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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jan 18 '12 at 4:53
  • $\begingroup$ @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. $\endgroup$ – Ben Crowell Jan 18 '12 at 5:41
  • $\begingroup$ @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. $\endgroup$ – Ben Crowell Jan 18 '12 at 16:32
  • $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jan 18 '12 at 16:33
  • $\begingroup$ @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. $\endgroup$ – 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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jan 18 '12 at 7:35
  • 1
    $\begingroup$ @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 $a$.) $\endgroup$ – Pete L. Clark Jan 18 '12 at 13:19
  • 1
    $\begingroup$ 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. $\endgroup$ – Brian M. Scott Jan 18 '12 at 13:55
  • $\begingroup$ 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. $\endgroup$ – Ben Crowell Jan 18 '12 at 16:09
  • $\begingroup$ @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? $\endgroup$ – 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: $$\frac{f}{g} = \frac{df}{dg} = \frac{df/dx}{dg/dx}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.