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.