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I have the function $f(x) = \dfrac{x^3}{e^x}$ and I'm trying to find its limit as x tends to negative infinity so that I can sketch the graph.

I can see just looking at the function that if I were to sub in any negative number for x it will give me a negative functional value, so I would expect to get a limit somewhere in the third quadrant. I have also plotted it in Graph and it tends to negative infinity.

However, when I try to evaluate the limit formally by repeated application of L'Hopital's rule I end up with positive infinity instead. I have included my process below, can someone please tell me where I'm going wrong?

$$\begin{align} &\lim_{x \to -\infty} \dfrac{x^3}{e^x} & [\text{evaluates to } \frac{-\infty}{0} \text{ so apply L'H}]\\ &\lim_{x \to -\infty} (3x^2 / e^x) &[\text{evaluates to} \frac{+\infty}{0} \text{so apply L'H}] \\ &\lim_{x \to -\infty} (6x / e^x) &[\text{evaluates to} \frac{-\infty}{0} \text{so apply L'H}] \\ &\lim_{x \to -\infty} (6 / e^x) &[\text{evaluates to} \frac{6}{\text{tiny positive number}}] \\ &= +\infty & \end{align}$$

Am I misapplying the rule, or making an algebraic error that I can't for the life of me pick up on?

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Yes, you are misapplying it. l'Hôpital is intended for indeterminate forms, where the numerator and denominator both approach zero or infinity, to give the most famous case. – J. M. Oct 23 '11 at 9:07
up vote 4 down vote accepted

From your formulation it's unclear whether x goes to infinity or minus infinity. First you say it goes to infinity, but the rest suggests you meant minus infinity.

  1. Assume x goes to infinity. Then your function definitely goes to 0 as x goes to infinity. Obviously exponent in the denominator goes to infinity faster than any polynomial.

You may indeed verify that by doing l'Lopital several times. Eventually nominator becomes finite, whereas the denominator remains exp(x).

  1. Assume x goes to -infinity. In this case you fraction tends to -infinity / +0. Hence it tends to -infinity, and there is no uncertainty here.

The l'Hopital rule applies only to uncertainties of kinds 0/0 and inf/inf. You may not use it elsewhere.

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I left out "negative" in my question, I've corrected it. Thanks for your answer, I can see that e^x evaluates as +0 not 0. – Katherine Rix Oct 23 '11 at 9:35

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