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Calculus-I-Lim-x-0-1-cos-xcos-x-2-cos-x-4-cos-x-8-x-2-

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Question Number 144450 by mnjuly1970 last updated on 25/Jun/21
.........Calculus(I)......... Lim_( x → 0) ((1 −cos(xcos((x/2)).cos((x/4))cos((x/8))))/x^( 2) )=?
$$ \\ $$$$\:\:\:\:\:\:\:\:………\mathrm{C}{alculus}\left(\mathrm{I}\right)……… \\ $$$$\:\:\mathrm{Lim}_{\:\:{x}\:\rightarrow\:\mathrm{0}} \frac{\mathrm{1}\:−{cos}\left({xcos}\left(\frac{{x}}{\mathrm{2}}\right).{cos}\left(\frac{{x}}{\mathrm{4}}\right){cos}\left(\frac{{x}}{\mathrm{8}}\right)\right)}{{x}^{\:\mathrm{2}} }=? \\ $$
Answered by Dwaipayan Shikari last updated on 25/Jun/21
lim_(x→0) ((1−cos(xcos(x/2)cos(x/4)cos(x/8)))/x^2 )=y log(1−x^2 y)=log(cos(xcos((x/2))..cos((x/8)))) lim_(x→0) log(1+x)≈x lim_(x→0) −x^2 y≈−((x^2 cos^2 ((x/2))cos^2 ((x/4))cos^2 ((x/8)))/(2!)) y=(1/2)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\left({xcos}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{4}\right){cos}\left({x}/\mathrm{8}\right)\right)}{{x}^{\mathrm{2}} }={y} \\ $$$${log}\left(\mathrm{1}−{x}^{\mathrm{2}} {y}\right)={log}\left({cos}\left({xcos}\left(\frac{{x}}{\mathrm{2}}\right)..{cos}\left(\frac{{x}}{\mathrm{8}}\right)\right)\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{log}\left(\mathrm{1}+{x}\right)\approx{x} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:−{x}^{\mathrm{2}} {y}\approx−\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{4}}\right){cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{8}}\right)}{\mathrm{2}!} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 25/Jun/21
grateful...
$${grateful}… \\ $$
Answered by mathmax by abdo last updated on 25/Jun/21
L=lim_(x→0) ((1−cosx.cos((x/4)).cos((x/8)))/x^2 ) we have cosx∼1−(x^2 /2) and cos((x/4))∼1−(x^2 /(32)) cos((x/8))∼1−(x^2 /(128)) ⇒L∼((1−(1−(x^2 /2))(1−(x^2 /(32)))(1−(x^2 /(128))))/x^2 ) =((1−(1−(x^2 /(32))−(x^2 /2)+(x^4 /(64)))(1−(x^2 /(128))))/x^2 ) =((1−(1−((17)/(32))x^2 +(x^4 /(64)))(1−(x^2 /(128))))/x^2 ) =((1−(1−(x^2 /(128))−((17)/(32))x^2 +((17)/(32.128))x^4 +(x^4 /(64))−(x^6 /(64.128))))/x^2 ) =(1/(128))+((17)/(32)) +(...)x^2 +(...)x^4 ⇒ Lim_(x→0) =(1/(128))+((68)/(128))=((69)/(128))
$$\mathrm{L}=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−\mathrm{cosx}.\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{4}}\right).\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{8}}\right)}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{and}\:\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{4}}\right)\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{32}} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{8}}\right)\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{128}}\:\Rightarrow\mathrm{L}\sim\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{32}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{128}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{32}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{64}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{128}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{17}}{\mathrm{32}}\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{64}}\right)\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{128}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{128}}−\frac{\mathrm{17}}{\mathrm{32}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{32}.\mathrm{128}}\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{64}}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{64}.\mathrm{128}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}+\frac{\mathrm{17}}{\mathrm{32}}\:+\left(…\right)\mathrm{x}^{\mathrm{2}} \:+\left(…\right)\mathrm{x}^{\mathrm{4}} \:\Rightarrow \\ $$$$\mathrm{Lim}_{\mathrm{x}\rightarrow\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{128}}+\frac{\mathrm{68}}{\mathrm{128}}=\frac{\mathrm{69}}{\mathrm{128}} \\ $$
Answered by mnjuly1970 last updated on 26/Jun/21
solution: Lim_( x → 0) ((1−cos (x{ cos((x/2)) cos((x/4)) cos((x/8) )}=f(x)))/x^( 2) ) := ((Lim_(x→ 0) ( 1 )−cos (Lim_(x→0) x f(x)))/(Lim_(x→ 0) (x^( 2) ))) :=_(as x→ 0) ^(f (x)→ 0) Lim_(x→0) ((1−cos (x))/x^( 2) ) =Lim_(x→0) ((2 sin^( 2) ((x/2) ))/x^( 2) ) := (1/2) Lim_( x → 0) (((sin ((x/2)))/(x/2)) )^( 2) = 1 ....m.n.july.1970....
$$\:\:\:{solution}: \\ $$$$\:\:\:\:\mathrm{Lim}_{\:{x}\:\rightarrow\:\mathrm{0}} \:\frac{\mathrm{1}−{cos}\:\left({x}\left\{\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:{cos}\left(\frac{{x}}{\mathrm{4}}\right)\:{cos}\left(\frac{{x}}{\mathrm{8}}\:\right)\right\}={f}\left({x}\right)\right)}{{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\::=\:\frac{\mathrm{Lim}_{{x}\rightarrow\:\mathrm{0}} \left(\:\mathrm{1}\:\right)−{cos}\:\left(\mathrm{Lim}_{{x}\rightarrow\mathrm{0}} \:{x}\:{f}\left({x}\right)\right)}{\mathrm{Lim}_{{x}\rightarrow\:\mathrm{0}} \:\left({x}^{\:\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\::\underset{{as}\:{x}\rightarrow\:\mathrm{0}} {\overset{{f}\:\left({x}\right)\rightarrow\:\mathrm{0}} {=}}\:\mathrm{Lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\:\left({x}\right)}{{x}^{\:\mathrm{2}} }\:=\mathrm{Lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{2}\:{sin}^{\:\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\:\right)}{{x}^{\:\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Lim}_{\:{x}\:\rightarrow\:\mathrm{0}} \:\left(\frac{{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\frac{{x}}{\mathrm{2}}}\:\right)^{\:\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{m}.{n}.{july}.\mathrm{1970}…. \\ $$

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