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L-lim-x-0-1-cos-xcos-2xcos-3x-sin-2-2x-

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Question Number 107747 by ajfour last updated on 12/Aug/20
L=lim_(x→0) (((1−cos xcos 2xcos 3x)/(sin^2 2x))) = ?
$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\right)\:=\:? \\ $$
Answered by bobhans last updated on 12/Aug/20
((Bobhans)/Π) L=lim_(x→0) (((1−(1−(x^2 /2))(1−2x^2 )(1−((9x^2 )/2)))/(4x^2 ))) L=lim_(x→0) (((1−(1−(5/2)x^2 +x^4 )(1−(9/2)x^2 ))/(4x^2 ))) L=lim_(x→0) (((1−(1−7x^2 +((45)/4)x^4 −(9/2)x^4 ))/(4x^2 )))=(7/4)
$$\:\:\:\:\:\:\:\:\frac{\mathcal{B}\mathrm{obhans}}{\Pi} \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{4x}^{\mathrm{2}} }\right) \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \right)\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{4x}^{\mathrm{2}} }\right) \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{7x}^{\mathrm{2}} +\frac{\mathrm{45}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} −\frac{\mathrm{9}}{\mathrm{2}}\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{4x}^{\mathrm{2}} }\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 12/Aug/20
Thanks Sir, most convinient way this is..
$${Thanks}\:{Sir},\:{most}\:{convinient}\:{way} \\ $$$${this}\:{is}.. \\ $$
Commented by malwaan last updated on 12/Aug/20
Sir Bobhans you can delet x^4 terms because we need only x^2 so L=lim_(x→0) (((1−(1−7x^2 +...))/(4x^2 )))=(7/4) thank you Sir
$${Sir}\:{Bobhans} \\ $$$${you}\:{can}\:{delet}\:{x}^{\mathrm{4}} \:{terms}\:{because} \\ $$$${we}\:{need}\:{only}\:{x}^{\mathrm{2}} \:{so} \\ $$$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\mathrm{7x}^{\mathrm{2}} +…\right)}{\mathrm{4x}^{\mathrm{2}} }\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$${thank}\:{you}\:{Sir} \\ $$
Answered by john santu last updated on 12/Aug/20
((⋇JS⋇)/Σ) in generrally lim_(x→0) ((1−cos px cos qx cos rx cos tx)/(sin^2 ax)) = (((1/2)(p^2 +q^2 +r^2 +t^2 ))/a^2 ) now lim_(x→0) ((1−cos x cos 2x cos 3x)/(sin^2 2x))= (((1/2)(1+4+9))/4) = (((1/2).14)/4) = (7/4)
$$\:\:\:\:\:\:\frac{\divideontimes\mathcal{JS}\divideontimes}{\Sigma} \\ $$$${in}\:{generrally}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{px}\:\mathrm{cos}\:{qx}\:\mathrm{cos}\:{rx}\:\mathrm{cos}\:{tx}}{\mathrm{sin}\:^{\mathrm{2}} {ax}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} } \\ $$$${now}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{4}+\mathrm{9}\right)}{\mathrm{4}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{14}}{\mathrm{4}}\:=\:\frac{\mathrm{7}}{\mathrm{4}} \\ $$
Commented by bemath last updated on 12/Aug/20
joss
$${joss} \\ $$
Commented by ajfour last updated on 12/Aug/20
Thanks for the general formula, can i please have a proof of it..
$${Thanks}\:{for}\:{the}\:{general}\:{formula}, \\ $$$${can}\:{i}\:{please}\:{have}\:{a}\:{proof}\:{of}\:{it}.. \\ $$
Commented by malwaan last updated on 12/Aug/20
Sir John santu is this right lim_(x→0) ((1−cos a_1 x cos a_2 x....cos a_n x)/(sin^2 bx)) = (((1/2)(a_1 ^2 + a_2 ^2 + ... + a_n ^2 ))/b^2 ) n ≥ 1 thank you so much
$${Sir}\:{John}\:{santu} \\ $$$${is}\:{this}\:{right} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{1}−{cos}\:{a}_{\mathrm{1}} {x}\:\:{cos}\:{a}_{\mathrm{2}} {x}….{cos}\:{a}_{{n}} {x}}{{sin}^{\mathrm{2}} \:{bx}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:…\:+\:{a}_{{n}} ^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} } \\ $$$${n}\:\geqslant\:\mathrm{1} \\ $$$${thank}\:{you}\:{so}\:{much} \\ $$
Answered by mathmax by abdo last updated on 12/Aug/20
let f(x) =((1−cosx cos(2x)cos(3x))/(sin^2 (2x))) we hsve cosx cos(2x)cos(3x) =(1/2)(cos(3x)+cos(x)cos(3x) =(1/2)cos^2 (3x) +(1/4)(cos(4x) +cos(2x)) =(1/4)(1+cos(6x)+cos(4x)+cos(2x)) ∼ (1/4)(1+1−(((6x)^2 )/2) +1−(((4x)^2 )/2) +1−(((2x)^2 )/2)) =(1/4)(4 −18x^2 −8x^2 −2x^2 ) =(1/4)(4−28x^2 )=1−7x^2 sin^2 (2x) =((1−cos(4x))/2) =(1/2)−(1/2)cos(4x)∼(1/2)−(1/2)(1−(((4x)^2 )/2)) =4x^2 and 1−cosx cos(2x)cos(3x) ∼ 7x^2 ⇒ f(x)∼((7x^2 )/(4x^2 )) =(7/4) ⇒lim_(x→0) f(x) =(7/4)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}−\mathrm{cosx}\:\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{cos}\left(\mathrm{3x}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)}\:\:\mathrm{we}\:\mathrm{hsve} \\ $$$$\mathrm{cosx}\:\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{cos}\left(\mathrm{3x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{3x}\right)+\mathrm{cos}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{3x}\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \left(\mathrm{3x}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}\left(\mathrm{4x}\right)\:+\mathrm{cos}\left(\mathrm{2x}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{6x}\right)+\mathrm{cos}\left(\mathrm{4x}\right)+\mathrm{cos}\left(\mathrm{2x}\right)\right)\:\sim \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{1}−\frac{\left(\mathrm{6x}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{1}−\frac{\left(\mathrm{4x}\right)^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{1}−\frac{\left(\mathrm{2x}\right)^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{4}\:−\mathrm{18x}^{\mathrm{2}} −\mathrm{8x}^{\mathrm{2}} −\mathrm{2x}^{\mathrm{2}} \right)\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{4}−\mathrm{28x}^{\mathrm{2}} \right)=\mathrm{1}−\mathrm{7x}^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\:=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{4x}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{4x}\right)\sim\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\left(\mathrm{4x}\right)^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\mathrm{4x}^{\mathrm{2}} \:\:\mathrm{and}\:\mathrm{1}−\mathrm{cosx}\:\mathrm{cos}\left(\mathrm{2x}\right)\mathrm{cos}\left(\mathrm{3x}\right)\:\sim\:\mathrm{7x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{7x}^{\mathrm{2}} }{\mathrm{4x}^{\mathrm{2}} }\:=\frac{\mathrm{7}}{\mathrm{4}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$$ \\ $$

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