L’Hôpital’s Rule

L’Hôpital’s rule says that if lim f(x)/g(x) has the indeterminate form 0/0 or ∞/∞ and lim f′(x)/g′(x) exists (or is infinite), then lim f(x)/g(x) = lim f′(x)/g′(x). In brief: lim_x→0 (e^x − 1)/x is 0/0, and the derivatives give e^x/1 → 1. Other indeterminate forms (0 · ∞, ∞ − ∞, 1^∞, 0⁰, ∞⁰) are first transformed into a quotient.

The rule replaces algebraic tricks with differentiation: instead of factoring or multiplying by conjugates, you differentiate the numerator and the denominator separately. It does not apply to every limit, and it is not a derivative of the quotient. This guide states the rule with conditions, shows how to apply it repeatedly, explains how to convert the other indeterminate forms, covers the cases where it fails and gives a table of forms and strategies.

What does the rule say?

Let f and g be differentiable near a (except possibly at a), with g′(x) ≠ 0 near a. If lim_x→a f(x) = lim_x→a g(x) = 0, or both limits are ±∞, and the limit lim_x→a f′(x)/g′(x) exists (finite or infinite), then

lim_x→a f(x)/g(x) = lim_x→a f′(x)/g′(x).

The same holds for one-sided limits and for x → ±∞. The rule is based on the Cauchy mean value theorem. Near a, f(x) ≈ f′(a)(x − a) and g(x) ≈ g′(a)(x − a), so the ratio of the functions behaves like the ratio of their rates of change.

Worked examples with 0/0 and ∞/∞

Example 1

lim_x→0 sin x/x: 0/0. Derivatives: cos x/1 → 1. (But note that sin x/x → 1 is used to derive (sin x)′ = cos x, so using the rule here is circular in a first course.)

Example 2

lim_x→2 (x³ − 8)/(x² − 4): 0/0. Derivatives: 3x²/(2x) → 12/4 = 3. Check by factoring: (x − 2)(x² + 2x + 4)/((x − 2)(x + 2)) → 12/4 = 3 ✓.

Example 3: applying the rule twice

lim_x→0 (e^x − 1 − x)/x²: 0/0. First: (e^x − 1)/(2x), again 0/0. Second: e^x/2 → 1/2.

Example 4

lim_x→0 (1 − cos x)/x²: first sin x/(2x) → 1/2 (0/0 again, and either the standard limit or a second application gives cos x/2 → 1/2).

Example 5: ∞/∞

lim_x→∞ ln x/x: derivatives (1/x)/1 = 1/x → 0. And lim_x→∞ x³/e^x: three applications give 6/e^x → 0. In general, an exponential beats any power, and a power beats a logarithm.

Example 6

lim_x→∞ (3x² + 5x)/(2x² − 1): ∞/∞. Derivatives: (6x + 5)/(4x), again ∞/∞; second: 6/4 = 3/2. Dividing by x² gives the same answer more quickly.

Example 7

lim_x→1 (ln x)/(x − 1): 0/0, so (1/x)/1 → 1.

How do you handle other indeterminate forms?

0 · ∞: rewrite as a quotient

lim_x→0⁺ x ln x: 0 · (−∞). Write ln x/(1/x): (−∞)/∞. Derivatives: (1/x)/(−1/x²) = −x → 0. So x ln x → 0.

∞ − ∞: combine into a fraction

lim_x→0⁺ (1/x − 1/sin x) = lim (sin x − x)/(x sin x): 0/0. Derivatives: (cos x − 1)/(sin x + x cos x), still 0/0; again: −sin x/(2cos x − x sin x) → 0/2 = 0.

1^∞, 0⁰, ∞⁰: take logarithms

If y = f(x)^g(x), then ln y = g(x) ln f(x), which has the form 0 · ∞. Find lim ln y = L, then lim y = e^L.

Example 8

lim_x→0⁺ x^x (0⁰): ln y = x ln x → 0, so y → e⁰ = 1.

Example 9

lim_x→∞ (1 + 1/x)^x (1^∞): ln y = x ln(1 + 1/x) = ln(1 + 1/x)/(1/x), 0/0. Derivatives: [(−1/x²)/(1 + 1/x)]/(−1/x²) = 1/(1 + 1/x) → 1. So y → e.

Example 10

lim_x→0⁺ (sin x)^x: ln y = x ln sin x = (ln sin x)/(1/x), (−∞)/∞. Derivatives: (cos x/sin x)/(−1/x²) = −x² cos x/sin x → 0 (since x/sin x → 1 and x cos x → 0). So y → 1.

Example 11

lim_x→∞ x^1/x (∞⁰): ln y = (ln x)/x → 0, so y → 1.

When does the rule fail or not apply?

• The form is not indeterminate: lim_x→0 (x + 1)/(x + 2) = 1/2, but the derivatives give 1/1 = 1, a wrong answer. Always check that the form is 0/0 or ∞/∞.
• The limit of the derivative quotient does not exist: lim_x→∞ (x + sin x)/x = 1 (by dividing by x), but f′/g′ = 1 + cos x has no limit. The rule gives no information here, though the original limit exists.
• The derivatives lead in circles: lim_x→∞ e^x/(e^x + e^−x): the derivatives (e^x)/(e^x − e^−x) and again return to the same type. Dividing by e^x gives 1/(1 + e^−2x) → 1.
• Simpler methods exist: for rational functions at infinity, the degree rule is faster.

Indeterminate forms and strategies

Where does the rule come from?

The rule is named after Guillaume de l’Hôpital, who published it in 1696 in the first calculus textbook, Analyse des infiniment petits. The result is due to Johann Bernoulli, who had taught calculus to l’Hôpital and was paid for his lessons, with an agreement that l’Hôpital could use his discoveries. Bernoulli’s authorship was later documented from his letters, and in some countries the rule is called the Bernoulli–l’Hôpital rule.

Where is it used?

The rule is used to prove the growth hierarchy of functions, to find asymptotic behaviour and to evaluate limits in physics and economics, such as the limit of an average rate when the interval shrinks. In numerical analysis, it is a way to understand the loss of precision in expressions like (e^x − 1)/x for small x. In statistics, limits of ratios of tail probabilities are computed with it.

Which mistakes are common?

Using the quotient rule

Differentiate the numerator and the denominator separately. Do not differentiate the quotient f/g.

Using the rule when the form is not indeterminate

Substitute first. If the limit is a number or ±∞ (like 5/0), the rule does not apply.

Concluding when f′/g′ has no limit

Then the rule says nothing about f/g. Try another method.

Forgetting to convert other forms

For 0 · ∞ or 1^∞, first rewrite the expression so that the rule applies.