Mathematics Advanced • Year 11 • Module 3 • Lesson 2

Limits

Apply limit evaluation to multi-step problems involving secant gradients, removable discontinuities, piecewise pricing models and limit laws.

Apply · Problem Set

Problem 1 — Bridging a hole in the graph

Consider the function

f(x) = (x² − 4) / (x − 2), x ≠ 2.

Set up: What are we solving for?

(i) Show that direct substitution gives 0/0, and use factor-and-cancel to evaluate lim_x→2 f(x). 2 marks

(ii) Sketch (in the space) the graph of f near x = 2, marking the hole at x = 2 with an open circle and the limit value with a small horizontal arrow. 2 marks

(iii) Define a new function g(x) that agrees with f for x ≠ 2 but is continuous everywhere. State the single extra rule needed and justify in one sentence. 2 marks

Stuck? Revisit lesson § Removable discontinuity.

Problem 2 — A piecewise pricing model

An electricity provider charges a unit price per kWh that changes at usage x = 200 kWh per month:

P(x) = { 0.32 x if x ≤ 200 ; 64 + 0.40(x − 200) if x > 200 }

(P in dollars per month, x in kWh.)

Set up: What are we solving for?

(i) Compute lim_x→200⁻ P(x) and lim_x→200⁺ P(x). 2 marks

(ii) Hence state whether lim_x→200 P(x) exists, and whether P is continuous at x = 200. 2 marks

(iii) The company wants the bill to be continuous at x = 200 (no sudden jump). They keep the 0.32 rate below 200 kWh but propose changing the upper-bracket constant. What value should replace the 64 in the upper formula to guarantee continuity, and why? 2 marks

Problem 3 — A secant gradient as a limit

For f(x) = x², the gradient of the secant joining the points (3, 9) and (3 + h, f(3 + h)) where h ≠ 0 can be written

m(h) = ( f(3 + h) − f(3) ) / h.

Set up: What are we solving for?

(i) Show that m(h) simplifies to 6 + h (for h ≠ 0), showing every line of algebra. 2 marks

(ii) Hence evaluate lim_h→0 m(h). State what this limit represents geometrically. 2 marks

(iii) Tabulate m(h) for h = 0.1, 0.01, 0.001 and confirm the numerical values match your algebraic answer to (ii). 2 marks

Stuck? You did the same algebra in Lesson 1 Worksheet 2 — but now framed as a limit.

Problem 4 — Limit laws and a quotient with a hole

You are given lim_x→2 f(x) = 3 and lim_x→2 g(x) = 5.

Set up: What are we solving for?

(i) Using only the limit laws (no specific f, g formula), evaluate each of:

(a) lim_x→2 [ f(x) + 2 g(x) ] = ____________ (b) lim_x→2 [ f(x) · g(x) ] = ____________

(ii) Why does the quotient law fail to apply to lim_x→3 [ (x − 3) / (x² − 9) ] by direct substitution? Show how to rearrange before applying limit laws, then evaluate. 3 marks

(iii) A student writes lim_x→0 [ x · (1/x) ] = lim_x→0 x · lim_x→0 (1/x) and concludes the limit is "0 × ∞ = 0". Explain in one sentence why this is wrong and state the correct value. 2 marks

Problem 5 — Choose the constant to fit the function

Consider the piecewise function

f(x) = { x² + k if x ≤ 1 ; 3x if x > 1 }.

Set up: What are we solving for?

(i) Find lim_x→1⁻ f(x) and lim_x→1⁺ f(x) in terms of k. 2 marks

(ii) Hence find the value of k for which lim_x→1 f(x) exists, and state the value of the limit for that k. 2 marks

(iii) Using that value of k, is f continuous at x = 1? Justify by checking both: (a) the limit exists and (b) the limit equals f(1). 2 marks

Stuck on (iii)? f(1) uses the top branch (≤ 1). Compare against the limit you found in (ii).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Problem 1 — Removable hole at x = 2

Set up. We confirm that f has a removable discontinuity at x = 2, find the natural value of the limit there, and patch the hole to make a continuous extension.

(i) Substituting x = 2: (4 − 4)/(2 − 2) = 0/0. Factor: x² − 4 = (x − 2)(x + 2); cancel (x − 2) (for x ≠ 2): expression = x + 2. lim_x→2 (x + 2) = 4.

(ii) The graph is the line y = x + 2 everywhere except an open circle at (2, 4). Sketch should show a straight line of gradient 1 passing through (0, 2), with a small open circle at (2, 4) and arrows from both sides converging to that y-value.

(iii) Define g(x) = f(x) for x ≠ 2 and g(2) = 4. Justification: this single redefinition is exactly the limit value, so the left limit, right limit and function value all agree — continuity restored.

Problem 2 — Electricity tariff piecewise

Set up. Check whether the two branches of the tariff meet at the boundary x = 200. If they do, the limit exists and the bill is continuous; if not, the customer sees a jump in the bill at the bracket boundary.

(i) Left limit: lim_x→200⁻ 0.32x = 0.32(200) = $64. Right limit: lim_x→200⁺ [64 + 0.40(x − 200)] = 64 + 0.40(0) = $64.

(ii) Left = right = 64, so lim_x→200 P(x) exists and equals $64. Since P(200) = 0.32(200) = 64 (top branch, x ≤ 200), the limit equals the function value, so P is continuous at x = 200.

(iii) The 64 in the upper formula was chosen specifically to equal 0.32 × 200, the value of the lower branch at x = 200. Any other constant would break the matching at the boundary and produce a jump discontinuity in the bill (the customer would see a sudden jump in their monthly charge). So the constant must remain 64; any change would cause a discontinuity.

Problem 3 — Secant gradient of x² at x = 3

Set up. Compute the gradient of the secant joining (3, 9) and (3 + h, f(3 + h)) as a function of h, then take the limit as h shrinks.

(i) f(3 + h) = (3 + h)² = 9 + 6h + h². m(h) = ((9 + 6h + h²) − 9) / h = (6h + h²) / h = h(6 + h) / h = 6 + h (h ≠ 0).

(ii) lim_h→0 (6 + h) = 6. Geometrically: this is the gradient of the tangent to y = x² at the point (3, 9) — i.e., the instantaneous rate of change of x² at x = 3.

(iii) m(0.1) = 6.1; m(0.01) = 6.01; m(0.001) = 6.001 — all clearly approaching 6, confirming the algebraic answer in (ii).

Problem 4 — Limit laws

Set up. Apply each law term-by-term, then identify where direct substitution fails.

(i) (a) 3 + 2(5) = 13. (b) 3 × 5 = 15. (c) 5 / 3 = 5/3. (d) 3³ = 27.

(ii) Direct substitution: (3 − 3) / (9 − 9) = 0/0. The quotient law requires lim of denominator ≠ 0, so it does not apply directly. Rearrange: (x − 3) / (x² − 9) = (x − 3) / ((x − 3)(x + 3)) = 1 / (x + 3) (for x ≠ 3). Now lim_x→3 1 / (x + 3) = 1 / 6 (quotient law applies, denominator is 6 ≠ 0). Answer: 1/6.

(iii) The product law lim[f · g] = lim f · lim g requires both individual limits to exist. lim_x→0 (1/x) does not exist (it heads to +∞ on one side, −∞ on the other), so the law does not apply. The actual value of x · (1/x) is 1 for every x ≠ 0, so lim_x→0 [x · (1/x)] = lim_x→0 1 = 1.

Problem 5 — Choose k for the piecewise function

Set up. Match the two one-sided limits at x = 1, then choose k to make them agree (existence). Finally compare against f(1) to check continuity.

(i) Left: lim_x→1⁻ (x² + k) = 1 + k. Right: lim_x→1⁺ 3x = 3.

(ii) Limit exists ⇔ 1 + k = 3, so k = 2. With k = 2, lim_x→1 f(x) = 3.

(iii) With k = 2, f(1) = 1² + 2 = 3 (top branch). Both conditions: (a) limit exists and equals 3 ✓; (b) limit equals f(1) = 3 ✓. Hence f is continuous at x = 1.