Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 28

Given to solve ,
lim_(x->oo) x^3/e^(x^2)
upon x tends to oo we get x^3/e^(x^2) = oo/oo
so, by applying the L'Hopital Rule we get
as for the general equation it is as follows
lim_(x->a) f(x)/g(x) is = 0/0 or (+-oo)/(+-oo) then by using the L'Hopital Rule we get the solution with the below form.
lim_(x->a) (f'(x))/(g'(x))

so , now evaluating
lim_(x->oo) x^3/e^(x^2)
upon using the L'Hopital Rule
=lim_(x->oo) ((x^3)')/((e^(x^2))')
=lim_(x->oo) (3x^2)/(e^(x^2)(2x))
=>lim_(x->oo) (3x)/(e^(x^2)(2))
now on x-> oo we get (3x)/(e^(x^2)(2)) =oo/oo
so,again by applying the L'Hopital Rule we get
lim_(x->oo) (3x)/(e^(x^2)(2))
=lim_(x->oo) ((3x)')/((e^(x^2)(2))')
=lim_(x->oo) (3)/(e^(x^2)(2)(2x))
=lim_(x->oo) (3)/(e^(x^2) (4x))
now as x-> oo
3/(e^(x^2) (4x)) =3/(e^(oo^2) (4(oo))) =0
so
lim_(x->oo) (3)/((e^(x^2) (4x))) =0
now we can state that
lim_(x->oo) x^3/e^(x^2) =0