# Am I misapplying L'Hopital's rule?

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

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.