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Question Number 138140 by liberty last updated on 10/Apr/21
lim_(x→0) ((cos (sin x)−cos x)/x^4 )=?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)−\mathrm{cos}\:{x}}{{x}^{\mathrm{4}} }=? \\ $$
Answered by EDWIN88 last updated on 10/Apr/21
recall cos A−cos B = 2sin (((B−A)/2))sin (((A+B)/2)) so lim_(x→0) ((2sin (((x−sin x)/2))sin (((x+sin x)/2)))/x^4 ) = lim_(x→0) ((2sin (((x−sin x)/2)))/((((x−sin x)/2))))((sin (((x+sin x)/2)))/((((x+sin x)/2)))).(((((x−sin x)/2))(((x+sin x)/2)))/x^4 ) = (1/2) lim_(x→0) (((x−sin x))/x^3 ) .lim_(x→0) ((x+sin x)/x) = (1/2) lim_(x→0) ((1−cos x)/(3x^2 )) .lim_(x→0) ((1+cos x)/1) = (1/2).2.(1/6) = (1/6).
$$\:{recall}\:\mathrm{cos}\:{A}−\mathrm{cos}\:{B}\:=\:\mathrm{2sin}\:\left(\frac{{B}−{A}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right) \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)}\frac{\mathrm{sin}\:\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)}.\frac{\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{sin}\:{x}\right)}{{x}^{\mathrm{3}} }\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}+\mathrm{sin}\:{x}}{{x}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{3}{x}^{\mathrm{2}} }\:.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}.\frac{\mathrm{1}}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}.\: \\ $$
Answered by mathmax by abdo last updated on 10/Apr/21
sinx∼x−(x^3 /6) ⇒cos(sinx)∼cos(x−(x^3 /6))∼1−(1/2)(x−(x^3 /6))^2 =1−(x^2 /2)(1−(x^2 /6))^2 =1−(x^2 /2)(1−(x^2 /3)+(x^4 /(36)))=1−(x^2 /2)+(x^4 /6)−(x^6 /(72)) cosx ∼1−(x^2 /2)+(x^4 /(4!)) ⇒f(x)∼((1−(x^2 /2)+(x^4 /6)−(x^6 /(72))−1+(x^2 /2)−(x^4 /(4!)))/x^4 ) ⇒ f(x)∼((1/6)−(1/(4.3.2)))−(x^2 /(72)) ⇒lim_(x→0) f(x)=(1/6)−(1/(24)) =((4−1)/(24)) =(3/(24))=(1/8)
$$\mathrm{sinx}\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{cos}\left(\mathrm{sinx}\right)\sim\mathrm{cos}\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)\sim\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{36}}\right)=\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{6}}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{72}} \\ $$$$\mathrm{cosx}\:\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{6}}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{72}}−\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}}{\mathrm{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\left(\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{4}.\mathrm{3}.\mathrm{2}}\right)−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{72}}\:\:\:\:\:\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{24}}\:=\frac{\mathrm{4}−\mathrm{1}}{\mathrm{24}}\:=\frac{\mathrm{3}}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by malwan last updated on 11/Apr/21
lim_(x→0) (([1− (((sinx)^2 )/(2!)) +...]−[1− (x^2 /(2!)) + (x^4 /(4!)) −...])/x^4 ) =lim_(x→0) (([1− (1/2)(x− (x^3 /(3!)) +...)^2 ]−[1−(x^2 /2)+(x^4 /(24))−..])/x^4 ) = lim_(x→0) (((1−(1/2)(x^2 −2×(x^4 /6)+..)−(1−(x^2 /2)+(x^4 /(24))−..)))/x^4 ) =lim_(x→0) (((1− (x^2 /2) + (x^4 /6) −...)−(1− (x^2 /2) + (x^4 /(24)) −..))/x^4 ) = lim_(x→0) ((x^4 ((1/6) − (1/(24)))±...)/x^4 ) = ((4−1)/(24)) = (3/(24)) = (1/8)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left[\mathrm{1}−\:\frac{\left({sinx}\right)^{\mathrm{2}} }{\mathrm{2}!}\:+…\right]−\left[\mathrm{1}−\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}\:+\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}\:−…\right]}{{x}^{\mathrm{4}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left[\mathrm{1}−\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+…\right)^{\mathrm{2}} \right]−\left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}−..\right]}{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{2}×\frac{{x}^{\mathrm{4}} }{\mathrm{6}}+..\right)−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}−..\right)\right)}{{x}^{\mathrm{4}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\left(\mathrm{1}−\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{x}^{\mathrm{4}} }{\mathrm{6}}\:−…\right)−\left(\mathrm{1}−\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{x}^{\mathrm{4}} }{\mathrm{24}}\:−..\right)}{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{6}}\:−\:\frac{\mathrm{1}}{\mathrm{24}}\right)\pm…}{{x}^{\mathrm{4}} } \\ $$$$=\:\frac{\mathrm{4}−\mathrm{1}}{\mathrm{24}}\:=\:\frac{\mathrm{3}}{\mathrm{24}}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by malwan last updated on 11/Apr/21
another method lim_(x→0) ((cos(sinx)−cosx)/x^4 ) =lim_(x→0) ((−2sin(((sinx+x)/2))sin(((sinx−x)/2)))/x^4 ) = lim_(x→0) ((+2(((x−sinx)/2))(((sinx+x)/2)))/x^4 ) = lim_(x→0) (1/2) (((x−(x− (x^3 /(3!)) +...))/x^3 ))(((sinx+x)/x)) = (1/2)×(1/6)×2 =^(?) (1/6) What is wrong ????
$${another}\:{method} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\:\frac{{cos}\left({sinx}\right)−{cosx}}{{x}^{\mathrm{4}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{−\mathrm{2}{sin}\left(\frac{{sinx}+{x}}{\mathrm{2}}\right){sin}\left(\frac{{sinx}−{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{+\mathrm{2}\left(\frac{{x}−{sinx}}{\mathrm{2}}\right)\left(\frac{{sinx}+{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\frac{{x}−\left({x}−\:\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+…\right)}{{x}^{\mathrm{3}} }\right)\left(\frac{{sinx}+{x}}{{x}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${What}\:{is}\:{wrong}\:???? \\ $$

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