{the}\:{beginning}\:{of}\:{the}\:{denominator}. \\ $$$${thank}\:{you}\: \\ $$
Commented by Subhi last updated on 03/May/23
welcome
$${welcome} \\ $$
Answered by Subhi last updated on 03/May/23
apply L Hopitals^′ law lim_(x→0) ((sin(x).(cos(2x)..........cos(nx)−cos(x).(d/dx)(cos(2x)........cos(nx)))/(2x)) with same way lim_(x→0) ((cos(x).cos(2x).....cos(nx)+sin(x)(d/dx)(cos(2x)....cos(nx))+sin(x).(d/dx)(....cos(nx))−cos(x).(d^2 /dx^2 )(cos(2x)....cos(nx)))/2) when substituting with x=0 terms including sin(x)=0 (d^2 /dx^2 )(cos(2x)....cos(nx)) (d/dx)=−2sin(2x).....cos(nx)+cos(2x).(d/dx)(cos(3x)....cos(nx)) (d^2 /dx^2 )=−4cos(2x).....cos(nx)−2sin(2x).(d/dx)(....cos(nx))−2sin(2x).(d/dx)(....cos(nx))+cos(2x).(d^2 /dx^2 )(cos(3x)....cos(nx)) −4+cos(2x).(d^2 /dx^2 )(cos(3x)...cos(nx)) by the same way lim_(x→0) ((1−cos(x)......cos(nx))/x^2 )=((1+2^2 +3^2 ......+n^2 )/2) =((n(n+1)(2n+1))/(2×6))
$${apply}\:{L}\:{Hopitals}^{'} {law} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}\left({x}\right).\left({cos}\left(\mathrm{2}{x}\right)……….{cos}\left({nx}\right)−{cos}\left({x}\right).\frac{{d}}{{dx}}\left({cos}\left(\mathrm{2}{x}\right)……..{cos}\left({nx}\right)\right)\right.}{\mathrm{2}{x}} \\ $$$${with}\:{same}\:{way} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{cos}\left({x}\right).{cos}\left(\mathrm{2}{x}\right)…..{cos}\left({nx}\right)+{sin}\left({x}\right)\frac{{d}}{{dx}}\left({cos}\left(\mathrm{2}{x}\right)….{cos}\left({nx}\right)\right)+{sin}\left({x}\right).\frac{{d}}{{dx}}\left(….{cos}\left({nx}\right)\right)−{cos}\left({x}\right).\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left({cos}\left(\mathrm{2}{x}\right)….{cos}\left({nx}\right)\right)}{\mathrm{2}} \\ $$$${when}\:{substituting}\:{with}\:{x}=\mathrm{0} \\ $$$${terms}\:{including}\:{sin}\left({x}\right)=\mathrm{0} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left({cos}\left(\mathrm{2}{x}\right)….{cos}\left({nx}\right)\right) \\ $$$$\frac{{d}}{{dx}}=−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)…..{cos}\left({nx}\right)+{cos}\left(\mathrm{2}{x}\right).\frac{{d}}{{dx}}\left({cos}\left(\mathrm{3}{x}\right)….{cos}\left({nx}\right)\right) \\ $$$$\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }=−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)…..{cos}\left({nx}\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right).\frac{{d}}{{dx}}\left(….{cos}\left({nx}\right)\right)−\mathrm{2}{sin}\left(\mathrm{2}{x}\right).\frac{{d}}{{dx}}\left(….{cos}\left({nx}\right)\right)+{cos}\left(\mathrm{2}{x}\right).\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left({cos}\left(\mathrm{3}{x}\right)….{cos}\left({nx}\right)\right) \\ $$$$−\mathrm{4}+{cos}\left(\mathrm{2}{x}\right).\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left({cos}\left(\mathrm{3}{x}\right)…{cos}\left({nx}\right)\right) \\ $$$${by}\:{the}\:{same}\:{way} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\left({x}\right)……{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }=\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} ……+{n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}×\mathrm{6}} \\ $$
Answered by qaz last updated on 04/May/23
lim_(x→0) ((1−cos x∙cos (2x)∙...∙cos (nx))/x^2 ) =−lim_(x→0) ((cos x∙cos (2x)∙...∙cos (nx)−1)/x^2 ) =−lim_(x→0) ((ln(1+cos x∙cos (2x)∙...∙cos (nx)−1))/x^2 ) =−lim_(x→0) ((lncos x+lncos (2x)+...+lncos (nx))/x^2 ) =−lim_(x→0) ((−(1/2)x^2 −(1/2)(2x)^2 −...−(1/2)(nx)^2 )/x^2 ) =((n(n+1)(2n+1))/(12))
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\centerdot\mathrm{cos}\:\left(\mathrm{2}{x}\right)\centerdot…\centerdot\mathrm{cos}\:\left({nx}\right)}{{x}^{\mathrm{2}} } \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{cos}\:{x}\centerdot\mathrm{cos}\:\left(\mathrm{2}{x}\right)\centerdot…\centerdot\mathrm{cos}\:\left({nx}\right)−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left(\mathrm{1}+\mathrm{cos}\:{x}\centerdot\mathrm{cos}\:\left(\mathrm{2}{x}\right)\centerdot…\centerdot\mathrm{cos}\:\left({nx}\right)−\mathrm{1}\right)}{{x}^{\mathrm{2}} } \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\mathrm{cos}\:{x}+{ln}\mathrm{cos}\:\left(\mathrm{2}{x}\right)+…+{ln}\mathrm{cos}\:\left({nx}\right)}{{x}^{\mathrm{2}} } \\ $$$$=−\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}\right)^{\mathrm{2}} −…−\frac{\mathrm{1}}{\mathrm{2}}\left({nx}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$

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