Given x→0limx4cos(sinx)−cosx=f(0)
⇒x→0limx42sin(2sinx+x)sin(2x−sinx)=k1
⇒x→0lim(2sinx+x)(2x−sinx)x42sin(2sinx+x)sin(2x−sinx)×(2sinx+x)(2x−sinx)=k1
⇒x→0lim2(2xsinx+x)(2x3x−sinx)=k1,(∵x→0limxsinx=1)
⇒2(x→0lim(2xsinx+x))(x→0lim(2x3x−sinx))=k1
Apply L'Hospital rule for an indeterminate form of 00, we get,
⇒2×2(1+1)×21×61=k1
⇒k=6