Menu Close

lim-x-0-1-cos-1-cos-x-x-x-x-x-

Tinku Tara 0 Comments
Pin




Question Number 32407 by saru53424@gmail.com last updated on 24/Mar/18
lim_(x→0) ((1−cos (1−cos x))/(x×x×x×x))
limx→01−cos(1−cosx)x×x×x×x
Commented by abdo imad last updated on 24/Mar/18
if you mean lim_(x→0) ((1−cos(1−cosx))/x^4 ) in this case its better to use hospital theorem let put u(x)=1−cos(1−cosx) and v(x)=x^4 ⇒ v^′ (x)=4x^3 ⇒v^(′′) (x)=12x^2 ⇒v^((3)) (x)=24x ⇒v^((4)) (x)=24 u^′ (x) =sinx sin(1−cosx)⇒u^(′′) (x)=cosx sin(1−cosx) +sinx.sinx cos(1−cosx)=cosxsin(1−cosx) +sin^2 x cos(1−cosx) u^((3)) (x)=−sinxsin(1−cosx) +cosx.sinx cos(1−cosx) +2sinx cosx.cos(1−cosx) −sin^2 x sin(1−cosx) =−sinxsin(1−cosx) +(1/2)sin(2x)cos(1−cosx) +sin(2x).cos(1−cosx)−sin^2 xsin(1−cosx) u^((4)) (x)=−cosx sin(1−cosx)−sinx sinxcos(1−cosx) +cos(2x)cos(1−cosx)−(1/2)sin(2x)sinx sin(1−cosx) +2cos(2x)cos(1−cosx)−sin(2x)sinx sin(1−cosx) −2sinx cosx sin(1−cosx)−sin^3 cos(1−cosx) u^((4)) (0)=1 +2 =3 ⇒lim_(x→0) ((1−cos(1−cosx))/x^4 ) =(3/(24)) =(1/8) +(perhaps).....
ifyoumeanlimx→01−cos(1−cosx)x4inthiscaseitsbettertousehospitaltheoremletputu(x)=1−cos(1−cosx)andv(x)=x4⇒v′(x)=4x3⇒v″(x)=12x2⇒v(3)(x)=24x⇒v(4)(x)=24u′(x)=sinxsin(1−cosx)⇒u″(x)=cosxsin(1−cosx)+sinx.sinxcos(1−cosx)=cosxsin(1−cosx)+sin2xcos(1−cosx)u(3)(x)=−sinxsin(1−cosx)+cosx.sinxcos(1−cosx)+2sinxcosx.cos(1−cosx)−sin2xsin(1−cosx)=−sinxsin(1−cosx)+12sin(2x)cos(1−cosx)+sin(2x).cos(1−cosx)−sin2xsin(1−cosx)u(4)(x)=−cosxsin(1−cosx)−sinxsinxcos(1−cosx)+cos(2x)cos(1−cosx)−12sin(2x)sinxsin(1−cosx)+2cos(2x)cos(1−cosx)−sin(2x)sinxsin(1−cosx)−2sinxcosxsin(1−cosx)−sin3cos(1−cosx)u(4)(0)=1+2=3⇒limx→01−cos(1−cosx)x4=324=18+(perhaps)…..

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *