What Is L'Hôpital's Rule?
L'Hôpital's rule is a technique for evaluating limits that result in the indeterminate forms 0/0 or ∞/∞. Named after the French mathematician Guillaume de l'Hôpital (though actually discovered by Johann Bernoulli), the rule states the following.
lim f(x) = 0 and lim g(x) = 0 (or both limits equal ±∞), then lim f(x)/g(x) = lim f'(x)/g'(x), provided the new limit exists.In plain language: when plugging in the limit point gives you 0/0 or ∞/∞, you can take the derivative of the top and the derivative of the bottom separately, then try the limit again.
Important: L'Hôpital's rule is not the quotient rule. The quotient rule differentiates a fraction; L'Hôpital's rule differentiates the numerator and denominator separately. Mixing these up is one of the most common mistakes students make.
The rule is powerful because many limits that look impossible with direct substitution or algebraic manipulation become trivial after one or two applications of L'Hôpital's rule.
When to Use L'Hôpital's Rule
L'Hôpital's rule applies only to the indeterminate forms 0/0 and ∞/∞. Before applying the rule, you must verify that direct substitution produces one of these forms. If substitution gives a defined number (like 5/2) or a form like 2/0 (which indicates a vertical asymptote, not an indeterminate), L'Hôpital's rule does not apply.
The seven indeterminate forms
Other indeterminate forms can be transformed into 0/0 or ∞/∞, then L'Hôpital's rule applies. The seven indeterminate forms are: 0/0, ∞/∞, 0 · ∞, ∞ − ∞, 0⁰, ∞⁰, and 1^∞. The last five need algebraic or logarithmic manipulation first.
Always check the form first. Applying L'Hôpital's rule to a non-indeterminate limit will give you a wrong answer. For example, lim (x→0) (x + 1)/(x + 2) = 1/2, but if you blindly differentiate you get lim 1/1 = 1 — completely wrong. The rule only works when direct substitution yields 0/0 or ∞/∞.
The 0/0 Form — Worked Examples
Evaluate lim (x→0) sin(x)/x.
Direct substitution: sin(0)/0 = 0/0. ✓ Indeterminate, apply L'Hôpital's rule.
Take derivatives: f'(x) = cos(x), g'(x) = 1.
New limit: lim (x→0) cos(x)/1 = cos(0)/1 = 1.
1This matches the famous limit you may have seen proved with the squeeze theorem.
Evaluate lim (x→0) (1 − cos(x))/x².
Direct substitution: (1 − 1)/0 = 0/0. ✓ Apply L'Hôpital's: lim (x→0) sin(x)/(2x).
Still 0/0, apply again: lim (x→0) cos(x)/2 = 1/2.
1/2This example shows you may need to apply L'Hôpital's rule multiple times in a row.
Evaluate lim (x→1) (x³ − 1)/(x − 1).
Direct substitution: 0/0. ✓ Apply L'Hôpital's: lim (x→1) 3x²/1 = 3(1)² = 3.
3This limit could also be evaluated by factoring x³ − 1 = (x − 1)(x² + x + 1), but L'Hôpital's is faster for higher powers.
The ∞/∞ Form — Worked Examples
Evaluate lim (x→∞) (2x + 3)/(5x − 1).
Direct substitution: ∞/∞. ✓ Apply L'Hôpital's: lim (x→∞) 2/5 = 2/5.
2/5You could also divide numerator and denominator by x, but L'Hôpital's is mechanical.
Evaluate lim (x→∞) ln(x)/x.
Direct substitution: ∞/∞. ✓ Apply L'Hôpital's: lim (x→∞) (1/x)/1 = lim (x→∞) 1/x = 0.
0This confirms that natural logarithm grows much slower than any power of x.
Evaluate lim (x→∞) eˣ/x².
Direct substitution: ∞/∞. ✓ Apply L'Hôpital's: lim (x→∞) eˣ/(2x).
Still ∞/∞, apply again: lim (x→∞) eˣ/2 = ∞.
∞This shows exponentials always beat polynomials in a growth race — no matter how large the polynomial degree, eˣ eventually dominates.
Converting Other Indeterminate Forms
The other five indeterminate forms need algebraic tricks to convert into 0/0 or ∞/∞ before L'Hôpital's rule applies.
- Form 0 · ∞ — Rewrite as a fraction
Example:
lim (x→0⁺) x · ln(x) = lim (x→0⁺) ln(x)/(1/x). Now it's−∞/∞. Apply L'Hôpital's:lim (1/x)/(−1/x²) = lim −x = 0. - Form ∞ − ∞ — Combine into a single fraction
Example:
lim (x→0) (1/sin(x) − 1/x) = lim (x − sin(x))/(x · sin(x)). Now it's0/0, apply L'Hôpital's twice to get0. - Forms 0⁰, ∞⁰, 1^∞ — Take the natural log first
Example: for
lim (x→0⁺) xˣ, lety = xˣ, thenln(y) = x · ln(x). From above, this limit is0, soy → e⁰ = 1. Answer:lim (x→0⁺) xˣ = 1.
The pattern is always the same: take ln of the expression to convert an exponential indeterminate into a product indeterminate, then convert the product to 0/0 or ∞/∞ for L'Hôpital's.
Common Mistakes to Avoid
Always check the form first. lim (x→0) cos(x)/x is 1/0, not 0/0. Applying L'Hôpital's gives you −sin(x)/1 = 0, which is wrong. The actual answer is that the limit does not exist (one-sided limits go to ±∞).
Differentiate the numerator and denominator separately, not as a quotient. If f(x) = x² + 1 and g(x) = e^x, then f'(x) = 2x and g'(x) = e^x — use these separately, not d/dx[(x² + 1)/eˣ].
If the new limit is still indeterminate, apply L'Hôpital's again. If it's no longer indeterminate, stop immediately and evaluate — don't keep differentiating.
L'Hôpital's rule requires that the derivatives exist near the limit point and that g'(x) ≠ 0 near the point (except possibly at the point itself). In practice this is almost always satisfied, but on rigorous exams it's worth mentioning.
For limits at infinity of rational functions, dividing by the highest power of x is often faster. For limits involving factoring or known limits like lim sin(x)/x = 1, direct methods may be cleaner. L'Hôpital's is a powerful tool, but not always the best one.
If you're stuck on a limit problem, scan it with Solver AI to see whether L'Hôpital's or another technique is the right approach, with every step explained.