Differentiability, mean value theorems, Taylor, L'Hôpital

In one sentence

Every theorem in this family carries hypotheses that look like formalities, and the questions are built almost entirely from removing one and seeing whether you notice.

Why the exam asks it

Because Rolle, the mean value theorem, Taylor and L'Hôpital are the four results every candidate can state, and stating them is not the skill being tested. Identifying which hypothesis an option has dropped is. Expect at least one option whose statement is true, one whose statement is a converse, and one that would be true on a closed interval and is offered on an open one.

The idea

The mean value family

Rolle: $f$ continuous on $[a,b]$, differentiable on $(a,b)$, $f(a) = f(b)$. Then $f'(c) = 0$ for some $c$ inside.

Note the asymmetry, because it is examined: continuity is required on the closed interval, differentiability only on the open one. $f(x) = |x|$ on $[-1,1]$ satisfies the continuity hypothesis and fails differentiability at one interior point — and the conclusion fails with it.

Mean value theorem: same hypotheses without $f(a) = f(b)$; then $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.

Almost everything else in this area is MVT in disguise. Bounded derivative gives Lipschitz. $f' \equiv 0$ gives constant. $f' > 0$ gives strictly increasing.

Darboux: a derivative has the intermediate value property, even when it is not continuous. This is the surprising one, and the paper likes it precisely because it is surprising: $f'$ need not be continuous, but it can never have a jump discontinuity.

$x^{2} \sin(1/x)$, the function this topic is built on

Define $f(x) = x^2 \sin(1/x)$ for $x \ne 0$, and $f(0) = 0$.

At the origin,

$$f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0,$$

because $|h \sin(1/h)| \le |h|$. So $f$ is differentiable everywhere.

Away from the origin, $f'(x) = 2x\sin(1/x) - \cos(1/x)$. As $x \to 0$ the first term vanishes and the second oscillates between $-1$ and $1$ forever. So $f'$ has no limit at $0$, and is therefore discontinuous there — while $f'(0)$ exists and equals $0$.

Differentiable everywhere, derivative not continuous. That single function answers a large fraction of the false options in this topic, and it is worth being able to produce it and its derivative from memory.

L'Hôpital, in the correct direction

If $f/g$ is $0/0$ or $\infty/\infty$ near $a$, and $\lim f'/g'$ exists, then $\lim f/g$ exists and equals it.

The hypothesis is that the derivative quotient converges. The rule does not say the two limits exist together, and it does not run backwards.

Where intuition breaks

"Differentiable, so continuously differentiable." $x^2\sin(1/x)$. The most productive counterexample in the topic, for exactly the reason above.

"$f/g$ has a limit, so $f'/g'$ has one." Take $f(x) = x^2 \sin(1/x)$ and $g(x) = x$ as $x \to 0$. Then $f/g = x\sin(1/x) \to 0$ comfortably. But $f'/g' = 2x\sin(1/x) - \cos(1/x)$ has no limit at all.

Why it is tempting: L'Hôpital is used so often left-to-right that it starts to feel like an equivalence. It is a one-way implication with a hypothesis, and this is a converse assumed. If $f'/g'$ fails to converge, L'Hôpital tells you nothing — it does not tell you $f/g$ diverges.

"$f'$ exists everywhere, so $f'$ is continuous." Same counterexample, and worth separating from the first because it is often phrased as a fact about derivatives rather than about $f$. Darboux is the correct consolation: $f'$ has the intermediate value property even when discontinuous, so it cannot have a jump.

"$f'(x_0) = 0$ and $f'$ does not change sign, so $f$ is constant near $x_0$." $f(x) = x^3$ at $x_0 = 0$. Here $f' = 3x^2 \ge 0$ with $f'(0) = 0$, and $f$ is strictly increasing. A vanishing derivative at an isolated point costs you nothing.

"Rolle needs differentiability on $[a,b]$." It needs continuity on $[a,b]$ and differentiability only on $(a,b)$. Options that tighten the hypothesis are not false statements about functions, but they are wrong answers about the theorem, and the paper counts them as wrong.

"$f$ has a Taylor series at $a$, so it equals its Taylor series near $a$." False, and the witness is $e^{-1/x^2}$ with $f(0) = 0$: every derivative at the origin is zero, so the Taylor series is identically $0$, while the function is not. Smooth is not analytic, and the gap between them is a real one.

The exam's angle

1. Name the theorem being invoked, then list its hypotheses. Almost every wrong option is a correct theorem with one hypothesis removed. Which one is missing is the question.
2. Check the interval, closed against open. Rolle and MVT distinguish them deliberately.
3. If L'Hôpital appears, check the direction. Does the option assume $f'/g'$ converges, or conclude it?
4. Reach for $x^2 \sin(1/x)$ early. It settles anything about differentiability implying continuity of the derivative, and anything about L'Hôpital's converse.
5. Reach for $e^{-1/x^2}$ for anything about Taylor series representing the function.

The night before

• Rolle and MVT: continuity on $[a,b]$, differentiability on $(a,b)$. The asymmetry is examined.
• $x^2\sin(1/x)$ with $f(0) = 0$: differentiable everywhere, $f'$ discontinuous at $0$. $f'(x) = 2x\sin(1/x) - \cos(1/x)$.
• Darboux: derivatives have the intermediate value property. No jump discontinuities.
• L'Hôpital runs one way: it needs $\lim f'/g'$ to exist. Its failure proves nothing.
• $x^3$ at $0$: derivative vanishes, function strictly increasing.
• $e^{-1/x^2}$: smooth, all derivatives zero at $0$, not equal to its Taylor series. Smooth $\ne$ analytic.
• Bounded derivative on an interval $\Rightarrow$ Lipschitz, straight from MVT.