The Problem It Solves

You are trying to find lim(x→2) (x²−4)/(x−2). Plug in x=2: you get (4−4)/(2−2) = 0/0. That is not zero. It is not undefined in the usual sense. It is indeterminate — the limit might be anything, and ordinary substitution gives you no information.

Algebra saves this particular case: factor the numerator to get (x−2)(x+2)/(x−2), cancel, get x+2, plug in 2, answer is 4. But what about lim(x→0) sin(x)/x? You cannot factor your way out of that. The numerator and denominator are fundamentally different kinds of functions and no algebraic manipulation simplifies it.

This is the gap L'Hôpital's Rule was invented to fill.

The Rule

If lim(x→a) f(x)/g(x) gives 0/0 or ∞/∞, and both f and g are differentiable near a, then:

limx→a f(x)/g(x) = limx→a f'(x) / g'(x)

You replace the fraction with a new fraction: derivative of top over derivative of bottom. Not the quotient rule. Two separate derivatives.

⚠ The mistake everyone makes

Students apply the quotient rule to f/g and get (f'g − fg')/g². That is wrong. L'Hôpital's says differentiate f and g separately, then form the new ratio f'(x)/g'(x). Two individual derivatives, not the quotient rule.

When It Applies — and When It Doesn't

L'Hôpital's Rule requires an indeterminate form. Check before applying. The qualifying forms are 0/0 and ∞/∞. That's it for the fraction version. If you get 0/3, the limit is just 0. If you get 5/0, the limit is ±∞ (or does not exist). L'Hôpital's does not apply to either.

Always try substitution first. Always try factoring or simplification first. L'Hôpital's is a fallback — powerful but not the first tool you reach for.

0/0 Examples

📋 lim(x→0) sin(x)/x
Checksin(0)/0 = 0/0 ✓
Differentiate topd/dx[sin x] = cos x
Differentiate bottomd/dx[x] = 1
New limitlim(x→0) cos(x)/1 = cos(0) = 1
📋 lim(x→0) (1 − cos x)/x² — needs two applications
First check0/0 ✓. Differentiate: sin(x)/(2x) → still 0/0 at x=0
Apply againd/dx[sin x] = cos x. d/dx[2x] = 2. New limit: cos(0)/2 = 1/2
📋 lim(x→1) (x³ − 1)/(x − 1)
Check0/0 ✓ (could also factor, but let's use L'Hôpital's)
Apply3x²/1. At x=1: 3
Verify by factoring(x−1)(x²+x+1)/(x−1) → x²+x+1 at x=1 = 3 ✓

∞/∞ Examples

📋 lim(x→∞) x²/eˣ — exponentials beat polynomials
Check∞/∞ ✓
1st application2x/eˣ → still ∞/∞
2nd application2/eˣ → 2/∞ = 0
Meaningeˣ grows faster than any polynomial. No matter what power of x you pick, eˣ eventually dominates.
📋 lim(x→∞) ln(x)/x
Check∞/∞ ✓
Apply(1/x)/1 = 1/x → 0 as x→∞
Meaningln(x) grows, but much slower than x. The ratio shrinks to zero.

Other Indeterminate Forms: 0·∞, ∞−∞, 1^∞, 0⁰, ∞⁰

These are not directly in the ratio form L'Hôpital's needs, so you convert them first.

0·∞: Rewrite as a fraction. If you have f·g where f→0 and g→∞, write it as f/(1/g) or g/(1/f) to get 0/0 or ∞/∞.

1^∞, 0⁰, ∞⁰: Take ln first. If L = lim f(x)^g(x), then ln L = lim g(x)·ln f(x), which is 0·∞ form. Convert to a fraction and apply L'Hôpital's. Then exponentiate at the end: L = e^(ln L).

📋 lim(x→∞) (1 + 1/x)^x — this gives e
Form1^∞. Take ln: x·ln(1 + 1/x). As x→∞: ∞·0.
Rewriteln(1+1/x)/(1/x). Both→0, apply L'Hôpital's.
ApplyTop: (−1/x²)/(1+1/x). Bottom: −1/x². Simplifies to 1/(1+1/x) → 1.
Answerln L = 1, so L = e. This is how e is defined.

When L'Hôpital's Rule Fails

Applying L'Hôpital's when the form is not indeterminate gives a wrong answer. lim(x→0) x/sin(x): both→0, so 0/0 ✓, apply: 1/cos(x) = 1. Correct. But lim(x→0) (2+x)/sin(x): numerator→2, denominator→0. This is 2/0, not 0/0. L'Hôpital's does not apply — the limit is ±∞, not 1/cos(0) = 1.

Also: sometimes L'Hôpital's produces a circular loop that never terminates. lim(x→∞) eˣ/eˣ = 1, but applying L'Hôpital's gives eˣ/eˣ again — you go in circles forever. Recognise when the answer is already obvious.

Frequently Asked Questions

Why differentiate separately rather than using the quotient rule?▾
Because the theorem literally says to form f'(x)/g'(x) — two separate derivatives divided. The quotient rule differentiates a ratio as a single function and gives something completely different. The proof of L'Hôpital's Rule (via Cauchy's Mean Value Theorem) shows why the separate-derivative ratio equals the original limit. It is not intuitive — just be precise about what the theorem actually says.
How many times can you apply it?▾
As many times as needed, provided the result remains indeterminate each time. lim(x→0) (1−cos x)/x² needs two applications. lim(x→0) (x − sin x)/x³ needs three. After each application, check whether the new form is still 0/0 or ∞/∞ before applying again.
Is L'Hôpital's the best method for sin(x)/x?▾
For the specific limit lim(x→0) sin(x)/x, the standard proof in most calculus courses uses the squeeze theorem, not L'Hôpital's — because L'Hôpital's requires knowing the derivative of sin, which is typically derived using lim sin(x)/x = 1. Using L'Hôpital's here is technically circular. In practice most instructors accept it, but the squeeze theorem proof is the rigorous one.