i'm trying to do this problem but somehow i get it...?

Question

lim x^4e^3x
x-> - infinity
x^4(e^3x) = infinity * o
(e^3x) / (1/ (x^4)) = 0 / infinity
but l'hospital's is supposed to be 0/0...so ?

sry i think i post the question wrong..i meant to say i didn't get it...

Answer 1

lim (x->-∞) x^4 e^(3x)
As you noted, x^4 -> ∞ but e^3x -> 0, so we need to rewrite it in 0/0 or ∞/∞ form.
The easiest way to do this is to write e^(3x) = 1/e^(-3x), giving us
lim (x->-∞) x^4 e^(3x)
= lim (x->-∞) x^4 / e^(-3x) which is in ∞/∞ form and therefore suitable: apply L'Hopital's rule to get
= lim (x->-∞) 4x^3 / (-3)e^(-3x), still in ∞/∞ form, apply it again:
= lim (x->-∞) 12x^2 / (9)e^(-3x), still in ∞/∞ form, apply it again:
= lim (x->-∞) 24x / (-27)e^(-3x), still in ∞/∞ form, apply it again:
= lim (x->-∞) 24 / (81)e^(-3x)
= lim (x->-∞) (8/27) e^(3x)
= 0.

Answer 2

Use L'Hospitals 4 times:
x^4/(e^-3x)
4x^3/(-3e^-3x)
12x^2/(9e^-3x)
24x/(-27e^-3x)
24/(81e^-3x)
=24/infinity=0