IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~10 min read

Introduction to Limits

A limit describes what a function GETS CLOSE TO as the input approaches a value — which may be a finite number or ±∞. The function does NOT have to be defined at the target value; that’s the whole point. Three techniques handle nearly every paper question: direct substitution when it’s nice, algebraic simplification (factor and cancel, or rationalise) when you hit 0/0, and divide by the highest power when x heads to infinity.

📘 What you need to know

Limit notation: lim_{x → a} f(x) = L means “f(x) gets arbitrarily close to L as x gets arbitrarily close to a” — even if f(a) is undefined.
Step 1 always: try direct substitution. If f(a) is a clean value, that’s the limit.
0/0 means “try harder”, not “no limit”. Factor and cancel, or multiply by the conjugate, to remove the common zero.
For lim_{x → ±∞}, divide top and bottom by the highest power of x. Each k/xⁿ term then vanishes.
For rational lim at infinity: degree(num) < degree(den) → 0; equal → ratio of leading coefficients; greater → diverges to ±∞.
One-sided limits: lim_{x → a⁻} (from below) and lim_{x → a⁺} (from above). The two-sided limit exists ⇔ both one-sided limits exist AND are equal.
For exponential limits: e^−x → 0 as x → ∞. Dividing by e^x turns large exponentials into tame constants.
Indeterminate forms like 0/0 and ∞/∞ aren’t answers — they’re a flag that more work (or l’Hôpital, later) is needed.

The three core techniques

The vast majority of paper questions yield to one of these three approaches. Pick the right tool from the form of the function.

Direct substitution lim_{x → a} f(x) = f(a) (when f(a) is defined and the function has no jump there)

When direct substitution gives 0/0, the limit might still exist — but you need to clean up the function first.

Factor & cancel

e.g. x² − 9x − 3 = (x + 3) for x ≠ 3

when both top and bottom share a factor of (x − a), cancel it, then substitute

Rationalise

e.g. multiply by √(x+a) + √a√(x+a) + √a

when there’s a surd in numerator or denominator, multiply by the conjugate to remove it

Limits at infinity — divide by highest power lim_{x → ±∞} kxⁿ = 0 for all k ∈ ℝ and n > 0

For rational functions at infinity, the answer falls out instantly from comparing degrees: if deg(num) < deg(den), limit is 0; if equal, limit is the ratio of leading coefficients; if greater, the function diverges. Always know which case you’re in before doing algebra.

One-sided limits — and when the limit doesn’t exist

The notation x → a⁺ means “as x approaches a from the right” (from above; x > a). Likewise x → a⁻ means “from the left” (x < a). The two-sided limit exists if and only if both one-sided limits exist AND agree.

Left: a smooth function where the limit from below and limit from above both approach the same value L — the two-sided limit exists. Right: a piecewise function with a JUMP at x = c; the two one-sided limits give different values, so there’s no single two-sided limit.

🧭 Recipe — evaluate a limit

Try direct substitution first. If f(a) gives a clean value, you’re done — that’s the limit.
If you get 0/0, try to simplify algebraically: factor numerator and denominator and cancel the common factor, or multiply by the conjugate of a surd expression.
If x → ±∞, divide numerator and denominator by the highest power of x appearing, then apply k/xⁿ → 0.
For one-sided limits, use only the relevant side of any piecewise definition. The two-sided limit exists only if both sides agree.
If you still get 0/0 or ∞/∞ after trying the above, the limit needs l’Hôpital’s Rule or Maclaurin series (later HL notes).

Worked examples

WE 1

Direct substitution — when the function is well-behaved

Find lim_{x → 2} x² − 3x + 72x + 1.

Step 1 — try direct substitution f(2) = (4 − 6 + 7)/(4 + 1) = 5/5 = 1 f(2) is well-defined, so the limit equals it lim = 1 when the function is continuous at the target, the limit is just the value. No fancy work needed — and that’s the whole point of direct substitution.

WE 2

Factor and cancel — escaping the 0/0 form

Find lim_{x → 3} x² − 5x + 6x² − 9.

Step 1 — direct substitution num(3) = 9 − 15 + 6 = 0 den(3) = 9 − 9 = 0 0/0 — indeterminate. Try simplification. Step 2 — factor both num: x² − 5x + 6 = (x − 2)(x − 3) den: x² − 9 = (x − 3)(x + 3) Step 3 — cancel the common (x − 3) f(x) = (x − 2)/(x + 3) for x ≠ 3 Step 4 — substitute into the simplified form at x = 3: (3 − 2)/(3 + 3) = 1/6 lim = 1/6 the original function is UNDEFINED at x = 3 (denominator is 0), but the limit exists because the shared factor (x − 3) cancels.

WE 3

Rationalisation — surds and the conjugate trick

Find lim_{x → 0} √(x + 4) − 2x.

Step 1 — direct substitution num: √4 − 2 = 0; den: 0 — gives 0/0 Step 2 — multiply by the conjugate (√(x+4) + 2)/(√(x+4) + 2) num: (√(x+4) − 2)(√(x+4) + 2) = (x + 4) − 4 = x den: x · (√(x+4) + 2) f(x) = x / [x · (√(x+4) + 2)] Step 3 — cancel x (valid for x ≠ 0) f(x) = 1 / (√(x+4) + 2) Step 4 — substitute x = 0 into the simplified form = 1/(√4 + 2) = 1/(2 + 2) = 1/4 lim = 1/4 the conjugate trick converts “√(stuff) − constant” into “stuff − constant²” using (a − b)(a + b) = a² − b² — and that polynomial form usually shares the cancellable factor you need.

WE 4

Limit at infinity — divide by the highest power

Find lim_{x → ∞} 7x³ − 2x + 53x³ + x² − 4.

Step 1 — identify the highest power: x³ degrees match (both cubic) → expect a finite, non-zero limit Step 2 — divide every term by x³ num: (7x³ − 2x + 5)/x³ = 7 − 2/x² + 5/x³ den: (3x³ + x² − 4)/x³ = 3 + 1/x − 4/x³ f(x) = (7 − 2/x² + 5/x³) / (3 + 1/x − 4/x³) Step 3 — apply k/xⁿ → 0 as x → ∞ = (7 − 0 + 0) / (3 + 0 − 0) = 7/3 lim = 7/3 shortcut: for a rational function at infinity with equal degrees, the limit is just the ratio of the leading coefficients — here, 7/3. The full algebra is shown for transparency.

WE 5

One-sided limits at a jump — piecewise function

The function f is defined piecewise as f(x) = 2x² + 1 for x < 1, and f(x) = 5 − 3x for x ≥ 1. Find (a) lim_{x → 1⁻} f(x), (b) lim_{x → 1⁺} f(x), and (c) state whether the two-sided limit lim_{x → 1} f(x) exists, justifying your answer.

(a) limit from below: use the x < 1 piece lim x→1⁻ (2x² + 1) = 2(1)² + 1 = 3 (b) limit from above: use the x ≥ 1 piece lim x→1⁺ (5 − 3x) = 5 − 3 = 2 (c) compare the two one-sided limits 3 ≠ 2 → the two-sided limit DOES NOT EXIST (a) 3 (b) 2 (c) does not exist (one-sided limits differ) the function jumps from 3 to 2 at x = 1, so although f(1) = 2 (using the right piece), the function is NOT continuous there. Two-sided limits demand agreement from both sides.

WE 6

Limit at infinity with an exponential

Find lim_{x → ∞} e^x − 32e^x + 1.

Step 1 — note that e^x → ∞ as x → ∞ direct substitution gives ∞/∞ — indeterminate Step 2 — divide top and bottom by e^x (the dominant term) num: (e^x − 3)/e^x = 1 − 3·e^(−x) den: (2e^x + 1)/e^x = 2 + e^(−x) f(x) = (1 − 3·e^(−x)) / (2 + e^(−x)) Step 3 — apply e^(−x) → 0 as x → ∞ = (1 − 0)/(2 + 0) = 1/2 lim = 1/2 for exponentials, dividing by the FASTEST-GROWING term (here e^x) is the analogue of “divide by highest power of x” — every other term decays to 0 against it.

💡 Top tips

Try direct substitution FIRST — it solves the easy cases instantly, and even when it fails, the form of the failure (0/0 vs ∞/∞ vs nonzero/0) tells you the next step.
0/0 ≠ “no limit”. It just means the function isn’t defined there. The limit often exists; you just have to simplify.
For limits at infinity, identify the dominant term (highest power of x, or fastest-growing exponential), then divide top and bottom by it. Every other term collapses to 0.
For rational functions at infinity, the answer is just degree comparison — memorise the three cases (less, equal, greater) and skip the algebra when allowed.
Always check both sides when computing a two-sided limit near a piecewise boundary or a vertical asymptote. They might disagree.

⚠ Common mistakes

Writing “0/0” as the final answer — that’s an INDETERMINATE form, not a value. It means you need to simplify or use a more advanced technique.
Saying “limit doesn’t exist” after only direct substitution failed — most 0/0 cases have a perfectly fine limit; you just haven’t simplified yet.
For limits at infinity, “plugging in ∞” without dividing first — gives nonsense like ∞ − ∞ or ∞/∞. Divide by the highest power BEFORE evaluating.
Using the wrong piece of a piecewise function for a one-sided limit — for x → a⁻ use the piece for x < a; for x → a⁺ use the piece for x > a (or x ≥ a).
Forgetting that f(a) and lim_{x → a} f(x) can differ — even when both exist. The limit ignores the actual value at a; it only cares about what f(x) approaches near it.

Up next: Continuity & Differentiability. Now that you can compute limits, you can FORMALLY say what it means for a function to be continuous at a point: lim_{x → a⁻} f(x) = lim_{x → a⁺} f(x) = f(a). And differentiability is stronger still — a corner (like the vertex of |x|) is continuous but NOT differentiable. The implications go one way only: differentiable ⟹ continuous, but the converse is false.

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