Is it possible / allowed to use L'Hôpitals rule for products? In our readings, we had L'Hôpitals rule and defined it like that:

$\lim_{x\rightarrow x_{0}}\frac{f'(x)}{g'(x)}$

Because we had it in our readings, we are allowed to use this to find limit of functions.
Now my question is, is it possible to use this rule for products? If yes, do you think I would be allowed to do it (since we have dicussed this rule in our reading...)?
Actually, a fraction is a product at the same time, isn't it?
Because we can also write:
$\lim_{x\rightarrow x_{0}}f'(x)*\frac{1}{g'(x)}$
and it would be called product, or am I totally wrong here?
How would you use L'Hôpitals rule for products? Possible at all?
I could imagine it has something to do with fraction and reciprocal.
But not sure about that.
 A: That's a fairly standard rewrite.
If you need to find something like
$$
\lim_{x\to a} f(x)g(x)
$$
where $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=\infty$, you can define $h(x)=\frac{1}{g(x)}$ and consider $\lim_{x\to a} \frac{f(x)}{h(x)}$. You should be able to use the form of l'Hôpital you know on that. 
Of course that requires $g$ to be well-behaved around $a$.
A: You can rewrite the product as a quotient, but you have to do this before you do the derivatives.
So if you've got a limit $\lim_{x\to x_0}(f(x)g(x))$  and $f(x)\to 0$ and $g(x)\to\infty$, then you can for example decide to move $g$ to the denominator giving $f(x)/(g(x))^{-1}$. And now both numerator and denominator go to zero, thus if also the other conditons are fulfilled, you can apply l'Hôpital to get
$$\lim_{x\to x_0}\frac{f(x)}{(g(x))^{-1}}
= \lim_{x\to x_0}\frac{f'(x)}{-(g(x))^{-2}g'(x)}
= -\lim_{x\to x_0}\frac{f'(x)}{g'(x)}(g(x))^{2}$$
Note that this is not the same as $\lim_{x\to x_0}f'(x)g'(x)$.
A: You know that you can use $\frac 0 0$ and $\frac \infty \infty$ in L'Hôpital's rule. Thus, if you have $0\cdot\infty$ where $f(k)\rightarrow0$ and $g(k)\rightarrow\infty$, you can rewrite $\lim_{k\rightarrow c}{f(k)\cdot g(k)}$ as $\lim_{k\rightarrow c}{\frac {f(k)}{\frac 1 {g(k)}}}$. Now the problem is no longer in the form $0\cdot \infty$ but $\frac 0 0$, and thus you can apply L'Hôpital's rule.
A: To answer your question you should consider what L. Hospitals rule says. I will break up the theorem to two parts: condition and conclusion. I will highlight conditions only.
Conditions: 1. $ g(x)$ and $f(x)$ should be continuously differentiable in the deleted neighborhood of the real number $a$.
 2. $g'(x)$ should not be zero for all values of $x$ in the deleted neighborhood of $a$.
3. $ \displaystyle \lim _{x \rightarrow a} \frac{f(x)}{g(x)}$ should be either $\left[ 0/0 \right]$ or $ \left[ \pm \infty / \infty \right] $ indeterminate.
Now to answer your question  on whether to use the theorem on products or not, I should say you must convert the products especially of the form $0 \cdot \infty $ to quotients of the form $ 0 / \frac{1}{\infty} $ or $ \infty / \frac{1}{0} $ to conform to the conditions of the L Hospitals rule.
