Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 35

Given to solve,
lim_(x->oo) e^x/(x^4)
as x->oo then the e^x/(x^4) =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) (e^x)/(x^4)
= lim_(x->oo) ((e^x)')/((x^4)')
= lim_(x->oo) ((e^x))/((4x^3))
again ((e^x))/((4x^3)) is of the form oo/oo so , we can apply again L'Hopital Rule .
=lim_(x->oo) ((e^x)')/((4x^3)')
=lim_(x->oo) ((e^x))/(((4*3)x^2))
=lim_(x->oo) ((e^x))/((12x^2))
again ((e^x))/((12x^2)) is of the form oo/oo so , we can apply again L'Hopital Rule . =lim_(x->oo) ((e^x)')/((12x^2)')
=lim_(x->oo) ((e^x))/(((12*2)x))
= lim_(x->oo) ((e^x))/(((24)x))
again ((e^x))/(((24)x)) is of the form oo/oo so , we can apply again L'Hopital Rule .
= lim_(x->oo) ((e^x)')/(((24)x)')
=lim_(x->oo) ((e^x))/(24)
on plugging the valuex= oo , we get
=((e^(oo)))/(24)
=oo