Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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))

limx01cos(1cosx)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).....

ifyoumeanlimx01cos(1cosx)x4inthiscaseitsbettertousehospitaltheoremletputu(x)=1cos(1cosx)andv(x)=x4v(x)=4x3v(x)=12x2v(3)(x)=24xv(4)(x)=24u(x)=sinxsin(1cosx)u(x)=cosxsin(1cosx)+sinx.sinxcos(1cosx)=cosxsin(1cosx)+sin2xcos(1cosx)u(3)(x)=sinxsin(1cosx)+cosx.sinxcos(1cosx)+2sinxcosx.cos(1cosx)sin2xsin(1cosx)=sinxsin(1cosx)+12sin(2x)cos(1cosx)+sin(2x).cos(1cosx)sin2xsin(1cosx)u(4)(x)=cosxsin(1cosx)sinxsinxcos(1cosx)+cos(2x)cos(1cosx)12sin(2x)sinxsin(1cosx)+2cos(2x)cos(1cosx)sin(2x)sinxsin(1cosx)2sinxcosxsin(1cosx)sin3cos(1cosx)u(4)(0)=1+2=3limx01cos(1cosx)x4=324=18+(perhaps).....

Terms of Service

Privacy Policy

Contact: info@tinkutara.com