# How did Bernoulli prove L'Hôpital's rule?

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?

• 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. Jan 10, 2011 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. 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.

• That looks pretty authoritative; thanks. (By the way, your second expression should have g in the denominator, not f.) Jan 10, 2011 at 20:29
• @John Palmieri: Thank you, I got it corrected. Jan 10, 2011 at 20:33

Regarding the above answers, it is important to state what is considered (see this link) 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.

• Excellent point! It's precisely this subtlety that makes de l'Hospital rule one of the more difficult results of basic analysis. Jan 12, 2011 at 2:05
• @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$of cases that come up in practice (and certainly 100% of the cases that show up in freshman calculus). Jan 12, 2011 at 3:41 • @Andy: Maybe you are right... But how does one gives a calculus 1 proof that$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$by induction. Jan 23, 2011 at 4:43 • @Andy: that's right, except that you are assuming that$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. Jan 30, 2011 at 23:47 • Just commenting in case people find this thread by googling (like I did). @Lucas: I think the main advantage of this approach is avoiding Cauchy's MVT (but still using the usual MVT). If$f'(h) = o(h)$, then by MVT,$f(k)/k = [f(k)-f(0)] / [k-0] = f'(t) = o(t) = o(k)$for some$0<t<k$, so$f(k) = o(k^2)$as desired. Actually, using a change of variables (and the monotonicity of$g$near$a$), we get the 3rd proof of general L'Hopital here: www.jstor.org/stable/2307183 Jan 20, 2014 at 21:52 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. • 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. Jan 10, 2011 at 20:25 • 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. Jan 10, 2011 at 20:27
• Sorry, I see. I was misled by your use of the word "smooth". Jan 10, 2011 at 20:45
• By smooth, I just meant "has enough derivatives for the following to work" :). Jan 10, 2011 at 20:46
• @Daniel Litt: I don't think the derivatives work out right in that case. Jan 12, 2011 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. 193-205.