What is the limit of #(x^2)(e^x)# as x goes to negative infinity?

2 Answers
Oct 8, 2015

#lim_(x rarr -oo) x^2e^x#

Attempting to evaluate this limit by simply "plugging in" #-oo# will cause the indeterminate form #oo * 0#

However, if rewrite this as

#lim_(x rarr -oo) x^2/e^-x#

We can apply L'Hôpital and go on our merry way

#lim_(x rarr -oo) x^2/e^-x = lim_(x rarr -oo)(2x)/(-e^-x) = lim_(x rarr -oo) 2/e^-x = 0#

Not sure how one would go around doing it without L'Hôpital though.

Oct 8, 2015

#=lim_{x to -infty} 2x e^{x} = 0#

Explanation:

Use L'Hôpital's Rule:

Since:
#e^{x} = 1/{e^{-x}}#

We can re-write this as:
#lim_{x to -infty}x^2e^{x} =lim_{x to -infty}{x^2}/{e^{-x}}#

by L'Hôpital's Rule:
#=lim_{x to -infty}{2x}/{-e^{-x}}#

apply L'Hôpital's Rule again:
#=lim_{x to -infty}{2}/{e^{-x}} = {2}/{infty} = 0#

We can see this by graphing #y=2x e^{x}# :

graph{x^2 e^x [-10, 10, -5, 5]}