Question

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

Answer 1

`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.

Answer 2

`=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]}