Question

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?

Answer 1

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. 159-160 of the 2nd edition) that L'Hôpital was interested in calculating
$$\lim\limits_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}=\frac{16}{9}a.$$

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 $$\frac{a^2-a^2}{a-a}=2a,$$ and in September of the same year he writes:

'Je vous avoue que je me suis fort appliqué à résoudre l'équation $$\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}=y$$ 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.

Answer 2

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.

Answer 3

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 $x-a$ 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.