L’Hôpital’s Rule and Indeterminate Forms

You’ll be able to

Verify that a limit has an indeterminate form before applying L’Hôpital’s Rule
Apply the rule repeatedly when the quotient of derivatives is still indeterminate
Convert the forms 0·∞, ∞ − ∞ and 1^∞ into a quotient the rule can handle

The rule has a precondition, and it is checkable

If lim f(x)/g(x) has the form 0/0 or ∞/∞, then that limit equals lim f′(x)/g′(x), provided the second limit exists. The precondition is not decoration. Applied to a limit that is not indeterminate, the rule produces a confidently wrong answer: lim(x→0) (x + 2)/(x + 1) is plainly 2, but differentiating top and bottom gives 1/1 = 1. Always substitute first and name the form.

Differentiate separately, never as a quotient

L’Hôpital replaces f/g with f′/g′ — the derivatives of the numerator and denominator taken independently. It is not the quotient rule and produces nothing like it. This confusion is common enough that it is worth saying out loud while working: "top derivative over bottom derivative." If the new quotient is still 0/0 or ∞/∞, apply the rule again; some limits need three passes before the form resolves.

Forms that are not quotients yet

The rule only accepts quotients, so the other indeterminate forms have to be rewritten. A product 0·∞ becomes a quotient by moving one factor into the denominator as its reciprocal: x·ln x = ln x / (1/x). A difference ∞ − ∞ usually yields to a common denominator. An exponential form 1^∞, 0⁰ or ∞⁰ is handled by taking ln of the expression, finding the limit of the logarithm, and exponentiating at the end — the last step is the one students forget, leaving the answer as ln L instead of L.

L’Hôpital’s Rule

If lim f/g is 0/0 or ∞/∞, then lim f(x)/g(x) = lim f′(x)/g′(x)

Check the form on every pass, including after each application. Stop as soon as the form is no longer indeterminate.

Worked example

Evaluate lim(x→0) (e^(2x) − 1 − 2x)/x².

1.Substitute x = 0: the numerator is e⁰ − 1 − 0 = 0 and the denominator is 0, so the form is 0/0 and the rule applies.
2.Differentiate top and bottom separately: (2e^(2x) − 2)/(2x).
3.Substitute again: (2 − 2)/0 = 0/0, still indeterminate — apply the rule a second time.
4.Differentiate again: (4e^(2x))/2 = 2e^(2x).
5.Now substitute x = 0: 2e⁰ = 2. The form is no longer indeterminate, so this is the limit.

Answer: The limit is 2. Two applications were needed, and checking the form after the first one is what signals that.

Watch out

Applying the rule to a form that is not indeterminate is a silent error — it produces a number, just not the right one. Substitution and a named form are the first two lines of every one of these problems.

Checkpoint

For which limit is L’Hôpital’s Rule applicable as written?

Checkpoint

To evaluate lim(x→0⁺) x·ln x, which is the form 0·(−∞), the first step is to:

On the exam

BC uses L’Hôpital far beyond limit questions: it is how you show a series term goes to zero, and how improper integrals are evaluated at a bound. The form-checking habit pays off in Units 6 and 10 as much as here.

Answer the 2 checkpoints as you read.