AP Calculus AB and BC

When L'Hopital's Rule Fails: Two Counterexamples

L'Hopital's Rule needs two things: the quotient has to be an indeterminate form at the point, and the quotient of derivatives has to have a limit. Apply it to a determinate form and it returns the wrong number. Apply it where the derivative quotient oscillates and it returns nothing at all.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   The quotient is an indeterminate form at a

   $$\frac{f}{g} \to \frac{0}{0} \ \text{ or } \ \frac{\infty}{\infty}$$

   This is the hypothesis people forget, because differentiating the top and bottom is a mechanical move that always produces a number. If the quotient is determinate, direct substitution already gave the right answer and the rule is simply not talking about this problem.

   Drop it and the theorem fails

   $$\lim_{x \to 0} \frac{x+1}{x+2}$$

   The limit of (x+1)/(x+2) as x approaches 0

   Direct substitution gives $1/2$, and that is the answer. Differentiating top and bottom gives $1/1 = 1$, which is not. The quotient was never indeterminate: it evaluated cleanly to $1/2$, so the rule never applied. Notice that the wrong answer looks perfectly respectable, which is what makes this failure worth checking for rather than trusting.

2. 2

   The limit of the derivative quotient exists

   $$\lim_{x \to a} \frac{f'(x)}{g'(x)} \text{ exists}$$

   The rule transfers a limit from one quotient to another. If the second quotient has no limit, there is nothing to transfer, and crucially this does NOT mean the original limit fails to exist. The rule is silent, not negative.

   Drop it and the theorem fails

   $$\lim_{x \to \infty} \frac{x + \sin x}{x}$$

   The limit of (x + sin x)/x as x tends to infinity

   This is a genuine $\infty/\infty$ form, so the first hypothesis holds. The real limit is 1, because $\sin x / x \to 0$. But differentiating gives $(1 + \cos x)/1$, which swings between 0 and 2 forever and has no limit however far out you look. L'Hopital returns nothing here, while a one-line algebraic split returns the answer immediately.

3. 3

   g'(x) is nonzero near a, except possibly at a itself

   $$g'(x) \ne 0 \text{ for } 0 < |x - a| < \delta$$

   The new quotient has to make sense on a whole punctured neighbourhood, not just at scattered points. If $g'$ keeps returning to zero as you approach $a$, the expression $f'/g'$ is undefined infinitely often on the way in and no limit can be taken. Constructing a clean elementary example of this needs a function whose derivative oscillates through zero, so none is claimed here; in practice this hypothesis holds automatically for the functions on the AP exam.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

All 2 counterexamples above are checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. For the $0/0$ case at a finite $a$, extend $f$ and $g$ by defining $f(a) = g(a) = 0$, which makes both continuous at $a$.
2. For $x$ near $a$, apply the Cauchy Mean Value Theorem to $f$ and $g$ on the interval between $a$ and $x$: there is a point $\xi$ strictly between them with $\frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(\xi)}{g'(\xi)}$.
3. The left side is just $f(x)/g(x)$, since both values at $a$ are 0. This is where the indeterminate form is used, and where a determinate form would break the argument immediately.
4. As $x \to a$, the point $\xi$ is squeezed toward $a$ as well, so if $f'/g'$ has a limit, $f(x)/g(x)$ inherits it. If $f'/g'$ has no limit, the argument produces nothing.

What it does not say