L'Hopital's Rule Calculator

L’Hopital’s Rule resolves indeterminate limits, specifically the 0/0 and ∞/∞ forms that trip up direct substitution. The rule: if lim f(x)/g(x) is indeterminate as x approaches a, then the limit equals lim f’(x)/g’(x) at the same point, provided that second limit exists.

The rule can be applied repeatedly if the first application still gives an indeterminate form. A polynomial fraction that factors out a common root can require two or three applications before yielding a finite answer.

A worked example. Take lim(x→1) of (x² − 1)/(x − 1). Direct substitution gives 0/0. Apply the rule: derivative of numerator is 2x, derivative of denominator is 1. At x = 1: 2(1)/1 = 2. This matches what you get by factoring: (x − 1)(x + 1)/(x − 1) = x + 1 → 2 at x = 1.

A bit of history. The rule was published in 1696 by Guillaume de l’Hôpital in his textbook Analyse des Infiniment Petits, the first calculus textbook. The mathematical work was almost entirely Johann Bernoulli’s, who had a private contract with l’Hôpital giving him exclusive rights to publish Bernoulli’s discoveries in exchange for a regular salary. Bernoulli was furious after l’Hôpital died and the rule kept his name. Modern mathematicians often spell it “L’Hopital’s Rule” to honor the original published name, while acknowledging the credit really belongs to Bernoulli.

Common mistakes. Two errors come up over and over:

• Applying the rule when the limit is not indeterminate. If direct substitution gives 1/0 or 2/3, you’re done — the rule does not apply and gives wrong answers if you force it.
• Using the quotient rule instead of differentiating numerator and denominator independently. The rule is f’/g’, not (f/g)'.

Other indeterminate forms. The rule directly handles only 0/0 and ∞/∞. For 0 × ∞, 1^∞, 0^0, ∞^0, and ∞ − ∞, you first rearrange algebraically into 0/0 or ∞/∞ form. For example, lim(x→0⁺) x · ln(x) is 0 × (−∞). Rewrite as ln(x) / (1/x), which is −∞/∞, then apply the rule: derivative of numerator is 1/x, derivative of denominator is −1/x². The ratio is −x, which goes to 0. So lim(x→0⁺) x · ln(x) = 0.

About this calculator. It works with polynomial numerator and denominator of up to degree 3. For each application of the rule, it takes the polynomial derivative: f’(x) has coefficients [3a, 2b, c] from [a, b, c, d].

When direct substitution gives 0/0, the rule is applied. If the result is still 0/0, it applies again. After a third application, if still indeterminate, the limit is reported as undefined for these inputs. In practice that would mean the limit does not exist or requires different techniques (Taylor series, factoring, or substitution).