# L'Hospital's Rule and indeterminate form $\frac{\infty}{-\infty}$

Suppose I have a limit of the form \begin{align*} \lim\limits_{x \to -\infty} \frac{x}{e^{x^2}}. \end{align*} As $$x \to -\infty$$, $$x \to -\infty$$ and $$e^{x^2} \to \infty$$. Now, if we were subtracting this limit (suppose, for example, we're evaluating some term in the integration by parts formula from $$\infty$$ to $$-\infty$$), it doesn't quite matter. We can move constants outside a limit, and can surely move them back in as well. $$-x \to \infty$$ as $$x \to -\infty$$, so that is our $$\frac{\infty}{\infty}$$ indeterminate form which allows us to apply L'Hospital's rule. But, what if this weren't the case? Is $$\frac{\infty}{-\infty}$$ an indeterminate form? I suppose I could multiply the limit by $$1 = \frac{-1}{-1}$$ and pull one of $$-1$$'s outside the limit to turn this limit into the form $$\frac{\infty}{\infty}$$, though this feels like cheating. I'm really concerned with whether this is valid.

I'd appreciate any insights on this.

• It is my understanding that L'Hopital applies to the infinite over infinite case without restriction. – Chris Leary Mar 19 at 22:12

Yes, $$\frac\infty{-\infty}$$ and $$\frac{-\infty}\infty$$ are indeterminate forms. And you can apply L'Hopital's Rule to $$\lim_{x\to-\infty}\frac x{e^{x^2}}$$. So, compute$$\lim_{x\to-\infty}\frac1{2xe^{x^2}}=0.$$So, $$\lim_{x\to-\infty}\frac x{e^{x^2}}=0$$.
Yes,You can really bring $$-1$$ outside so that it changes to $$\frac\infty{\infty}$$ form . It's not 'cheating'