To prove L'Hôpital's rule, the standard method is to use use Cauchy's Mean Value Theorem (and note that once you have Cauchy's MVT, you don't need an $\epsilon$$\delta$ definition of limit to complete the proof of L'Hôpital). I'm assuming that Cauchy was responsible for his MVT, which means that Bernoulli didn't know about it when he gave the first proof. So what did he do instead?

L'Hôpital's rule was first published in Analyse des Infiniment Petits. According to The Historical Development of The Calculus by Edwards (p. 269),
Edit. J.L. Coolidge explains in The Mathematics of Great Amateurs (see pp. 159160 of the 2nd edition) that L'Hôpital was interested in calculating



I've not read the 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. 


Regarding the above answers, it is important to state what is considered (see http://planetmath.org/encyclopedia/LHospitalsRule.html) to be L'Hôpital rule: $$ \lim_{x\to a} f(x)/g(x) = \lim_{x\to a} f'(x)/g'(x) $$ whenever $f(a)=g(a)=0$ and the righmost limit make sense. Note that the weaker rule stated in the answer above $$ \lim_{x\to a} f(x)/g(x) = f'(a)/g'(a) $$ is an easy consequence of the definition of the derivative, dividing both $f(x)$ and $g(x)$ by $xa$ and taking limits. Despite the temptation to state and prove L'Hôpital in this weaker form, this form becomes useless whenever you have to use L'Hôpital rule twice to obtain an indefinite limit. 

