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Question Number 32407 by saru53424@gmail.com last updated on 24/Mar/18
limx→01−cos(1−cosx)x×x×x×x
Commented by abdo imad last updated on 24/Mar/18
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).....
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