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I'm trying to solve this limit:

$$\lim_{x\to-\infty} xe^x$$

I'm trying to solve using the l'Hôpital rule. My question is can I use this rule in the last limt below?

$\lim_{x\to-\infty} xe^x=\lim_{x\to-\infty} \frac{x}{e^{-x}}$

Note the numerator tends to $-\infty$, while the denominator tends to $\infty$, can I use the l'Hôpital rule in this case?


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marked as duplicate by Hakim, Cookie, ᴡᴏʀᴅs, RecklessReckoner, Kirill Jul 20 '14 at 2:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes---------------- – Peter Franek Jul 19 '14 at 18:57
You can use it as long as the denominator is infinite regardless of the numerator (assuming the limit exists). – Brad Jul 19 '14 at 19:04
@brad which limit? numerator limit? or the limit itself? – user42912 Jul 19 '14 at 19:10
up vote 7 down vote accepted

Try to compute $\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{-x}}$. Numerator and denominator tend to $\infty$, so you can use l'Hospital rule.

$$\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{-x}}=\lim_{x \to -\infty}\frac{-1}{-e^{-x}}=\lim_{x \to -\infty}\frac{1}{e^{-x}}=0$$, so

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

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This is right but you should probably say why f(lim g(x)) = lim f(g(x)), with f(x) = -x in this case. – Frank Conry Jul 19 '14 at 22:23
@FrankConry I didn't understand why we have to use this fact in this solution. – user42912 Jul 21 '14 at 10:32

Hint: Your concern is that the top and bottom seem to go to different infinities. Could you multiply in such a way as to get rid of that difference?

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How is this a hint? (There's no constant factor in this problem.) And, in any event, constant factors do make a difference: $\lim_{x\to\infty} \frac{2x}{x} \ne \lim_{x\to\infty} \frac{x}{x}$ – apnorton Jul 19 '14 at 18:58
@anorton let me phrase it a bit differently – Semiclassical Jul 19 '14 at 18:59
@anorton: Constant factors make a difference in the numerical result, but not in whether or not the limit is finite/zero/infinity. – Semiclassical Jul 19 '14 at 19:10
@anorton You can pull arbitrary constant factors to the outside of a convergent limit. – Tim Seguine Jul 19 '14 at 22:32

You could also try a substitution:

$$y=-x\implies \left(x\to -\infty\implies y\to\infty\right)$$

and your limit becomes

$$\lim_{x\to-\infty} xe^{x}=\lim_{y\to\infty}-ye^{-y}=-\lim_{y\to\infty}\frac y{e^y}$$

and you don't need to worry about signs anymore.

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