Mathematics Advanced • Year 11 • Module 3 • Lesson 2
Limits
Build procedural fluency in evaluating limits by direct substitution, factor-and-cancel, and one-sided checks for piecewise functions.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the sentence:
limx→a f(x) = L means: as x gets arbitrarily close to ________, the value of f(x) gets arbitrarily close to ________.
Q1.2 State the condition for the two-sided limit at x = a to exist:
limx→a f(x) exists if and only if limx→a− f(x) ________ limx→a+ f(x) and both are ________.
Q1.3 Match the form to the method (write a/b/c next to each):
(i) Substitution gives a real number ____ (a) Check one-sided limits separately.
(ii) Substitution gives 0/0 ____ (b) Answer is that value of f(a).
(iii) Function changes formula at x = a ____ (c) Factor numerator and denominator, cancel, then substitute.
2. Worked example — evaluating limx→2 (x² − 4) / (x − 2)
Follow each line of algebra. Every step has a reason on the right.
Problem. Evaluate limx→2 (x² − 4) / (x − 2).
Step 1 — Try direct substitution.
Substitute x = 2: (4 − 4) / (2 − 2) = 0 / 0 (indeterminate)
Reason: 0/0 is not a value — we cannot stop here. Must simplify algebraically.
Step 2 — Factor the numerator.
x² − 4 = (x − 2)(x + 2) (difference of squares)
Reason: factor out the troublesome (x − 2).
Step 3 — Cancel the common factor.
(x − 2)(x + 2) / (x − 2) = x + 2 (for x ≠ 2)
Reason: cancelling is valid because for the limit we only care about x near 2, not x = 2.
Step 4 — Take the limit of the simplified expression.
limx→2 (x + 2) = 2 + 2 = 4
Conclusion. limx→2 (x² − 4) / (x − 2) = 4. The graph has a removable hole at x = 2.
3. Faded example — fill in the missing steps
Evaluate limx→3 (x² − 9) / (x − 3). Fill in each blank. 3 marks
Step 1 — Substitute x = 3:
(__² − __) / (__ − __) = ____ / ____ (indeterminate form)
Step 2 — Factor the numerator:
x² − 9 = ( ____ )( ____ )
Step 3 — Cancel and take the limit:
After cancelling, the expression becomes ____ (for x ≠ 3).
limx→3 ( ____ ) = ____ .
Conclusion. limx→3 (x² − 9) / (x − 3) = ____ .
4. Graduated practice — evaluate each limit
For each limit, identify which of the three methods to use (substitution / factor-and-cancel / one-sided check). Show the algebra.
Foundation — direct substitution (4 questions)
| Q | Limit | Working / value |
|---|---|---|
| 4.1 1 | limx→1 (3x + 2) | |
| 4.2 1 | limx→2 (x² + 3x − 1) | |
| 4.3 1 | limx→0 (5 − x²) | |
| 4.4 1 | limx→3 (x² + 1) / (x − 1) |
Standard — factor and cancel (6 questions)
Each gives 0/0 on direct substitution. Show the factoring step.
4.5 limx→−1 (x² − 1) / (x + 1) 2 marks
4.6 limx→4 (x² − 16) / (x − 4) 2 marks
4.7 limx→3 (x² − 5x + 6) / (x − 3) 2 marks
4.8 limh→0 (h² + 4h) / h 2 marks
4.9 limh→0 ( (2 + h)² − 4 ) / h 2 marks
4.10 limx→2 (x³ − 8) / (x − 2) (Hint: x³ − 8 = (x − 2)(x² + 2x + 4).) 2 marks
Extension — piecewise / one-sided (2 questions)
4.11 For the piecewise function f(x) = { x + 1 if x < 2 ; 5 if x = 2 ; 7 − x if x > 2 }, find the left-hand limit, the right-hand limit, and determine whether limx→2 f(x) exists. 3 marks
4.12 Explain in two lines why limx→0 |x| / x does not exist. Include the left and right limits in your reasoning. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Limit notation
As x gets arbitrarily close to a, f(x) gets arbitrarily close to L.
Q1.2 — Existence of a two-sided limit
limx→a f(x) exists ⇔ left-hand limit equals right-hand limit and both exist (are finite).
Q1.3 — Method matching
(i) real number → b (substitution gives f(a)). (ii) 0/0 → c (factor-and-cancel). (iii) piecewise at boundary → a (check one-sided).
Q3 — Faded example limx→3 (x² − 9) / (x − 3)
Step 1: (9 − 9) / (3 − 3) = 0/0. Step 2: x² − 9 = (x − 3)(x + 3). Step 3: cancel (x − 3): expression becomes x + 3 (for x ≠ 3); limx→3 (x + 3) = 6.
Q4.1 — limx→1 (3x + 2)
Linear polynomial → substitution: 3(1) + 2 = 5.
Q4.2 — limx→2 (x² + 3x − 1)
Polynomial → substitution: 4 + 6 − 1 = 9.
Q4.3 — limx→0 (5 − x²)
Substitution: 5 − 0 = 5.
Q4.4 — limx→3 (x² + 1) / (x − 1)
Check denominator at x = 3: 3 − 1 = 2 ≠ 0, so substitute directly: (9 + 1) / (3 − 1) = 10 / 2 = 5.
Q4.5 — limx→−1 (x² − 1) / (x + 1)
Substitution gives 0/0. Factor: x² − 1 = (x − 1)(x + 1). Cancel (x + 1) (for x ≠ −1): (x − 1). Then limx→−1 (x − 1) = −1 − 1 = −2.
Q4.6 — limx→4 (x² − 16) / (x − 4)
0/0 → factor: (x − 4)(x + 4) / (x − 4) = x + 4 (for x ≠ 4). Limit = 4 + 4 = 8.
Q4.7 — limx→3 (x² − 5x + 6) / (x − 3)
0/0 → factor: x² − 5x + 6 = (x − 2)(x − 3). Cancel (x − 3): (x − 2). Limit = 3 − 2 = 1.
Q4.8 — limh→0 (h² + 4h) / h
0/0 → factor h: h(h + 4) / h = h + 4 (for h ≠ 0). Limit = 0 + 4 = 4.
Q4.9 — limh→0 ((2 + h)² − 4) / h
Expand: (2 + h)² − 4 = 4 + 4h + h² − 4 = 4h + h². Divide: (4h + h²)/h = 4 + h (for h ≠ 0). Limit = 4. (Foreshadows the derivative of x² at x = 2!)
Q4.10 — limx→2 (x³ − 8) / (x − 2)
0/0 → factor using hint: (x − 2)(x² + 2x + 4) / (x − 2) = x² + 2x + 4 (for x ≠ 2). Limit = 4 + 4 + 4 = 12.
Q4.11 — Piecewise limit at x = 2
Left limit: limx→2− f(x) = limx→2− (x + 1) = 3. Right limit: limx→2+ f(x) = limx→2+ (7 − x) = 5. Since 3 ≠ 5, the two-sided limit limx→2 f(x) does not exist. (The value f(2) = 5 is irrelevant to the limit.)
Q4.12 — limx→0 |x| / x
For x > 0: |x| = x, so |x|/x = 1. Hence limx→0+ |x|/x = 1. For x < 0: |x| = −x, so |x|/x = −1. Hence limx→0− |x|/x = −1. Since 1 ≠ −1, the two-sided limit does not exist. (The graph has a jump discontinuity at x = 0.)