Mathematics Advanced • Year 11 • Module 3 • Lesson 2

Limits

HSC-style writing on limits — substitution, factor-and-cancel, one-sided checks and an extended response on continuity.

Master · Past-Paper Style

1. Short-answer questions

1.1 Evaluate lim_x→4 (x² − 16) / (x − 4). Show full algebraic working. 2 marks Band 3

1.2 For the piecewise function f(x) = { 2x + 1 if x ≤ 1 ; 4 − x if x > 1 }, find each of:
(a) lim_x→1⁻ f(x), (b) lim_x→1⁺ f(x), (c) state whether lim_x→1 f(x) exists, and (d) state whether f is continuous at x = 1. 3 marks Band 3-4

1.3 Consider lim_x→0 ( (x + 2)² − 4 ) / x.
(a) Show that direct substitution gives the indeterminate form 0/0.
(b) Expand and simplify to evaluate the limit algebraically. 4 marks Band 4

Stuck on 1.3(b)? Expand (x + 2)² = x² + 4x + 4, then subtract 4 and divide every term by x.

2. Extended response

2.1 Consider the function

f(x) = { (x² − 9) / (x − 3) if x ≠ 3 ; k if x = 3 }.

(a) State the indeterminate form encountered when x = 3 is substituted into (x² − 9) / (x − 3).
(b) Evaluate lim_x→3 (x² − 9) / (x − 3), showing the factor-and-cancel step explicitly.
(c) Hence find the value of k that makes f continuous at x = 3.
(d) Sketch the graph of f for the value of k from (c), labelling clearly the value at x = 3 and the surrounding linear behaviour.
(e) Explain in 2 or 3 sentences why this discontinuity is called "removable", and what (if anything) goes wrong if we instead pick k = 5. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 1 mark

• 1 mark — explicitly states 0/0 with substitution shown.

Part (b) — 2 marks

• 1 mark — factors x² − 9 = (x − 3)(x + 3) and cancels (x − 3) with "(for x ≠ 3)" annotation.

• 1 mark — evaluates lim_x→3 (x + 3) = 6.

Part (c) — 1 mark

• 1 mark — states k = 6, justified by "we need lim = f(3) for continuity".

Part (d) — 1 mark

• 1 mark — sketch shows the line y = x + 3 (gradient 1, y-intercept 3) with a solid dot at (3, 6) marking f(3) = 6.

Part (e) — 2 marks

• 1 mark — explains "removable" means the limit exists, so we can patch the gap by defining f(3) equal to the limit.

• 1 mark — explains that with k = 5 the limit value (6) does not equal f(3), so the graph has a single isolated dot at height 5 sitting off the line, breaking continuity (the limit exists but is not equal to f(3)).

Your response:

Stuck on the sketch? You found in (b) that the function equals x + 3 for all x ≠ 3 — start there.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — lim_x→4 (x² − 16) / (x − 4) (2 marks)

Sample response. Substitution: (16 − 16) / (4 − 4) = 0/0 (indeterminate). Factor numerator as difference of squares: x² − 16 = (x − 4)(x + 4). Cancel (x − 4) (for x ≠ 4): expression = x + 4. lim_x→4 (x + 4) = 8.

Marking notes. 1 mark — explicit 0/0 followed by factor-and-cancel with "for x ≠ 4". 1 mark — correct final value. Bare answer "8" with no working scores 1/2; missing the "for x ≠ 4" annotation does not lose marks at Band 3 but is expected at Band 4+.

1.2 — Piecewise function at x = 1 (3 marks)

Sample response.

(a) Left limit uses top branch (x ≤ 1): lim_x→1⁻ (2x + 1) = 2(1) + 1 = 3.

(b) Right limit uses bottom branch (x > 1): lim_x→1⁺ (4 − x) = 4 − 1 = 3.

(d) f(1) uses top branch: f(1) = 2(1) + 1 = 3. Since lim_x→1 f(x) = 3 = f(1), f is continuous at x = 1.

Marking notes. 1 mark — left and right limits both correct with branch identified. 1 mark — correct existence statement (must explicitly say "exists and equals 3"). 1 mark — correct continuity check (must compare limit with f(1), not just state "yes"). Common error: stating limit = 3 without checking f(1).

1.3 — lim_x→0 ( (x + 2)² − 4 ) / x (4 marks)

(a) Sample response. At x = 0: ((0 + 2)² − 4) / 0 = (4 − 4) / 0 = 0/0, indeterminate.

(b) Sample response. Expand (x + 2)² = x² + 4x + 4. So the numerator = x² + 4x + 4 − 4 = x² + 4x = x(x + 4). Divide by x (for x ≠ 0): expression = x + 4. lim_x→0 (x + 4) = 4.

Marking notes. (a) 1 mark — explicit 0/0 with substitution. (b) 1 mark — correct expansion; 1 mark — factor out x and cancel with "(for x ≠ 0)"; 1 mark — correct limit value 4. Common errors: forgetting to subtract 4 from the expansion; cancelling x without noting x ≠ 0 (loses 0.5 at Band 5).

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). Substituting x = 3 into (x² − 9)/(x − 3) gives (9 − 9)/(3 − 3) = 0/0, which is an indeterminate form. [1 mark.]

Part (b). Factor the numerator as a difference of squares: x² − 9 = (x − 3)(x + 3). Cancel the common factor (x − 3) (valid for x ≠ 3, which is exactly what we need for a limit at x = 3): (x² − 9)/(x − 3) = x + 3. [1 mark — factor and cancel with note.] Then lim_x→3 (x + 3) = 3 + 3 = 6. [1 mark — evaluation.]

Part (c). For continuity at x = 3 we need lim_x→3 f(x) = f(3) = k. Hence k = 6. [1 mark — value and justification.]

Part (d). For x ≠ 3, the graph of f is the line y = x + 3 (gradient 1, y-intercept 3). With k = 6, the point (3, 6) sits exactly on this line, so the sketch is a single unbroken straight line through (0, 3) with gradient 1, with a solid dot drawn at (3, 6) emphasising f(3) = 6. [1 mark — correct sketch with labelled point.]

Part (e). The discontinuity is called "removable" because the limit at x = 3 exists (equals 6) — so the gap in the original (x ≠ 3) function can be "removed" simply by defining f(3) equal to the limit. [1 mark.] If we instead chose k = 5, the limit would still be 6 but f(3) would be 5. The graph would then show the line y = x + 3 missing the point (3, 6), replaced by an isolated solid dot at (3, 5). Since lim ≠ f(3), f would not be continuous: the discontinuity remains, even though the limit exists. [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: Identifies 0/0 and attempts factor-and-cancel but does not always note "for x ≠ 3"; finds limit = 6 but may stop without addressing continuity. ≈ 3-4 marks.

Band 4: Parts (a)-(c) clean. Sketch in (d) shows the line but omits the labelled solid dot at (3, 6); (e) explains "removable" but does not analyse the k = 5 case. ≈ 4-5 marks.

Band 5: All parts attempted. (e) addresses both why k = 6 works and what goes wrong with k = 5, but may not explicitly invoke "limit = f(3)" as the continuity criterion. ≈ 5-6 marks.

Band 6: Every part complete with correct mathematical reasoning, "(for x ≠ 3)" annotation written explicitly, sketch carefully labelled, and the k = 5 contrast presented as a clear example of "limit exists but ≠ function value". 7/7.