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

However, if rewrite this as

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

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

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Use L'Hôpital's Rule:

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

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

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

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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.

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.

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.

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