L'Hôpital's Rule

A haunted tour of one of calculus's spookiest limits. Warning: you may be interrupted by a ghost with strong opinions. 👻

🦇🦇🦇

L'Hôpital's Rule

The spell that rescues limits from the land of 00
(say it: “loh-pee-TAHL”)  •  a Halloween story of math, money & a ghostwriter 👻

🧙 The Rule

Some limits give you a nonsense answer when you plug in: 00 or ∞∞. These are indeterminate forms – monsters hiding behind the limit. L'Hôpital's Rule is the wand that banishes them.

If limx→af(x)g(x) is 00 or ±∞±∞, then
limx→af(x)g(x)  =  limx→af′(x)g′(x)
…as long as the limit on the right exists (or is ±∞). Differentiate top and bottom separately.

⚰️ Rules of the haunted house

🔦 Why it works: zoom into the haunting

At the point a, both functions hit 0, so fg is undefined – a ghost lives there 👻. But zoom in and every smooth curve looks like a straight line. The ratio of the heights is then the ratio of the slopes: f′(a)g′(a).

Zoom window

Heights at the probe (bars)

Ratio of slopes at a

Casting the spell

■ orange = f(x), ■ green = g(x), dashed lines = tangent lines at a. Slide the zoom right and watch the curves straighten and the heights’ ratio lock onto the slopes’ ratio.

🏃 Ghost race to the door

Two ghosts float toward the same door. Ghost f moves at speed f′ and ghost g at speed g′. At the door, both distances are 0 – so 00. But just before arriving, how far away each one is depends only on its speed:

Distance to door

The ratio no matter how close

Distance = speed × time, so the time cancels: fg = f′·tg′·t = f′g′. That's the whole idea of the rule.

🕯️ Who was L'Hôpital? The ghostwriter scandal

The rule bears the name of a French nobleman… but he may not have discovered it.

1661

Guillaume de l'Hôpital is born in Paris into a noble family – he becomes the Marquis de l'Hôpital. He starts out as a cavalry officer, but poor eyesight ends his army career, so he turns to his real love: mathematics.

1691–92

A young Swiss genius, Johann Bernoulli (b. 1667), visits Paris and tutors the Marquis in the brand-new calculus invented by Leibniz.

1694

The deal 💰: l'Hôpital pays Bernoulli a regular salary (reported as 300 livres a year) in return for his math discoveries – and for keeping quiet about it. Bernoulli sends him the trick for “00” limits in a letter.

1696

l'Hôpital publishes Analyse des infiniment petits – the first calculus textbook ever. The rule is inside, under his name. (He thanked Bernoulli and Leibniz in the preface, but didn't say how much was Bernoulli's.)

1704

l'Hôpital dies at about 42. With the Marquis gone, Bernoulli starts claiming the rule was his. Many people think he's just jealous and bitter…

1921

Plot twist 👻 Historian Paul Schafheitlin finds Bernoulli's old lecture notes in the library at Basel. They match the textbook – the lessons Bernoulli taught the Marquis. Bernoulli was telling the truth.

👻 The ghostwriter: The most famous rule in limit-land was written by Bernoulli, published by l'Hôpital, and named for the wrong man. Mathematicians even have a name for this: Stigler's Law of Eponymy – “no scientific discovery is named after its original discoverer.”
🦇 Spooky spelling: In his own day the name was spelled L'Hospital. That silent “s” was a ghost – the French later replaced it with the little hat (^) on the o: Hôpital. Both spellings appear in textbooks, so don't panic on either.

🧟 The disguised monsters

Other indeterminate forms aren't fractions yet. Rewrite them as 00 or ∞∞ first, then cast the spell.

🧛

0 · ∞

Turn the product into a quotient:
f·g = f1g

Example: x ln x → ln x1x

🧟

∞ − ∞

Combine into one fraction with a common denominator, then check what you get.

🕷️

00, 1∞, ∞0

Let y = fg. Take the log: ln y = g · ln f (a 0·∞ again). Find the limit L of ln y, then the answer is eL.

🎃 Trick or treat?

For each limit, decide: is it a Treat (0/0 or ∞/∞ – use the rule), a Trick (not indeterminate – the rule is a trap!), or a Disguise (rewrite first)?

⚰️ Your turn: dig up the answers

Try each on paper, then open the tomb.

1. limx→0tan xx
Open the tomb

0/0 ✔ → limx→0sec2x1 = 1

2. limx→∞exx2
Open the tomb

∞/∞ ✔ → ex2x (still ∞/∞) → ex2 → ∞. Exponentials beat polynomials!

3. limx→0+x ln x
Open the tomb

0·(−∞) disguise → rewrite: ln x1x (−∞/∞) → 1x−1x2 = −x → 0

4. limx→0+xx
Open the tomb

00 disguise. Let y = xx, so ln y = x ln x → 0 (problem 3). So y → e0 = 1.