==>lim(h->0) [(3x^2+6xh+3h^2-12x-12h-5/(x+h)^3)-3x^2+…
as you can see 3x^2 cancel and 12x cancel
==>lim(h->0)[6xh+3h^2-12h-(5/(x+h)^3)-5…
==>lim(h->0) 6xh/h +3h^2/h-12h/h-(5/(x+h)^3)/h+(5/x^3)h
the h's cancel
==>lim(h->0) 6x+3h-12-(5/(x+h)^3)/h+(5/x^3)h
as you can see, we're getting there. it's just those pesky fractions at the end. it just takes some expansion and simplification
==>first, multiply both numerator and denominator by 1/h, then multiply out the denominator:
==>lim(h->0) lim(h->0) 6x+3h-12-5/h (x^3+3x^2h+3xh^2+h^3)+5/h x^3
Then you combine the two fractions by getting a common denominator. multiple the first one by x^3/x^3 and the second one by (x^3+3x^2h+3xh^2+h^3)/(x^3+3x^2h+3xh^2+h… this gives you you
==>lim(h->0) 6x+3h-12-
(5x^3-15x^2h-15xh^2-5h^3+5h^3)/(x^6+3x^…
as you can see, in the numerator, 5x^3 cancels and an h can be pulled out of the numerator, which then cancels with the h in the denominator
==>lim(h->0) 6x+3h-12-
(h(15x^2-15xh-5h^2+5h^2))/(h(x^6+3x^5h+…
(15x^2-15xh-5h^2+5h^2)/(x^6+3x^5h+3x^4h…
from here, we can simply plug in h=0
==>lim(h->0) 6x+3(0)-12-(5x^3-15^2(0)-
15x(0)^2-5(0)^3+5^3)/(x^6+3x^5(0)+3x^4(…
=6x-12+15x^2/x^6
=6x-12+15x^-4
sorry if that seemed convoluted, but there ya go. also, on my computer the full equations are not shown on yahoo answers. i hope that's not the case on your computer, if it is, i hope it helped nontheless.