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Question Number 58530 by maxmathsup by imad last updated on 24/Apr/19

find lim_(x→0) ((1−cos(2x) cos(3x^3 ))/x^2 )

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\:{cos}\left(\mathrm{3}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} } \\ $$

Answered by tanmay last updated on 24/Apr/19

cos2x=1−(((2x)^2 )/(2!))+others term ignored because those terms contain x^r when r>2 cos(3x^3 )=1−(((3x^3 )^2 )/(2!))+others terms ignored lim_(x→0) ((1−{(1−((4x^2 )/2))(1−((9x^6 )/2))})/x^2 ) =lim_(x→0) ((1−(1−((9x^6 )/2)−2x^2 +9x^8 ))/x^2 ) =lim_(x→0) ((1−1+2x^2 +others terms ignored)/x^2 ) =2

$${cos}\mathrm{2}{x}=\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}!}+{others}\:{term}\:{ignored}\:{because} \\ $$$${those}\:{terms}\:{contain}\:{x}^{{r}} \:\:{when}\:{r}>\mathrm{2} \\ $$$${cos}\left(\mathrm{3}{x}^{\mathrm{3}} \right)=\mathrm{1}−\frac{\left(\mathrm{3}{x}^{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{2}!}+{others}\:{terms}\:{ignored} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left\{\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{9}{x}^{\mathrm{6}} }{\mathrm{2}}\right)\right\}}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{9}{x}^{\mathrm{6}} }{\mathrm{2}}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{9}{x}^{\mathrm{8}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} +{others}\:{terms}\:{ignored}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2} \\ $$

Commented by malwaan last updated on 25/Apr/19

method name? please sir !

$${method}\:{name}? \\ $$$${please}\:{sir}\:! \\ $$

Commented by tanmay last updated on 25/Apr/19

we know that cosθ=1−(θ^2 /(2!))+(θ^4 /(4!))−(θ^6 /(6!))+...∞ i have just put the value of cosθ in expansion form

$${we}\:{know}\:{that}\:{cos}\theta=\mathrm{1}−\frac{\theta^{\mathrm{2}} }{\mathrm{2}!}+\frac{\theta^{\mathrm{4}} }{\mathrm{4}!}−\frac{\theta^{\mathrm{6}} }{\mathrm{6}!}+...\infty \\ $$$${i}\:{have}\:{just}\:{put}\:{the}\:{value}\:{of}\:{cos}\theta\:{in}\:{expansion}\:{form} \\ $$

Commented by maxmathsup by imad last updated on 25/Apr/19

limited developpement or taylor series .

$${limited}\:{developpement}\:{or}\:{taylor}\:{series}\:. \\ $$

Commented by malwaan last updated on 27/Apr/19

thank you

$${thank}\:{you} \\ $$

Answered by kaivan.ahmadi last updated on 25/Apr/19

lim_(x→0) ((1−(1/2)(cos(2x+3x^3 )+cos(2x−3x^3 )))/x^2 )= lim_(x→0) (((1−cos(2x+3x^3 ))+(1−cos(2x−3x^3 )))/(2x^2 ))∼ lim_(x→0) (((2x+3x^3 )^2 +(2x−3x^3 )^2 )/(4x^2 ))= lim_(x→0) ((8x^2 +18x^6 )/(4x^2 ))=lim_(x→0) ((8+18x^4 )/4)=(8/4)=2

$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{3}} \right)+{cos}\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right)\right)}{{x}^{\mathrm{2}} }= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{3}} \right)\right)+\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right)\right)}{\mathrm{2}{x}^{\mathrm{2}} }\sim \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{3}} \right)^{\mathrm{2}} +\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{8}{x}^{\mathrm{2}} +\mathrm{18}{x}^{\mathrm{6}} }{\mathrm{4}{x}^{\mathrm{2}} }={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{8}+\mathrm{18}{x}^{\mathrm{4}} }{\mathrm{4}}=\frac{\mathrm{8}}{\mathrm{4}}=\mathrm{2} \\ $$

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