We have
#f (x)=(x (1+acosx)-bsinx)/x^3#
Given #f (0)=1#
And as #f (x)# is continuous so
#lim f (x)=1#
#xrarr0#
Now applying L'hopitals rule we get by differentiating numerator and denominator separately
#limx→0 f (x)=(x×(0+a (-sinx))+(1+acosx)-bcosx)/(3x^2)#
#=>limx→0f (x)=(1-axsinx+(a-b)cosx)/(3x^2)#
Now as #xrarr0# denominator approaches zero hence numerator must also approach zero.
#:.1+a-b=0#
#=>b-a=1# ........(1)
Now again applying L'hospital rule we get
#limxrarr0f (x)=(-axcosx-asinx+(a-b)(-sinx))/(6x)#
=#(-axcosx-(2a-b)sinx)/(6x)#
Now using #limxrarro(sinx/x)=1# as #xrarr0# we have
#limxrarr0 f (x)=(-ax/xcosx-(2a-b)sinx/x)/6=-(3a-b)/6#
Now #-(3a-b)/6=1#
#=>3a-b=-6# .........(2)
From equation (1) and (2)
#a=-5/2#
#b=-3/2#
