IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~10 min read
l’Hôpital’s Rule
When a limit lim x→a f(x)/g(x) gives you the indeterminate form 00 or ±∞±∞, l’Hôpital’s rule lets you replace the quotient by f′(x)/g′(x) and try the limit again. If it’s still indeterminate, you differentiate again — and keep going until the limit settles. Calculation-heavy, mechanical, exam-friendly. Here’s exactly how to drive it.
📘 What you need to know
When the rule applies: ONLY when direct substitution gives the indeterminate form 00 or ±∞±∞. Other forms like 0·∞ or ∞ − ∞ must be rewritten first into one of these.
The rule: if lim x→a f(x)/g(x) is 00 or ±∞±∞ and the new limit lim x→a f′(x)/g′(x) exists, then they’re equal.
Differentiate top and bottom SEPARATELY — this is NOT the quotient rule. You’re differentiating f and g individually, not the whole fraction.
Always check the indeterminate form first: substitute x = a into the original quotient and confirm you get 0/0 or ±∞/±∞. If the limit isn’t indeterminate, l’Hôpital doesn’t apply (and you’d get a wrong answer).
Apply repeatedly: if f′(a)/g′(a) is still indeterminate, differentiate again. Continue until the limit can be evaluated directly. Each application requires the indeterminate-form check to be re-confirmed.
Works for x → ∞ too: not just for finite a. The same procedure applies — check the form, differentiate, retry.
Handle hidden indeterminate forms: 0·∞ converts to 0/0 or ∞/∞ via rewriting (e.g., x·ln x = ln x / (1/x)); ∞ − ∞ converts via common denominator.
State each step in exam answers: write “indeterminate form 0/0, apply l’Hôpital’s rule” before each differentiation. Examiners want to see the form-check explicitly.
The rule and why it works geometrically
l’Hôpital’s rule — formula booklet
If lim x→af(x)g(x) = 00 or ±∞±∞, then lim x→af(x)g(x) = lim x→af′(x)g′(x)
Both curves vanish at x = a. Near a, each curve hugs its tangent line, so f(x) ≈ f′(a)(x−a) and g(x) ≈ g′(a)(x−a). The (x−a) factors cancel, leaving f′(a)/g′(a). That’s the rule.
Indeterminate forms — what counts
Direct cases
00 or ±∞±∞
Apply l’Hôpital straight away: differentiate top and bottom, take the limit.
Hidden cases — rewrite first
0·∞ or ∞ − ∞
Convert algebraically into 0/0 or ∞/∞, THEN apply l’Hôpital. Example: x·ln x = ln x / (1/x) is now ∞/∞ form.
Forms l’Hôpital does NOT cover directly: 1∞, 00, ∞0. For these, take ln of the expression, evaluate the limit, then exponentiate back.
Repeated applications — when one differentiation isn’t enough
After differentiating once, if the new quotient f′/g′ still gives an indeterminate form at x = a, differentiate AGAIN — taking the derivative of f′ to get f″, and the derivative of g′ to get g″ — and check the form one more time. The classic x²/ex → ∞/∞ needs two passes; (sin x − x)/x³ needs three. Each round requires a fresh form-check before you’re allowed to differentiate again.
🧭 Recipe — applying l’Hôpital’s rule
Check the form: substitute x = a (or take the limit as x → ∞) and confirm you get 0/0 or ±∞/±∞. If not indeterminate, l’Hôpital does NOT apply — go back to algebraic methods.
Convert hidden forms: if you have 0·∞ or ∞−∞, rewrite as a single quotient that is 0/0 or ∞/∞ before differentiating.
Differentiate top and bottom separately: compute f′(x) and g′(x). Do NOT use the quotient rule — top and bottom are independent.
Take the new limit: evaluate lim x→a f′(x)/g′(x). If it’s a finite number or ±∞, that’s your answer.
Repeat if still indeterminate: if f′(a)/g′(a) is still 0/0 or ±∞/±∞, differentiate again (now f″, g″) and re-check. Continue until the limit can be evaluated directly.
Worked examples
WE 1
Single application — 0/0
Use l’Hôpital’s rule to evaluate lim x→0 e3x − 12x.
Step 1 — check the format x = 0: (e⁰ − 1)/(2·0) = (1 − 1)/0 = 0/0 ⟶ indeterminate ✓Step 2 — differentiate top and bottom separatelyd/dx (e^(3x) − 1) = 3 e^(3x)d/dx (2x) = 2Step 3 — take the new limitlim x→0 3e^(3x) / 2 = 3·e⁰ / 2 = 3/2limit exists — done.lim x→0 e3x − 12x = 32single application suffices because the new quotient gives 3/2 at x = 0 (no longer indeterminate). Sanity check via series: e^(3x) − 1 ≈ 3x + (3x)²/2 + …, so (e^(3x)−1)/(2x) ≈ 3/2 + 9x/4 + … → 3/2 at x = 0 ✓.
WE 2
Two applications — trig limit, 0/0
Use l’Hôpital’s rule to evaluate lim x→0 1 − cos 2xx².
Step 1 — check the format x = 0: (1 − cos 0)/0² = (1 − 1)/0 = 0/0 ✓ indeterminateStep 2 — differentiate top and bottomd/dx (1 − cos 2x) = 2 sin 2xd/dx (x²) = 2xStep 3 — new limitlim x→0 (2 sin 2x)/(2x) = (sin 2x)/xat x = 0: sin 0 / 0 = 0/0 ⟶ STILL indeterminateStep 4 — apply l’Hôpital AGAINd/dx (sin 2x) = 2 cos 2xd/dx (x) = 1lim x→0 (2 cos 2x)/1 = 2 cos 0 = 2lim x→0 1 − cos 2xx² = 2two applications needed. Cross-check via Maclaurin: 1 − cos 2x = 1 − (1 − (2x)²/2! + …) = 2x² − …, so the quotient ≈ 2x²/x² = 2 ✓. Compare with the standard limit (1 − cos x)/x² → 1/2 — the factor of 4 from “2x” inside cos gives the answer 2.
WE 3
Limit at infinity — ∞/∞ form
Use l’Hôpital’s rule to evaluate lim x→∞ x²ex.
Step 1 — check the formas x → ∞: x² → ∞, e^x → ∞⟶ ∞/∞ indeterminate ✓Step 2 — differentiate top and bottomd/dx (x²) = 2xd/dx (e^x) = e^xStep 3 — new limitlim x→∞ 2x / e^xstill ∞/∞ ⟶ apply l’Hôpital againStep 4 — differentiate againd/dx (2x) = 2d/dx (e^x) = e^xlim x→∞ 2 / e^x = 2/∞ = 0lim x→∞ x²ex = 0classic “exponential beats polynomial” result. By induction, lim x→∞ x^n / e^x = 0 for ANY positive integer n — keep applying l’Hôpital until the polynomial reduces to a constant, while e^x stays e^x. This generalises to ANY polynomial divided by ANY exponential going to infinity.
WE 4
Three applications — a classic case
Use l’Hôpital’s rule to evaluate lim x→0 sin x − xx³.
Step 1 — check the format x = 0: (0 − 0)/0 = 0/0 ✓Step 2 — first differentiationd/dx (sin x − x) = cos x − 1; d/dx (x³) = 3x²at x = 0: (1 − 1)/0 = 0/0 ⟶ STILL indeterminateStep 3 — second differentiationd/dx (cos x − 1) = −sin x; d/dx (3x²) = 6xat x = 0: 0/0 ⟶ STILL indeterminateStep 4 — third differentiationd/dx (−sin x) = −cos x; d/dx (6x) = 6at x = 0: (−cos 0)/6 = −1/6 ✓ existslim x→0 sin x − xx³ = −16three applications because the numerator’s zero is “third-order”: sin x − x = −x³/6 + x⁵/120 − … vanishes to order x³ at 0. Via Maclaurin: (sin x − x)/x³ ≈ (−x³/6)/x³ = −1/6 ✓. The general lesson: the number of applications matches the order of the zero in the numerator (when the denominator is a single power).
WE 5
Limit at a finite a ≠ 0
Use l’Hôpital’s rule to evaluate lim x→2 x³ − 8x² − 4.
Step 1 — check the form (NOT at 0!)at x = 2: (8 − 8)/(4 − 4) = 0/0 ✓Step 2 — differentiate top and bottomd/dx (x³ − 8) = 3x²d/dx (x² − 4) = 2xStep 3 — new limit at x = 2lim x→2 3x² / 2x = 3(2)² / [2(2)] = 12/4 = 3lim x→2 x³ − 8x² − 4 = 3algebraic cross-check: factor — (x³ − 8) = (x − 2)(x² + 2x + 4); (x² − 4) = (x − 2)(x + 2). Cancel (x − 2): quotient → (x² + 2x + 4)/(x + 2) → (4 + 4 + 4)/4 = 12/4 = 3 ✓. Either method works; l’Hôpital is faster when factoring is hard.
WE 6
Hidden form — 0·∞ rewritten as ∞/∞
Use l’Hôpital’s rule to evaluate lim x→0+x ln x.
Step 1 — identify the indeterminate formas x → 0⁺: x → 0, ln x → −∞⟶ 0 · (−∞) ⟶ NOT a direct l’Hôpital formStep 2 — rewrite as a quotientx ln x = (ln x) / (1/x)as x → 0⁺: ln x → −∞, 1/x → +∞⟶ −∞/∞ ✓ now l’Hôpital appliesStep 3 — differentiate top and bottomd/dx (ln x) = 1/xd/dx (1/x) = −1/x²Step 4 — simplify and take limit(1/x) / (−1/x²) = (1/x) · (−x²) = −xlim x→0⁺ (−x) = 0lim x→0+x ln x = 0key trick: 0·∞ becomes 0/0 or ∞/∞ by putting one factor in the denominator (use 1/x or 1/f(x) for whichever is easier to differentiate). Here putting ln x on top and 1/x on bottom gave the simpler algebra. Same idea works for limits like x² · e^(−x) → 0 and √x · ln x → 0.
💡 Top tips
Always announce the form first: write “form 0/0, apply l’Hôpital” before each round of differentiation. Examiners want to see the check, and it stops you from accidentally applying the rule when you shouldn’t.
Simplify between applications: after the first differentiation, cancel common factors in the new quotient before applying l’Hôpital again. Often a factor of x, sin x, or ex cancels and saves a derivative.
Compare with the Maclaurin-series method: if you can write Maclaurin series for f and g, dividing the leading terms often gives the answer instantly — sometimes faster than l’Hôpital. Both methods are in the syllabus.
Polynomial / exponential: any polynomial divided by ex (or any exponential with base > 1) tends to 0 as x → ∞. The number of l’Hôpital applications needed equals the polynomial degree.
For 0·∞: rewrite as a quotient. Which factor to put on top? Pick the one that’s easier to differentiate (usually the algebraic factor, putting log/trig in the numerator).
⚠ Common mistakes
Applying the quotient rule instead of differentiating separately: l’Hôpital is NOT (f/g)′ = (f′g − fg′)/g². You differentiate top and bottom INDEPENDENTLY: top becomes f′, bottom becomes g′. Two separate derivatives, no quotient rule.
Applying l’Hôpital when the form isn’t indeterminate: if the limit gives a number like 5/2 or 0/3 or 7/0, l’Hôpital is INVALID. Substituting directly is already the answer (or the limit is ±∞). Always check the form first.
Forgetting to re-check after each application: once the new quotient is no longer 0/0 or ±∞/±∞, STOP differentiating. Applying l’Hôpital one extra time will give a WRONG answer.
Mishandling 0·∞ and ∞−∞ forms: these need ALGEBRAIC rewriting before l’Hôpital. You can’t differentiate a product or difference using the rule — only a quotient that’s 0/0 or ∞/∞.
Sign slips in trig derivatives: cos differentiates to −sin (a sign appears); sin differentiates to cos (no sign). When applying l’Hôpital multiple times to trig limits, track signs carefully — one slip flips the answer.
Up next: Limits using Maclaurin Series. The same indeterminate-form limits can often be cracked faster by substituting Maclaurin series for f(x) and g(x), then simplifying. For many limits you can read off the answer just by looking at the leading terms — no repeated differentiation required. This is the final note of the Calculus topic.
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