# L'Hospital's rule (products)

## Main Question or Discussion Point

To deal with the indeterminate form $0⋅\pm \infty$, we write the product $f(x)g(x)$ as $\frac{f(x)}{1/g(x)}$ or $\frac{g(x)}{1/f(x)}$, before applying L'Hospital's rule to one of these forms.
However, on occasion, applying L'Hospital's rule to one of these forms gets us nowhere ($\lim_{x \rightarrow -\infty} x e^x$ for instance), despite working for the other (equivalent) quotient.
Is there a way to know beforehand whether $f(x)g(x)$ should be expressed as $\frac{f(x)}{1/g(x)}$ or $\frac{g(x)}{1/f(x)}$? Or is it always based on trial and error?

RUber
Homework Helper
I think there is some intuition needed. Usually, any function that can be reduced away should be in a position to allow that to happen, as in your example x exp(x).
By leaving the polynomial part on top, you can be sure that application of L'Hopital's rule will eventually reduce that away.
Other than that, I think guess and check until you've done enough...then educated guess and check.

Mark44
Mentor
To deal with the indeterminate form $0⋅\pm \infty$, we write the product $f(x)g(x)$ as $\frac{f(x)}{1/g(x)}$ or $\frac{g(x)}{1/f(x)}$, before applying L'Hospital's rule to one of these forms.
However, on occasion, applying L'Hospital's rule to one of these forms gets us nowhere ($\lim_{x \rightarrow -\infty} x e^x$ for instance), despite working for the other (equivalent) quotient.
You can write $x e^{x}$ as $\frac x {e^{-x}}$. In that form you have the form $[\frac{-\infty}{\infty}]$, so L'Hopital's Rule can be applied.
Is there a way to know beforehand whether $f(x)g(x)$ should be expressed as $\frac{f(x)}{1/g(x)}$ or $\frac{g(x)}{1/f(x)}$? Or is it always based on trial and error?