Solve this fast........!!!!?

Lt x---->0 [cos(sinx) - cosx]/x^4

This limit is of the indeterminate form 0/0, so we
can apply L'Hopital's rule to find that the limit equals

lim(x->0)[(-sin(sin(x))*cos(x) + sin(x)) / (4*x^3)],

which is still of the indeterminate form 0/0, so apply
L'Hopital's rule again to get

lim(x->0)[(-cos(sin(x))*cos^2(x) + sin(sin(x))*sin(x) + cos(x))/(12*x^2)],

which is still of the indeterminate form 0/0,
so apply L'Hopital's rule yet again to get

lim(x->0)[(sin(sin(x))*cos^3(x) + 2*cos(sin(x))*cos(x)*sin(x) +

cos(sin(x))*cos(x)*sin(x) + sin(sin(x))*cos(x) - sin(x)) / (24*x)] =

lim(x->0)[(1/24)*(sin(x)/x)*

(3*cos(sin(x))*cos(x) - 1 + (sin(sin(x))/sin(x))*(cos(x) + cos^3(x)))] =

(1/24)*1*(3 - 1 + 1*2) = 4/24 = 1/6.

Edit: Note that in the last step I made use of the well-known
limit lim(x->0)(sin(x)/x) = 1, and the fact that

lim(x->0)(sin(sin(x)) / sin(x)) = 1.

To see how this is the case, let u = sin(x). Then u -> 0
as x -> 0, and so the limit becomes lim(u->0)(sin(u)/u) = 1.

1/6