L'Hôpital's Rule — Mathematics Knowledge System

Overview

L'Hôpital's Rule provides a way to evaluate limits that produce indeterminate forms like 0/0 or ∞/∞ by differentiating the numerator and denominator separately (not using the quotient rule) and taking the limit of the resulting ratio. It converts a difficult limit problem into a (hopefully simpler) derivative problem.

Intuition

When both f(x) and g(x) approach 0 (or both approach infinity) as x→a, the ratio f(x)/g(x) is a 'race' between how fast the numerator and denominator vanish (or blow up). L'Hôpital's Rule says that race is decided by their derivatives — the rates at which they're approaching those limiting values — rather than their raw values, which are both uninformatively 0 or ∞.

Interactive Graph

Formal Definition

Definition

If f and g are differentiable near a (except possibly at a), g'(x) ≠ 0 near a, and the limit of f/g produces the indeterminate form 0/0 or ∞/∞, then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Valid provided the right-hand limit exists (or is ±∞)

L'Hôpital's Rule

Derivation

For the 0/0 case with f(a)=g(a)=0, L'Hôpital's Rule follows from the Cauchy Mean Value Theorem applied on an interval [a,x]:

\frac{f(x)}{g(x)} = \frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)} \text{ for some } c \text{ between } a \text{ and } x

Cauchy's Mean Value Theorem gives this equality for some c strictly between a and x

\text{As } x \to a,\ c \to a \text{ as well (squeezed between them), so } \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{c\to a}\frac{f'(c)}{g'(c)}

Taking the limit as x→a forces c→a too, giving the result

Applications

Evaluating limiting behavior of physical quantities (e.g. small-angle approximations, relativistic limits as v→c) often produces indeterminate forms resolved by L'Hôpital's Rule.

Worked Examples

Direct substitution gives 0/0, an indeterminate form — apply L'Hôpital's Rule.
$\lim_{x\to 0}\frac{\sin x}{x} \overset{0/0}{=} \lim_{x\to 0}\frac{\cos x}{1}$
Evaluate the new limit directly.
$= \cos(0) = 1$

Answer: 1

Practice Problems

Difficulty 4/10

Evaluate lim(x→0) (e^x - 1 - x)/x².

Difficulty 5/10

Evaluate lim(x→π/2) (cos x)/(x - π/2).

Common Mistakes

Common Mistake

Applying L'Hôpital's Rule to a limit that is not actually an indeterminate form (e.g. a finite/nonzero ratio).

L'Hôpital's Rule only applies to 0/0 or ∞/∞ (and forms reducible to these). Applying it to, say, a limit that evaluates directly to 3/2 gives a wrong (often coincidentally different) answer.

Common Mistake

Using the quotient rule on f/g instead of differentiating numerator and denominator separately.

L'Hôpital's Rule replaces f(x)/g(x) with f'(x)/g'(x) — differentiate top and bottom independently, do NOT apply the quotient rule to the original ratio.

Summary

L'Hôpital's Rule resolves 0/0 and ∞/∞ indeterminate limits by replacing f/g with f'/g' and taking the limit again.
It may be applied repeatedly if the new ratio is still indeterminate.
Other indeterminate forms (0·∞, ∞-∞, 0⁰, 1^∞, ∞⁰) must first be algebraically rewritten as a 0/0 or ∞/∞ quotient.
Always verify the limit is genuinely indeterminate before applying the rule.

References

WebsiteWikipedia — L'Hôpital's rule