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 

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

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thanks a lot