lim x-->0 [cos(sinx)-cosx] / x4



ans: 1/6

3 years ago

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Answers : (3)

                                        

Using L Hopitals rule


lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d/dx[cos(sin(x))-cos(x)]/x^4 if ([cos(sin(x))-cos(x)]=0 and x=4)


Again since the expression is yielding 0/0 appyling L hopitals rule, 


lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d2/dx2[cos(sin(x))-cos(x)]/x^4 at x=0


Again since the expression is yielding 0/0 appyling L hopitals rule, 


lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d3/dx3[cos(sin(x))-cos(x)]/x^4 at x=0


Again since the expression is yielding 0/0 appyling L hopitals rule, 


lim x tends to 0 [cos(sin(x))-cos(x)]/x^4 = d4/dx4[cos(sin(x))-cos(x)]/x^4 at x=0


Now Since the expression 4/24= 1/6 .


Hence 1/6 is the answer 


 



3 years ago
                                        

lim[cos(sinx)-cosx]/x^4


=lim[-sin(sinx)cosx+sinx]/(4x^3)


=lim[sin(sinx)sinx-cos(sinx)cos2x+cosx]/12x2


=lim[cosx-cos(sinx)cos2x]/12x2


=lim[1-cos(sinx)cosx]/12x2


=lim[sin(sinx)cos2x+cos(sinx)sinx]/24x


differentiate once more w.r.t. to get the answer


Dear Moderator: If I need to provide any more details, please contact me at <mn_chetan@yahoo.com

3 years ago
                                        

thanks a lot

3 years ago

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