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?

7$\begingroup$ When I was young I found L'Hopital rule, 2 weeks before it was taught in class. I asked my math teacher, if this rule was true, and he was amazed that I found it by myself. I found the rule, just by looking at the plots of f and g. So, my notion of the rule, was pure graphical. I think that Bernoulli found it the same way and then tried to prove it by limits. $\endgroup$ – Lucas K. Jan 10 '11 at 20:32
L'Hôpital's rule was first published in Analyse des Infiniment Petits.
According to The Historical Development of The Calculus by Edwards (p. 269),
L'Hospital's argument, which is stated verbally without functional notation (see the English translation included in Struik's source book, pp. 313  316), amounts simply to the assertion that $$\frac{f(a+dx)}{g(a+dx)}= \frac{f(a) + f'(a) dx}{g(a) + g'(a)dx}=\frac{f'(a) dx}{g'(a) dx} =\frac{f'(a)}{g'(a)}$$ provided that $f(a) = g(a) = 0$. He concludes that, if the ordinate $y$ of a given curve "is expressed by a fraction, the numerator and denominator of which do each of them become 0 when $x = a$," then "if the differential of the numerator be found, and that is divided by the differential of the denominator, after having made $x = a$, we shall have the value of [the ordinate $y$ when $x = a$]."
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
$$\lim\limits_{x\to a}\frac{\sqrt{2a^3xx^4}a\sqrt[3]{a^2x}}{a\sqrt[4]{ax^3}}=\frac{16}{9}a.$$
As a matter of fact this particular problem had worried him a good deal. We find him writing in July 1693 to John Bernoulli suggesting that we should substitute directly in the original equation, getting $$\frac{a^2a^2}{aa}=2a,$$ and in September of the same year he writes:
'Je vous avoue que je me suis fort appliqué à résoudre l'équation $$\frac{\sqrt{2a^3xx^4}a\sqrt[3]{a^2x}}{a\sqrt[4]{ax^3}}=y$$ lorsque $x=a$, car ne voyant point de jour pour у réussir puisque toutes les solutions qui se présentent d'abord ne sont pas exactes.'
All this suggests that L'Hospital learnt the correct solution from Bernoulli, but did not give him the specific credit, with the unfortunate result that the method came to be known as L'Hospital's method.

$\begingroup$ That looks pretty authoritative; thanks. (By the way, your second expression should have g in the denominator, not f.) $\endgroup$ – John Palmieri Jan 10 '11 at 20:29

$\begingroup$ @John Palmieri: Thank you, I got it corrected. $\endgroup$ – Andrey Rekalo Jan 10 '11 at 20:33
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.

1$\begingroup$ Excellent point! It's precisely this subtlety that makes de l'Hospital rule one of the more difficult results of basic analysis. $\endgroup$ – Theo Buehler Jan 12 '11 at 2:05

$\begingroup$ @Lucas : It's not worthless  if the first $k$ derivatives of $f(x)$ and $g(x)$ of $x$ vanish at $0$ but the $k+1$st derivative of $g(x)$ doesn't vanish at $0$, then you can factor out an $x^k$ from both $f(x)$ and $g(x)$ and perform the above analysis. It still gives a weaker theorem, but it gives a theorem that applies in 90$ of cases that come up in practice (and certainly 100% of the cases that show up in freshman calculus). $\endgroup$ – Andy Putman Jan 12 '11 at 3:41

1$\begingroup$ @Andy: Maybe you are right... But how does one gives a calculus 1 proof that $f(x) = x^k F(x)$ when the derivatives of order $< k$ of $f$ vanishes? For $k=1$ it is an easy consequence of the definition of derivative, but for $k=2$ already I am not able to give a simple proof... Can you give me some indications? I imagine that an appropriate $k=2$ argument can be adapted to give a proof for any $k$ by induction. $\endgroup$ – Lucas Seco Jan 23 '11 at 4:43

$\begingroup$ @Lucas : If $f(0)=0$, then we can write $f(x) = x F(x)$. If in addition $f'(0)=0$, then $F(x)=0$ so we can write $F(x) = x G(x)$ and $f(x)=x^2 G(x)$. Etc. $\endgroup$ – Andy Putman Jan 23 '11 at 15:17

1$\begingroup$ @Andy: that's right, except that you are assuming that $G(x)$ has a derivative at $0$, that is, you are assuming that the limit $\lim_{x\to 0} f(x)/x^2$ exists. Do you know how to prove this? I've seen approximations to second derivatives that looks like this in numerical Calculus textbooks, but I don't know if they justify it rigorously. I really like your point of view about L'Hopital and would consider using it in my Calculus 1 classes if I am able to follow it throughly. $\endgroup$ – Lucas Seco Jan 30 '11 at 23:47
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.

$\begingroup$ Many text books cover L'Hôpital's rule well before they get to Taylor series, which might explain why they don't use this approach. I also don't know enough about the history of mathematics to know if Bernoulli would have used smoothness and Taylor series like this, but it looks plausible. $\endgroup$ – John Palmieri Jan 10 '11 at 20:25

3$\begingroup$ There's no need to know anything about Taylor series to do this. All you have to observe is that if $f(x)$ is differentiable at $0$, then you can define the function $F(x) = f(x)/x$, which has a limit at $0$ by assumption. In any case, it looks like Andrey's quotation about confirms my guess. $\endgroup$ – Andy Putman Jan 10 '11 at 20:27

$\begingroup$ Sorry, I see. I was misled by your use of the word "smooth". $\endgroup$ – John Palmieri Jan 10 '11 at 20:45

1$\begingroup$ By smooth, I just meant "has enough derivatives for the following to work" :). $\endgroup$ – Andy Putman Jan 10 '11 at 20:46

1$\begingroup$ @Daniel Litt: I don't think the derivatives work out right in that case. $\endgroup$ – John Palmieri Jan 12 '11 at 4:37
I think the important reference in which its author describes in detail the proof of L'Hospital's rule done by l'Hospital in his book but with todays language is the following
Lyman Holden, The March of the discoverer, Educational Studies in Mathematics, Vol. 5, No. 2 (Jun., 1973), pp. 193205.

$\begingroup$ JSTOR link: Holden  The march of the discoverer. $\endgroup$ – LSpice Sep 18 '18 at 22:00