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

Answer 1

Scarlett Allison

#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.

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Answer 2

Samantha Adkison

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

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]}

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Answer 3

Axel Alvarez

The limit of (x^2)(e^x) as x goes to negative infinity is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.