L’Hôpital’s Rule
- Recognize the indeterminate forms 0/0 and ∞/∞
- Apply L’Hôpital’s Rule to evaluate qualifying limits
- Avoid misapplying the rule to non-indeterminate limits
When substitution gives 0/0 or ∞/∞
Some limits produce an indeterminate form — 0/0 or ∞/∞ — that algebra alone struggles with, especially when the functions are not polynomials. L’Hôpital’s Rule provides a derivative-based tool: if lim f(x)/g(x) gives 0/0 or ∞/∞, then it equals lim f′(x)/g′(x), provided that second limit exists. You differentiate the top and bottom separately — this is not the quotient rule.
Check the form first, every time
L’Hôpital’s Rule applies only to the forms 0/0 and ∞/∞. Before using it you must verify by substitution that you actually have one of these. Applying it to a limit like 3/0 or 5/2 gives a wrong answer, because the rule’s hypothesis is not met. If a first application still yields 0/0, you may apply the rule again.
Evaluate lim(x→0) (e^(2x) − 1)/x.
- 1.Substitute x = 0: numerator e⁰ − 1 = 0, denominator 0 — the 0/0 indeterminate form, so L’Hôpital applies.
- 2.Differentiate the top: d/dx[e^(2x) − 1] = 2e^(2x).
- 3.Differentiate the bottom: d/dx[x] = 1.
- 4.Take the new limit: lim(x→0) 2e^(2x)/1 = 2e⁰ = 2.
L’Hôpital differentiates the numerator and denominator independently: (f/g)′ is not what you compute. And never apply it to a determinate form like 4/0 — first confirm the limit is genuinely 0/0 or ∞/∞.
For which limit is L’Hôpital’s Rule directly applicable?
Evaluate lim(x→0) sin(3x)/x using L’Hôpital’s Rule.
Always write the phrase "form is 0/0" (or ∞/∞) before applying L’Hôpital on free response. Stating the indeterminate form is part of a valid justification; jumping straight to differentiating the pieces can cost the point even with a correct answer.
Answer the 2 checkpoints as you read.
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