lim x-->∞ (x^3)*(e^-x)?

Question

lim x-->∞ (x^3)*(e^-x)
(indeterminate of type ∞ * 0???)
use fg = f/(1/g) OR g/(1/f) me thinks.

e^(-x)/(1/x^3) = 0/0 ???

(x^3)/(1/e^-x) = ∞/0 ???

Should I continue with l'hospitals, or did I goof somewhere?

Answer 1

You can rewrite this as (x^3) / (e^ x). This gives you ∞/∞, so NOW you can use l'Hopitals. (Remember the rule only works if you end up with ∞/∞ or 0/0. So we always have to check before applying and reapplying it.)

So taking the derivatives of the top and bottom, you get
(3x^2) / (e^x) this gives you ∞/∞ again. So once again apply l'Hopital's to get 6x / e^x. We still get ∞/∞, so applying it one more time gives 6/e^x. This limit goes to zero.

Answer 2

My answer could be a million, because of the fact some thing over a infinity is 0, so a million+0 is a million, and a million to the infinity is a million. If l'scientific institution rule is important then you definately wrote out the issue incorrect.

Answer 3

x^3 * (e^-x) = x^3 / (e^x)

Then apply l'hopital's rule three times.

Answer 4

lim(x^3)*(e^-x) =
x→∞
lim(x^3)/(e^x) =
x→∞
lim(3x^2)/(e^x) =
x→∞
lim(6x)/(e^x) =
x→∞
lim(6)/(e^x) = 0
x→∞