Verified answer

lim x³ e^x = - ∞ * e^- ∞ ===> - ∞ * 0 not determined

x--> - ∞

[ x^3 / e^-x ]

Applying L'hopital rule & flip the e^x

[ 3x^2 / -e^-x ]

Applying L'hopital rule

[ 6x / e^-x ]

Applying L'hopital rule

[ 6 / -e^-x ]

lim [ 6 / -e^-x ] = [ 6 / - e^-(-∞) ] = [ 6 / - e^(∞) ] = [ 6 / -∞ ] = 0

x--> - ∞

x^3 e^x

but that is not in l'hopital form f(x)/g(x)

so

x^3 e^x = x^3 / e^ - x

no use l'hopital repeatedly, differentiating top and bottom each time until we can get the limit, becasue we can use l'hopitaL again and again, as long as each overall term is indeterminate inf/inf or 0/0

3x^2 / -e^-x

6x / e^-x

6 / -e^-x = -6 e^x

at minus infinity that has limit 0