Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 27

Given to solve,
lim_(x->oo) x^3/(e^(x/2))
as x->oo then the x^3/(e^(x/2))=oo/oo form
so upon applying the L 'Hopital rule we get the solution as follows,
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))
=lim_(x->oo) ((x^3)')/((e^(x/2))')
= lim_(x->oo) (3x^2)/((e^(x/2))(1/2))
again (3x^2)/((e^(x/2))(1/2)) is of the form oo/oo so , applying the L'Hopital Rule we get
= lim_(x->oo) (3x^2)/((e^(x/2))(1/2))
= lim_(x->oo) ((3x^2)')/(((e^(x/2))(1/2))')
=lim_(x->oo) ((6x))/(((e^(x/2))(1/2)(1/2)))
=lim_(x->oo) ((6x))/(((e^(x/2))(1/4)))
again ((6x))/(((e^(x/2))(1/4))) is of the form oo/oo so , applying the L'Hopital Rule we get
=lim_(x->oo) ((6x))/(((e^(x/2))(1/4)))
=lim_(x->oo) ((6x)')/(((e^(x/2))(1/4))')
=lim_(x->oo) ((6))/(((e^(x/2))(1/4)(1/2)))
=lim_(x->oo) ((6))/(((e^(x/2))(1/8)))
=lim_(x->oo) ((6*8))/(((e^(x/2))))
=lim_(x->oo) ((48))/(((e^(x/2))))
upon plugging the value of x= oo

we get
=lim_(x->oo) ((48))/(((e^((oo)/2))))
=lim_(x->oo) ((48))/(oo)
=0