# proof of l'Hôpital's rule that minimizes special-casing

A simple form of l'Hôpital's rule looks like this: If $u$ and $v$ are functions with $u(0)=0$ and $v(0)=0$, the derivatives $\dot{v}(0)$ and $\dot{v}(0)$ are defined, and the derivative $\dot{v}(0)\ne 0$, then \begin{align*} \lim_{x\rightarrow 0} \frac{u}{v} &= \frac{\dot{u}(0)}{\dot{v}(0)} \qquad . \end{align*}

To me, the clearest way to arrive at this result uses a little nonstandard analysis: Since $u(0)=0$, and the derivative $d u/d x$ is defined at $0$, $u(d x)=d u$ is infinitesimal, and likewise for $v$. By the definition of the limit, the limit is the standard part of \begin{equation*} \frac{u}{v} = \frac{d u}{d v} = \frac{d u/d x}{d v/d x} \qquad , \end{equation*} where by assumption the numerator and denominator are both defined (and finite, because the derivative is defined in terms of the standard part). The standard part of a quotient like $p/q$ equals the quotient of the standard parts, provided that both $p$ and $q$ are finite (which we've established), and $q \ne 0$ (which is true by assumption). But the standard part of $d u/d x$ is the definition of the derivative $\dot{u}$, and likewise for $d v/d x$, so this establishes the result.

The generalizations to $x\rightarrow a$, where $a\ne 0$, and $x\rightarrow \infty$ are pretty trivial with the changes of variable $x\rightarrow x-a$ and $x\rightarrow 1/x$.

But there are a bunch of other cases of l'Hôpital's that seem to me to involve toxic doses of case-splitting. There are cases where you have to differentiate more than once, and cases where the indeterminate form is $\infty/\infty$ rather than $0/0$.

Is it possible to treat all of this in a unified way, possibly using ideas from projective geometry or inversions with respect to a circle in the complex plane?

• The case where you differentiate more than once follows from the case where you differentiate once, by induction. I think the case $\infty / \infty$ follows by the $0/0$ case by letting $f(x)/g(x) = (1/g(x))/(1/f(x))$ and go back to a $0/0$ case. When you differentiate this new case, you get $$\lim_{x \to 0} \frac{1/g(x)}{1/f(x)} = \lim_{x \to 0} \frac{ -g'(x)/g(x)^2 }{ -f'(x)/f(x)^2 } = \lim_{x \to 0} \frac{f(x)^2/g(x)^2}{f'(x)/g'(x)}.$$ I've tried to figure out the logic of it, but I'm a little tired (this is my second "tired" post today... gosh, no energy) maybe you can complete my arg. – Patrick Da Silva Jan 18 '12 at 4:53
• @PatrickDaSilva: Thanks for your comment. I'm sure there is an inductive proof of then nth-derivative case, but I don't think the most obvious method of induction works. One of the hypotheses of l'Hôpital's rule is that $\dot{v}\ne 0$, which fails in the cases where you need multiple differentiation. I think an NSA-style approach works, since the Leibniz notation $d^2u/dx^2$ can be taken as essentially a literal division of two infinitesimal numbers. It's not so much that I can't find proofs in books or on the web, it's that there seems to be more need for case-splitting than I'd like. – Ben Crowell Jan 18 '12 at 5:41
• @PatrickDaSilva: Your $(1/g)/(1/f)$ idea seems to work nicely. It requires lots of uses of $\lim(ab)=\lim a \lim b$ and $\lim(a/b)=\lim a/\lim b$, which may create some complication in the case where $\lim b=0$, or in the case where you want to use l'Hopital's rule to prove that a limit doesn't exist. – Ben Crowell Jan 18 '12 at 16:32
• You're a fan of non-standard analysis huh? It sure does make proofs seem cleaner, but it takes a lot of work to understand why it works. I hope you've read stuff about it. – Patrick Da Silva Jan 18 '12 at 16:33
• @Ben Crowell: Danger! It is not possible to use l'Hôpital's rule to prove that a limit does not exist!! See e.g. the example on page 4 of math.uga.edu/~pete/2400diffmisc.pdf. – Pete L. Clark Jan 18 '12 at 17:37

Last week I covered L'Hôpital's Rule in my Spivak Calculus course. I ended up preferring the proof given in Rudin's Principles to Spivak's treatment, because (i) he does compile several different cases in a rather efficient way and (ii) he proves a stronger result in the case where $\lim_{x \rightarrow a} g(x) = \infty$, namely that there is no hypothesis needed on the limiting behavior of the numerator $f$.

I wrote this up in $\S$ 1 of these notes. However I must confess that I found the proof itself not very interesting and ended up not covering it in class.

I must also confess that I don't really understand the OP's sketch proof. In particular I do not recognize the hypothesis $g'(a) \neq 0$ that the OP is assuming: who says that $f$ and $g$ are even defined at $a$? I am also slightly skeptical that NSA really helps here to give a shorter proof, but I would be very interested to be proven wrong about this.

Added: If you are willing to assume that $f$ and $g$ are defined and differentiable at $a$, that $f(a) = g(a) = 0$ and that $g'(a) \neq 0$, the proof becomes almost trivial:

$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{g(x)-g(a)} = \lim_{x \rightarrow a} \frac{ \frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{ \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}}{\lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)}$.

(However this version is inadequate for many of the standard applications of freshman calculus.) So I presume we're talking about a stronger version than this?

• If one’s willing to prove only the $0/0$ case, and that only in the special case when $f\,'(a)$ and $g'(a)$ exist, it really is possible to give a very short NSA proof; Jerry Keisler does it about three lines in his Elementary Calculus, using the OP’s argument with more detailed notation. – Brian M. Scott Jan 18 '12 at 7:35
• @Brian: Thanks for that information. I think though that by making the extra hypotheses you suggest the standard proof becomes essentially a three line calculation as well...What do you think? (Hmm, maybe for the easy proof I also want to assume $f'$ and $g'$ are continuous at $a$.) – Pete L. Clark Jan 18 '12 at 13:19
• It’s six of one, half a dozen of the other for that result, though the NSA version of the argument may be just a hair more intuitive. Some of the basic theorems really are a bit more straightforward in the NSA version, but I’d not say that this was really one of them. – Brian M. Scott Jan 18 '12 at 13:55
• I see. The stronger form $\lim f/g=\lim f'/g'$ allows proof by induction for the nth-derivative case, whereas my weaker form $\lim f/g=f'/g'$ doesn't. In a), Step 1, I assume you really mean $-\infty<A<\infty$, not just $A<\infty$? And is b) meant to be the case where $A=\pm\infty$? There does seem to be a big cost in complexity for the stronger form. You're quantifying over five variables, with quantifiers three deep. I guess the $b=\pm\infty$ case is handled properly by a) Step 1, since $b$ occurs only in $\ldots<b$ and in limits, so there is no assumption that $b$ is real. – Ben Crowell Jan 18 '12 at 16:09
• @Ben: No, I really mean what I wrote (which is the same as what Rudin wrote). If $A = -\infty$, we still need to bound the quotient from above. Also, doesn't every statement of the form $\lim_{x \rightarrow a} f(x) = L$ involve three quantifiers? – Pete L. Clark Jan 18 '12 at 17:23

Here's a very simple proof. We consider functions $$f(x)$$ and $$g(x)$$ which have derivatives $$df/dx$$ and $$dg/dx$$ respectively. In general $$dy$$ means the difference between $$y_1$$ and $$y_0$$, and L'Hopital's rule applies to cases where both $$f_0$$ and $$g_0$$ are actually $$0$$, therefore $$df$$ and $$dg$$ are the same as $$f$$ and $$g$$ respectively. So we have: $$\frac{f}{g} = \frac{df}{dg} = \frac{df/dx}{dg/dx}$$