I've not read Bernoullithe old sources (this first appeared in a textbook of L'Hopital, right?), so the following is just an educated guess. It gives a slightly weaker result than the usual proof, but people back then didn't worry too much about things like differentiability.
Assume that $f(x)$ and $g(x)$ are smooth and go to $0$ as $x$ goes to $0$. Also, assume that $g'(x)$ goes to something nonzero as $x$ goes to $0$. We can then write $f(x)=x F(x)$ and $g(x)=x G(x)$ for some $F(x)$ and $G(x)$ that are smooth at $0$. Moreover, we have $f'(x) = F(x) + x F'(x)$ and $g'(x) = G(x) + x G'(x)$, so $f'(x)$ and $g'(x)$ go to $F(0)$ and $G(0)$ as $x$ goes to $0$, respectively. Finally, $f(x)/g(x) = F(x)/G(x)$, so we conclude that $f(x)/g(x)$ goes to $F(0)/G(0) = f'(0)/g'(0)$ as $x$ goes to $0$.
I've never understood why the above proof doesn't appear in calculus textbooks. I've found that students understand it much better than the usual one; indeed, it is really just the "canceling common factors of $x$" thing they've been doing for polynomials since they first learned about limits.