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Question Number 79065 by Khyati last updated on 22/Jan/20

]. lim_(x→0)  (([sin(α+β)x+sin(α−β)x+sin2αx)/(cos2βx−cos2αx)).x

].limx0[sin(α+β)x+sin(αβ)x+sin2αxcos2βxcos2αx.x

Commented by jagoll last updated on 22/Jan/20

lim_(x→0) (([sin (α+β)x+sin (α−β)x+sin 2αx])/(cos 2βx−cos 2αx))× x?

limx0[sin(α+β)x+sin(αβ)x+sin2αx]cos2βxcos2αx×x?

Commented by mathmax by abdo last updated on 22/Jan/20

let consider  f(x)=((sin(α+β)x+sin(α−β)x+sin(2αx))/(cos(2βx)−cos(2αx)))×x   for x∼0 we have  f(x)∼(((α+β)x+(α−β)x+2αx)/(1−((4β^2 x^2 )/2)−(1−((4α^2 x^2 )/2))))×x ⇒f(x)∼((x^2 (α+β+α−β +2α))/(x^2 (2α^2 −2β^2 )))  ⇒f(x) ∼((4α)/(2α^2 −2β^2 )) ⇒lim_(x→0)  f(x) =((2α)/(α^2 −β^2 ))  (if α≠β)  if α=β   f is not defined!

letconsiderf(x)=sin(α+β)x+sin(αβ)x+sin(2αx)cos(2βx)cos(2αx)×xforx0wehavef(x)(α+β)x+(αβ)x+2αx14β2x22(14α2x22)×xf(x)x2(α+β+αβ+2α)x2(2α22β2)f(x)4α2α22β2limx0f(x)=2αα2β2(ifαβ)ifα=βfisnotdefined!

Commented by jagoll last updated on 23/Jan/20

lim_(x→0)  ((x[2sin αxcos βx+sin 2αx])/(−2sin (α+β)x sin (β−α)x)) =  lim_(x→0)  ((2xsin αx+xsin 2αx)/(−2sin (α+β)x.sin (β−α)x))=  ((4α)/((α+β)(α−β)))×(1/2)=((2α)/(α^2 −β^2 ))

limx0x[2sinαxcosβx+sin2αx]2sin(α+β)xsin(βα)x=limx02xsinαx+xsin2αx2sin(α+β)x.sin(βα)x=4α(α+β)(αβ)×12=2αα2β2

Commented by Khyati last updated on 23/Jan/20

but answer is ((2α)/(α^2 −β^2 ))

butansweris2αα2β2

Commented by jagoll last updated on 23/Jan/20

yes. i will correct my typo

yes.iwillcorrectmytypo

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