IB Maths AA HLTopic 5 — CalculusPaper 1 & 2HL~11 min read
Limits using Maclaurin Series
Same indeterminate-form limits as l’Hôpital — different attack. Replace f(x) and g(x) by their Maclaurin series, cancel leading powers of x, and the answer often appears in one line. For limits where l’Hôpital takes three or four applications, the series approach often takes one substitution. Either method is fair game in the exam — pick whichever is faster for the limit in front of you.
📘 What you need to know
When to use this method: when lim x→a f(x)/g(x) gives 0/0 or ±∞/±∞ AND you know (or can build) Maclaurin series for f and g around the limit point.
The core trick: substitute the series, cancel the common leading power of x in numerator and denominator, then read off the limit from the remaining constant term.
Limits at x = 0 are natural: Maclaurin series are expanded AROUND x = 0, so this method works perfectly there. The leading nonzero term of each series tells you the “order of vanishing”.
Limits at x → ∞: substitute u = 1/x so that u → 0, then use Maclaurin series in u. This handles ∞/∞ forms involving exponentials, trig, or logs at infinity.
Order matching: include enough terms in each series so the leading nonzero terms of the numerator and denominator both appear. Stopping too early gives a 0/0 result; going too far is wasted work.
The five booklet series are your starting kit: ex, ln(1+x), sin x, cos x, arctan x. For composites (e.g., sin 3x, ln(1+2x)) substitute the inner expression directly into the booklet series.
How it relates to l’Hôpital: both methods agree on every limit where they apply. Maclaurin is usually faster when the function has a clean booklet series; l’Hôpital wins when derivatives are simpler than the series.
Exam phrasing: write “expanding f(x) and g(x) as Maclaurin series gives …” then show the substituted form, simplification, and final limit.
The method — substitute, cancel, read off
The Maclaurin-series approach to limits
lim x→0 f(x)g(x) = lim x→0 [Maclaurin series of f][Maclaurin series of g] ⟶ cancel common power, then substitute x = 0
Both methods give 1/2 for the same limit. l’Hôpital needs to differentiate twice; Maclaurin substitutes the cos x series once and reads the answer off the leading coefficient. For higher-order zeros the Maclaurin advantage grows.
Two scenarios — limit at 0 vs limit at ∞
Limit at x = 0
substitute series directly
Use booklet series (or composite versions) for f and g. The leading powers of x in each series determine the limit’s value or its blow-up rate.
Limit at x → ∞
let u = 1/x
As x → ∞, u → 0, so the limit becomes a Maclaurin limit in u around 0. Re-express everything in terms of u before substituting series.
Order of vanishing — quick diagnostic: write the leading nonzero term of each series. If numerator leads with axm and denominator with bxn: m = n ⟹ limit = a/b; m > n ⟹ limit = 0; m < n ⟹ limit is ±∞.
🧭 Recipe — evaluating a limit using Maclaurin series
Confirm indeterminate form: substitute the limit point directly; verify you get 0/0 or ±∞/±∞. If not, no special method is needed.
For x → ∞ only: substitute u = 1/x so the limit becomes u → 0. Rewrite the whole expression in u.
Write Maclaurin series for f and g: use the booklet series. For composites like sin 3x or ln(1 + 2x), substitute the inner expression into the booklet template.
Simplify the quotient: combine constants, cancel the common leading power of x from numerator and denominator. The result should be a series with a finite (or zero) constant term.
Take the limit: substitute x = 0 (or u = 0). Higher-power terms vanish, leaving the constant term as the answer.
Worked examples
WE 1
Composite exponential series
Use Maclaurin series to evaluate lim x→0 e2x − 1x.
Step 1 — check the format x = 0: (e⁰ − 1)/0 = 0/0 ✓Step 2 — substitute 2x into the booklet e^x seriese^x = 1 + x + x²/2! + x³/3! + …⟹ e^(2x) = 1 + (2x) + (2x)²/2 + (2x)³/6 + … = 1 + 2x + 2x² + 4x³/3 + …Step 3 — subtract 1e^(2x) − 1 = 2x + 2x² + 4x³/3 + …Step 4 — divide by x and take limit(e^(2x) − 1)/x = 2 + 2x + 4x²/3 + …as x → 0: all xⁿ terms (n ≥ 1) vanish ⟹ limit = 2lim x→0 e2x − 1x = 2cross-check via l’Hôpital: d/dx(e^(2x) − 1) = 2e^(2x), d/dx(x) = 1, so limit = 2e⁰/1 = 2 ✓. Both routes agree; the series shows the FULL behaviour (the 2 is the leading slope, the 2x is the next correction).
WE 2
Classic trig limit — single series
Use Maclaurin series to evaluate lim x→0 1 − cos xx².
Step 1 — check the format x = 0: (1 − 1)/0² = 0/0 ✓Step 2 — booklet series for cos xcos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …Step 3 — compute numerator1 − cos x = 1 − (1 − x²/2 + x⁴/24 − …) = x²/2 − x⁴/24 + …Step 4 — divide by x² and take limit(1 − cos x)/x² = 1/2 − x²/24 + …as x → 0: higher powers vanish ⟹ limit = 1/2lim x→0 1 − cos xx² = 12would have needed TWO l’Hôpital applications; here one substitution does it. The leading nonzero coefficient of 1 − cos x is 1/2 (at order x²), so dividing by x² leaves 1/2.
WE 3
Using the arctan series
Use the Maclaurin series for arctan x to evaluate lim x→0 arctan x − xx³.
Step 1 — check the format x = 0: (arctan 0 − 0)/0³ = 0/0 ✓Step 2 — booklet series for arctan xarctan x = x − x³/3 + x⁵/5 − x⁷/7 + …Step 3 — subtract x from the seriesarctan x − x = (x − x³/3 + x⁵/5 − …) − x = −x³/3 + x⁵/5 − …Step 4 — divide by x³ and take limit(arctan x − x)/x³ = −1/3 + x²/5 − …as x → 0: limit = −1/3lim x→0 arctan x − xx³ = −13subtracting x kills the leading term of the series, exposing the next coefficient (−1/3 at order x³). Would have needed THREE l’Hôpital applications and computing the third derivative of arctan x — much messier.
WE 4
Composite ln series with corrections
Use Maclaurin series to evaluate lim x→0 ln(1 + 2x) − 2x + 2x²x³.
Step 1 — check the format x = 0: (ln 1 − 0 + 0)/0³ = 0/0 ✓Step 2 — substitute 2x into the ln(1+x) seriesln(1+x) = x − x²/2 + x³/3 − x⁴/4 + …⟹ ln(1+2x) = 2x − (2x)²/2 + (2x)³/3 − (2x)⁴/4 + … = 2x − 2x² + 8x³/3 − 4x⁴ + …Step 3 — combine with −2x + 2x²ln(1+2x) − 2x + 2x² = (2x − 2x²) + 8x³/3 − … − 2x + 2x² = 8x³/3 − 4x⁴ + …(the 2x and −2x² perfectly cancel the first two booklet terms)Step 4 — divide by x³ and take limit[ln(1+2x) − 2x + 2x²]/x³ = 8/3 − 4x + …as x → 0: limit = 8/3lim x→0 ln(1 + 2x) − 2x + 2x²x³ = 83the −2x + 2x² in the numerator was engineered to kill the first two ln(1+2x) terms, leaving the cubic. Three l’Hôpital applications would also work but require differentiating ln(1+2x) three times — slower.
WE 5
Series in BOTH numerator and denominator
Use Maclaurin series to evaluate lim x→0 sin 3xsin 2x.
Step 1 — check the format x = 0: sin 0 / sin 0 = 0/0 ✓Step 2 — substitute into the booklet sin seriessin x = x − x³/3! + x⁵/5! − …sin 3x = 3x − (3x)³/6 + … = 3x − 9x³/2 + …sin 2x = 2x − (2x)³/6 + … = 2x − 4x³/3 + …Step 3 — divide top and bottom by x (the common leading factor)sin 3x / sin 2x = (3x − 9x³/2 + …) / (2x − 4x³/3 + …) = (3 − 9x²/2 + …) / (2 − 4x²/3 + …)Step 4 — take limitas x → 0: → 3/2lim x→0 sin 3xsin 2x = 32both series lead with first-order terms (3x and 2x), so the cancelled limit is just the ratio of leading coefficients: 3/2. Generalises to lim x→0 sin(ax)/sin(bx) = a/b for any a, b ≠ 0.
WE 6
Limit at infinity via u = 1/x substitution
Use Maclaurin series to evaluate lim x→∞ x(e3/x − 1).
Step 1 — identify formas x → ∞: 3/x → 0, so e^(3/x) → 1, x → ∞⟶ ∞ · 0 indeterminate ✓Step 2 — substitute u = 1/x (so u → 0+)x = 1/u, so x(e^(3/x) − 1) = (1/u)(e^(3u) − 1) = (e^(3u) − 1) / uStep 3 — Maclaurin series for e^(3u)e^(3u) = 1 + 3u + (3u)²/2 + (3u)³/6 + … = 1 + 3u + 9u²/2 + 9u³/2 + …e^(3u) − 1 = 3u + 9u²/2 + 9u³/2 + …Step 4 — divide by u and take u → 0(e^(3u) − 1) / u = 3 + 9u/2 + 9u²/2 + …as u → 0: limit = 3lim x→∞ x(e3/x − 1) = 3substitution u = 1/x turns the infinity limit into a Maclaurin-friendly limit at 0. The “magic” factor of 3 is just the coefficient of u in e^(3u) − 1. Generalises: lim x→∞ x(e^(k/x) − 1) = k for any constant k.
💡 Top tips
Keep enough terms — and not too many: include terms up to and including the highest power that appears in the denominator. Anything past that cancels out as x → 0.
Spot the order-of-vanishing pattern: if f(x) leads with axm and g(x) with bxn, the limit is a/b if m = n, 0 if m > n, and ±∞ if m < n.
Composite substitutions: for sin(kx), ekx, ln(1 + kx), cos(x²), etc., substitute the inner expression directly into the booklet series — no need to re-derive from first principles.
For x → ∞: the substitution u = 1/x almost always works when the function involves e1/x, sin(1/x), etc. Just track that u → 0+ for x → ∞ and u → 0− for x → −∞.
Choose method by speed: if the function has a clean booklet series, Maclaurin usually wins. If derivatives are simple but series are messy (e.g., for rational functions), l’Hôpital wins. Both are valid.
⚠ Common mistakes
Truncating too early: if you stop the series before the leading nonzero term reaches the denominator’s power, you’ll get a 0/0 result and have to redo. Always check the order of vanishing before truncating.
Forgetting the chain rule when substituting: ln(1 + 2x) is NOT 2x − x²/2 + x³/3 − … — that’s ln(1+x) with x doubled. You must substitute 2x everywhere: ln(1+2x) = 2x − (2x)²/2 + (2x)³/3 − … = 2x − 2x² + 8x³/3 − …
Mishandling x → ∞ without substitution: Maclaurin series don’t converge for large x. You MUST substitute u = 1/x first so the new variable goes to 0, then use the series in u.
Algebra slips when expanding (kx)n: (2x)³ = 8x³, not 2x³; (3x)² = 9x², not 3x². Easy slips that completely change the answer.
Confusing the series for f with the series for f/xn: divide the FULL numerator series by the denominator AFTER you’ve expanded both. Don’t pre-divide individual terms without keeping track.
🎉 You’ve finished Topic 5 — Calculus! That’s all of HL Calculus done: limits, differentiation (first principles through implicit and parametric), integration (including parts and substitution), differential equations, Maclaurin series, l’Hôpital, and this final note tying it all together. The two limit methods (l’Hôpital and Maclaurin) are interchangeable on the exam — choose whichever is cleaner for each specific limit. Next up: tying together the remaining Topic 4 (Probability) opener and the AA HL hub page rebuild.
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