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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.

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    $\begingroup$ It is my understanding that L'Hopital applies to the infinite over infinite case without restriction. $\endgroup$ – Chris Leary Mar 19 at 22:12
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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, $\lim_{x\to-\infty}\frac x{e^{x^2}}=0$.

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Yes,You can really bring $-1$ outside so that it changes to $\frac\infty{\infty}$ form . It's not 'cheating'

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