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lim-xat-0-cos-2-4x-x-2-lim-x-at-0-cos-3-6x-x-2-

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Question Number 64539 by mmkkmm000m last updated on 19/Jul/19
lim_(xat 0) [cos^2 (4x)]/x^2 −lim_(x at 0) [cos^3 (6x)]/x^2
$${lim}_{{xat}\:\mathrm{0}} \left[{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)\right]/{x}^{\mathrm{2}} \:\:−{lim}_{{x}\:{at}\:\mathrm{0}} \left[{cos}^{\mathrm{3}} \left(\mathrm{6}{x}\right)\right]/{x}^{\mathrm{2}} \\ $$
Commented by kaivan.ahmadi last updated on 19/Jul/19
lim_(x→0) (([cos^2 4x])/x^2 )−lim_(x→0) (([cos^3 6x])/x^2 )= lim_(x→0 ) (([cos^2 4x]−[cos^3 6x])/x^2 )=(([1^− ]−[1^− ])/0^+ )=((0−0)/0^+ )=(0/0^+ )=0
$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\left[{cos}^{\mathrm{2}} \mathrm{4}{x}\right]}{{x}^{\mathrm{2}} }−{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\left[{cos}^{\mathrm{3}} \mathrm{6}{x}\right]}{{x}^{\mathrm{2}} }= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}\:\:} \frac{\left[{cos}^{\mathrm{2}} \mathrm{4}{x}\right]−\left[{cos}^{\mathrm{3}} \mathrm{6}{x}\right]}{{x}^{\mathrm{2}} }=\frac{\left[\mathrm{1}^{−} \right]−\left[\mathrm{1}^{−} \right]}{\mathrm{0}^{+} }=\frac{\mathrm{0}−\mathrm{0}}{\mathrm{0}^{+} }=\frac{\mathrm{0}}{\mathrm{0}^{+} }=\mathrm{0} \\ $$
Commented by MJS last updated on 19/Jul/19
(0/0)≠0
$$\frac{\mathrm{0}}{\mathrm{0}}\neq\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
let A(x) =((cos^2 (4x)−cos^3 (6x))/x^2 ) lim_(x→0) A(x)? we have cosu ∼1−(u^2 /2) (u→0) and cos^2 x =((1+cos(2x))/2) ∼(1/2) +(1/2){1−((4x^2 )/2)} =1 −x^2 ⇒cos^2 (4x) ∼1−16x^2 cos^3 (6x) =cos(6x)cos^2 (6x) ∼(1−18x^2 )(1−36x^2 ) =1−36x^2 −18x^2 +18.36 x^4 ∼1−54x^2 ⇒ A(x) ∼ ((1−16x^2 −1+54x^2 )/x^2 ) =38 ⇒lim_(x→0) A(x)=38
$${let}\:{A}\left({x}\right)\:=\frac{{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)−{cos}^{\mathrm{3}} \left(\mathrm{6}{x}\right)}{{x}^{\mathrm{2}} }\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)? \\ $$$${we}\:{have}\:{cosu}\:\sim\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\:\left({u}\rightarrow\mathrm{0}\right)\:{and}\:{cos}^{\mathrm{2}} {x}\:=\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\sim\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right\}\:=\mathrm{1}\:−{x}^{\mathrm{2}} \:\Rightarrow{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)\:\sim\mathrm{1}−\mathrm{16}{x}^{\mathrm{2}} \\ $$$${cos}^{\mathrm{3}} \left(\mathrm{6}{x}\right)\:={cos}\left(\mathrm{6}{x}\right){cos}^{\mathrm{2}} \left(\mathrm{6}{x}\right)\:\sim\left(\mathrm{1}−\mathrm{18}{x}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{36}{x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\mathrm{36}{x}^{\mathrm{2}} −\mathrm{18}{x}^{\mathrm{2}} +\mathrm{18}.\mathrm{36}\:{x}^{\mathrm{4}} \:\sim\mathrm{1}−\mathrm{54}{x}^{\mathrm{2}} \:\Rightarrow \\ $$$${A}\left({x}\right)\:\sim\:\frac{\mathrm{1}−\mathrm{16}{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{54}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:=\mathrm{38}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{A}\left({x}\right)=\mathrm{38} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 19/Jul/19
(0/ε)=0
$$\frac{\mathrm{0}}{\varepsilon}=\mathrm{0} \\ $$
Commented by kaivan.ahmadi last updated on 19/Jul/19
sir [ ] means integer part
$${sir}\:\left[\:\right]\:{means}\:{integer}\:{part} \\ $$
Commented by mathmax by abdo last updated on 19/Jul/19
perhaps sir but your answer is not correct...
$${perhaps}\:{sir}\:{but}\:{your}\:{answer}\:{is}\:{not}\:{correct}… \\ $$
Commented by kaivan.ahmadi last updated on 19/Jul/19
why?
$${why}? \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 19/Jul/19
(0/0) is undefined
$$\frac{\mathrm{0}}{\mathrm{0}}\:{is}\:{undefined}\: \\ $$
Commented by MJS last updated on 19/Jul/19
lim_(x→0) [cos^2 4x]=0 lim_(x→0) [cos^3 6x]=0 I think that in this case the difference of the given limits is undefined
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\right]=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{cos}^{\mathrm{3}} \:\mathrm{6}{x}\right]=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{limits}\:\mathrm{is}\:\mathrm{undefined} \\ $$
Answered by MJS last updated on 19/Jul/19
lim_(x→0) ((cos^2 4x −cos^3 6x)/x^2 ) = =lim_(x→0) ((2−cos 18x +2cos 8x −3cos 6x)/(4x^2 )) =lim_(x→0) (((d^2 /dx^2 )[2−cos 18x +2cos 8x −3cos 6x])/((d^2 /dx^2 )[4x^2 ])) = =(1/2)lim_(x→0) (81cos 18x −32cos 8x+27cos 6x)= =38
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}\:−\mathrm{cos}^{\mathrm{3}} \:\mathrm{6}{x}}{{x}^{\mathrm{2}} }\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}−\mathrm{cos}\:\mathrm{18}{x}\:+\mathrm{2cos}\:\mathrm{8}{x}\:−\mathrm{3cos}\:\mathrm{6}{x}}{\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left[\mathrm{2}−\mathrm{cos}\:\mathrm{18}{x}\:+\mathrm{2cos}\:\mathrm{8}{x}\:−\mathrm{3cos}\:\mathrm{6}{x}\right]}{\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left[\mathrm{4}{x}^{\mathrm{2}} \right]}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{81cos}\:\mathrm{18}{x}\:−\mathrm{32cos}\:\mathrm{8}{x}+\mathrm{27cos}\:\mathrm{6}{x}\right)= \\ $$$$=\mathrm{38} \\ $$

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