These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator, angles in radians, and exact answers unless stated. Notation: lim (x → a) f(x) means “the limit as x approaches a of f(x)”; x → a⁺ means from the right and x → a⁻ means from the left. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Direct substitution into lim (x → 5) (√(x + 11) − 4)/(x − 5) gives 0/0. Which is the most appropriate next step?
- (A) Conclude that the limit is 0.
- (B) Conclude that the limit does not exist, because the denominator is 0 at x = 5.
- (C) Factor the numerator and denominator as polynomials and cancel (x − 5).
- (D) Multiply the numerator and denominator by √(x + 11) + 4.
Answer and explanation
Answer: (D). The 0 on top comes from a square root minus a number, so the conjugate is the tool. The top becomes (x + 11) − 16 = x − 5, which cancels with the bottom for x ≠ 5, leaving 1/(√(x + 11) + 4). Substituting gives 1/(4 + 4) = 1/8.
- (A) treats 0/0 as the number 0. 0/0 is indeterminate: it only says more work is needed.
- (B) ignores the top. A zero bottom rules out a finite limit only when the top is nonzero.
- (C) chooses a polynomial method, but the numerator contains a square root, so it is not a polynomial and does not factor in the usual way.
Question 2 (multiple choice · core)
Which statement describes the behaviour of (x² − 2x)/(x² − 4x + 4) as x → 2?
- (A) The limit is 0.
- (B) The limit is 1.
- (C) The expression → +∞ from both sides, so the limit is +∞.
- (D) The expression → +∞ from the right and −∞ from the left, so the limit does not exist.
Answer and explanation
Answer: (D). Substitution gives (4 − 4)/(4 − 8 + 4) = 0/0, so factor: x(x − 2)/(x − 2)² = x/(x − 2) for x ≠ 2. Substituting again gives 2/0, a nonzero number over 0, so there is no finite limit. Check signs: for x slightly above 2, the top is near 2 and the bottom is small and positive, so → +∞. For x slightly below 2, the bottom is small and negative, so → −∞. The sides disagree.
- (A) stops at the first 0/0 and treats it as 0.
- (B) “cancels the x² terms”, leaving −2x/(−4x + 4), which gives (−4)/(−4) = 1 at x = 2. Only whole factors cancel, never terms.
- (C) checks only the right-hand side. After nonzero/0, you must check both sides.
Question 3 (multiple choice · foundation)
A function f is defined by f(x) = x² + 1 for x < 2, f(2) = 7, and f(x) = 3x − 1 for x > 2. What is lim (x → 2) f(x)?
- (A) 5
- (B) 7
- (C) 10
- (D) The limit does not exist.
Answer and explanation
Answer: (A). The rule changes at x = 2, so find each one-sided limit with its own piece. Left: substitute into x² + 1 to get 5. Right: substitute into 3x − 1 to get 5. Both sides agree, so the limit is 5.
- (B) is the value f(2). The value at the point does not decide the limit.
- (C) adds the two one-sided limits. They should be compared, not added.
- (D) assumes a change of formula means no limit. The limit exists whenever the two one-sided limits are equal.
Question 4 (multiple choice · core)
What is lim (x → 0) cos(π(x² + x)/(2x))?
- (A) −1
- (B) 0
- (C) 1
- (D) The limit does not exist.
Answer and explanation
Answer: (B). Work from the inside out. The inside gives 0/0 at x = 0. Factor: π x(x + 1)/(2x) = π(x + 1)/2 for x ≠ 0, which approaches π/2. Cosine is continuous at π/2, so the limit is cos(π/2) = 0.
- (A) drops the 2 in the denominator, so the inside seems to approach π, and cos π = −1.
- (C) substitutes x = 0 into the inside as if it gave 0, then uses cos 0 = 1.
- (D) gives up because the inside is undefined at x = 0. The inside only needs a limit, not a value.
Question 5 (constructed response · core)
Let g(x) = (x³ − x)/(x² − x − 2).
For each of (a) x → 1, (b) x → −1 and (c) x → 2:
- state what direct substitution gives,
- name the procedure you choose as a result, and
- find the limit, or describe the behaviour if there is no finite limit.
Worked solution and suggested Marlbridge rubric
Factor once for later use: top x(x − 1)(x + 1), bottom (x − 2)(x + 1).
(a) Substitution: top 1 − 1 = 0, bottom 1 − 1 − 2 = −2. The bottom is not 0, so substitution is enough: lim (x → 1) g(x) = 0/(−2) = 0. (A zero numerator over a nonzero bottom is simply 0. It is not indeterminate.)
(b) Substitution: top −1 + 1 = 0, bottom 1 + 1 − 2 = 0. That is 0/0 with a polynomial over a polynomial, so factor and cancel (x + 1), valid for x ≠ −1: g(x) = x(x − 1)/(x − 2). lim (x → −1) g(x) = (−1)(−2)/(−3) = −2/3.
(c) Substitution: top 8 − 2 = 6, bottom 4 − 2 − 2 = 0. That is nonzero/0, so there is no finite limit; check signs using x(x − 1)/(x − 2), whose top is near 2 when x is near 2. x → 2⁺: bottom small and positive, so g(x) → +∞. x → 2⁻: bottom small and negative, so g(x) → −∞. The one-sided behaviours differ, so lim (x → 2) g(x) does not exist.
| Point | What earns it |
|---|---|
| 1 | (a) Substitution gives 0/(−2), a nonzero bottom, so the limit is 0 |
| 1 | (b) Identifies 0/0 and factors to cancel (x + 1) |
| 1 | (b) Limit −2/3 |
| 1 | (c) Identifies nonzero/0 and states that there is no finite limit |
| 1 | (c) Correct sign on each side (+∞ right, −∞ left) and concludes the limit does not exist |
Do not award the (a) point for “0/(−2) is indeterminate”. Factoring in (a) before substituting is not wrong, but notice it was unnecessary.
Question 6 (constructed response · core)
A function f is defined by
- f(x) = kx + 1 for x < 2
- f(x) = (√(x + 7) − 3)/(x − 2) for x > 2
where k is a constant. f(2) is not defined.
(a) Find lim (x → 2⁺) f(x). Name the procedure and explain why it fits. (b) Find lim (x → 2⁻) f(x) in terms of k. (c) Find the value of k for which lim (x → 2) f(x) exists. (d) A student says: “f(2) is not defined, so lim (x → 2) f(x) cannot exist.” Is the student right? Explain.
Worked solution and suggested Marlbridge rubric
(a) Use the piece for x > 2. Substitution gives (√9 − 3)/0 = 0/0. The top is a square root minus a number, so multiply by the conjugate √(x + 7) + 3: top: (x + 7) − 9 = x − 2. So for x > 2, f(x) = 1/(√(x + 7) + 3), and lim (x → 2⁺) f(x) = 1/(3 + 3) = 1/6.
(b) The left piece is a polynomial, so substitute: lim (x → 2⁻) f(x) = 2k + 1.
(c) The two-sided limit exists when the one-sided limits are equal: 2k + 1 = 1/6, so 2k = −5/6 and k = −5/12.
(d) No. A limit depends only on values of f(x) for x near 2, not at 2. With k = −5/12, both sides approach 1/6, so the limit exists and equals 1/6, even though f(2) is undefined.
| Point | What earns it |
|---|---|
| 1 | (a) Identifies 0/0 from a root expression and uses the conjugate to reach 1/(√(x + 7) + 3) |
| 1 | (a) Right-hand limit 1/6 and (b) left-hand limit 2k + 1 by substitution |
| 1 | (c) Sets the one-sided limits equal and finds k = −5/12 |
| 1 | (d) Explains that a limit uses nearby values, so an undefined f(2) does not stop the limit existing |
Acceptable alternative for (a): let u = √(x + 7), so x − 2 = u² − 9 = (u − 3)(u + 3) and the expression becomes 1/(u + 3) → 1/6.
Question 7 (constructed response · stretch)
A student finds lim (x → 0) |x|/(x² + x) like this:
Substitution gives 0/0. Factor the bottom: x(x + 1). Cancel the x with the |x| on top to get 1/(x + 1). Substitute: 1/(0 + 1) = 1. So the limit is 1.
(a) Identify the step where the student chose the wrong procedure, and explain why it is wrong. (b) Find lim (x → 0⁺) and lim (x → 0⁻) of |x|/(x² + x), and state lim (x → 0) |x|/(x² + x). (c) You are told that, to 4 significant figures, the expression equals 0.9901 at x = 0.01 and −1.010 at x = −0.01. Explain how these values support your answer to (b). (d) A second limit, lim (x → 0) x cos(1/x), cannot be found by substitution or by the Topic 1.6 rewriting techniques. Explain why, and name the method that is used for it.
Worked solution and suggested Marlbridge rubric
(a) The error is cancelling x with |x|. The absolute value creates a split at x = 0: |x| = x only when x ≥ 0, and |x| = −x when x < 0. The student should have found the two one-sided limits separately.
(b) For x > 0: |x|/(x(x + 1)) = x/(x(x + 1)) = 1/(x + 1) → 1. For x < 0: |x|/(x(x + 1)) = −x/(x(x + 1)) = −1/(x + 1) → −1. The one-sided limits are different, so lim (x → 0) |x|/(x² + x) does not exist.
(c) At x = 0.01 (just right of 0) the value 0.9901 is close to 1. At x = −0.01 (just left of 0) the value −1.010 is close to −1. The two sides head to different values, which matches (b). A table supports the answer but does not prove it.
(d) At x = 0, 1/x is undefined, so substitution fails. There is no common factor to cancel, no root to rationalise and no stacked fraction: cos(1/x) oscillates between −1 and 1 infinitely often near 0. Because cos(1/x) is bounded while the factor x → 0, the right method is the squeeze theorem (Topic 1.8), which shows the limit is 0.
| Point | What earns it |
|---|---|
| 1 | (a) Identifies cancelling x with the absolute value of x as the error, because it equals −x for x < 0 |
| 1 | (b) Right-hand limit 1 with correct simplification |
| 1 | (b) Left-hand limit −1 and conclusion that the limit does not exist |
| 1 | (c) Links each table value to the correct side and limit |
| 1 | (d) Explains why substitution and rewriting fail and names the squeeze theorem |
For (d), the value 0 is not required; naming the method with a reason is enough.
How did you do?
- Q1, Q2 or Q5 wrong: reread “Step 3: substitute and read the result”, “Step 4” and Figure 1 in the study guide.
- Q3, Q6 or Q7(a)–(b) wrong: review “Step 2: is there a split at x = a?” and Worked example 2.
- Q4 wrong: see “Composite functions: work from the inside out”.
- Q7(d) wrong: see “When none of these works”, then look ahead to Topic 1.8.
- Rewriting steps shaky: revisit the Topic 1.6 study guide.
Then tick off the topic checklist.