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

Question

65392 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-08T22:08:19

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

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 · 2015-10-08T22:12:32

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

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]}

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