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?



How did Bernoulli prove L'Hôpital's rule?
L'Hôpital's rule was first published in [*Analyse des Infiniment Petits*](https://archive.org/details/infinimentpetits1716lhos00uoft).


According to [*The Historical Development of The Calculus*](https://books.google.com/books?id=D2SWE_iZjYsC&printsec=frontcover&dq=historical+development+of+calculus+books&hl=en&ei=R2orTZWIG4nX4gbw7OXmCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CDIQ6AEwAA#v=onepage&q=historical%2520development%2520of%2520calculus%2520books&f=false) 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*](https://books.google.com/books?id=iWaWbZHCw5YC&printsec=frontcover&dq=mathematics+of+great+amateurs&source=bl&ots=KsJUFE_QnF&sig=CRYrtLI9i6sL2CcexwLGz09BjPo&hl=en&ei=Fz0sTcvVJMGEOsnS6csK&sa=X&oi=book_result&ct=result&resnum=) (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.
> 
> 
> 


