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. Notation: lim (x → a) f(x) means “the limit as x approaches a of f(x)”; x → a⁻ and x → a⁺ mean approaching from the left and from the right. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
You are told that lim (x → 2) f(x) = 3 and lim (x → 2) g(x) = −4. What is
lim (x → 2) [ (f(x))² − 2g(x) ] / [ f(x) + g(x) ]?
- (A) 17/7
- (B) −14
- (C) −17
- (D) −1
Answer and explanation
Answer: (C). Top: (f(x))² → 3² = 9 (power) and 2g(x) → 2 × (−4) = −8 (constant multiple), so the top tends to 9 − (−8) = 17 (difference). Bottom: 3 + (−4) = −1 (sum). The bottom limit is not 0, so the quotient property gives 17/(−1) = −17.
- (A) uses 3 − (−4) = 7 in the bottom: the denominator is a sum, not a difference.
- (B) replaces (f(x))² by 2f(x), giving (6 + 8)/(−1) = −14. Squaring is not doubling.
- (D) loses the double negative in the top: 9 − 8 = 1, then 1/(−1) = −1. Subtracting −8 means adding 8.
Question 2 (multiple choice · foundation)
What is lim (x → 2) (x³ − 3x)/√(x² + 5)?
- (A) 2/9
- (B) 2/3
- (C) 2/7
- (D) 14/3
Answer and explanation
Answer: (B). The top is a polynomial, so its limit is 8 − 6 = 2. The inner part of the bottom tends to 4 + 5 = 9 > 0, so the root property gives √9 = 3. The bottom limit is 3, not 0, so the quotient property gives 2/3.
- (A) forgets the square root and divides by 9.
- (C) treats √(x² + 5) as √(x²) + 5 = 2 + 5 = 7. The root covers the whole of x² + 5.
- (D) adds instead of subtracting in the top: 8 + 6 = 14.
Question 3 (multiple choice · core)
The function f is defined by f(y) = 3 − y for y < 0 and f(y) = 2y + 1 for y ≥ 0. What is lim (x → 0) f(x²)?
- (A) 0
- (B) 1
- (C) 3
- (D) The limit does not exist, because lim (y → 0) f(y) does not exist.
Answer and explanation
Answer: (B). The inner function x² tends to 0. For every x ≠ 0, x² > 0, so the inner values approach 0 from above, from both sides of x = 0. So f(x²) uses only the piece 2y + 1, and f(x²) → lim (y → 0⁺) f(y) = 2(0) + 1 = 1.
- (A) gives the limit of the inner function, x² → 0, and forgets to apply f.
- (C) uses the left-hand limit of f at 0, which is 3. But x² is never negative, so that piece is never used.
- (D) is a tempting rule but it is false. f does jump at 0, yet x² only reaches 0 from one side, so the jump does not matter here.
Question 4 (multiple choice · core)
Suppose lim (x → c) p(x) = 6 and lim (x → c) q(x) = 0. Which statement must be true?
- (A) lim (x → c) p(x)/q(x) = 0
- (B) lim (x → c) p(x)/q(x) = 6
- (C) lim (x → c) p(x) · q(x) = 0
- (D) lim (x → c) q(x)/p(x) does not exist
Answer and explanation
Answer: (C). Both limits exist, so the product property gives 6 × 0 = 0.
- (A) and (B) use the quotient property with a bottom limit of 0, which is not allowed. In fact, a top tending to 6 over a bottom tending to 0 is unbounded, so p/q has no finite limit.
- (D) is false: for q/p the bottom limit is 6 ≠ 0, so the quotient property applies and the limit is 0/6 = 0.
Question 5 (graph · core)
The graphs of f and g are shown for −2 ≤ x ≤ 4.
(a) Find lim (x → 2⁻) f(x) and lim (x → 2⁺) f(x). Does lim (x → 2) f(x) exist? (b) Find lim (x → 2) [f(x) + g(x)], or explain why it does not exist. Compare it with f(2) + g(2). (c) Find lim (x → 2) f(x)g(x), or explain why it does not exist. (d) Find lim (x → 0) f(g(x)).
Worked solution
(a) From the left, f heads to the open circle at height 3. From the right, f heads to the open circle at height 1. The one-sided limits are 3 and 1. They differ, so lim (x → 2) f(x) does not exist. (The filled dot, f(2) = 2, is not either limit.)
(b) The sum property cannot be used, because lim f(x) does not exist at 2. Use one-sided limits. For g: from the left, g → 0; from the right, g → 2.
- Left: 3 + 0 = 3. Right: 1 + 2 = 3.
So lim (x → 2) [f(x) + g(x)] = 3. But f(2) + g(2) = 2 + 2 = 4. The limit and the value are different, and that is allowed.
(c) Left: 3 × 0 = 0. Right: 1 × 2 = 2. They differ, so the limit does not exist.
(d) As x → 0, g(x) → g(0) = 1 (g has no break at 0). Near y = 1, f has no break, and f(1) = 1/2 + 2 = 5/2 (the left piece passes through (0, 2) and (2, 3), so it rises by 1/2 per unit). By the composite property, lim (x → 0) f(g(x)) = 5/2.
Suggested mark points (4): 1 for both one-sided limits of f and “does not exist”; 1 for one-sided sums 3 and 3, limit 3, and noting f(2) + g(2) = 4 is different; 1 for products 0 and 2 and “does not exist”; 1 for g(x) → 1 and f(1) = 5/2.
Question 6 (constructed response · core)
For a constant k, the function q is defined by
- q(x) = x² − 2x + k for x < 3
- q(x) = 2k − x for x ≥ 3
(a) Find lim (x → 3⁻) q(x) and lim (x → 3⁺) q(x) in terms of k. (b) Find the value of k for which lim (x → 3) q(x) exists. State the limit. (c) Using your value of k, find lim (x → 3) [q(x) − x]/[q(x) + x]. Justify each step by naming a limit property.
Worked solution and suggested Marlbridge rubric
(a) For x < 3, q is the polynomial x² − 2x + k, so lim (x → 3⁻) q(x) = 9 − 6 + k = 3 + k. For x > 3, q(x) = 2k − x, so lim (x → 3⁺) q(x) = 2k − 3.
(b) The two-sided limit exists only if the one-sided limits are equal: 3 + k = 2k − 3, so k = 6. The limit is 3 + 6 = 9.
(c) lim (x → 3) q(x) = 9 (part (b)) and lim (x → 3) x = 3 (identity).
- Top: 9 − 3 = 6 (difference property).
- Bottom: 9 + 3 = 12 (sum property).
- The bottom limit is 12 ≠ 0, so by the quotient property the limit is 6/12 = 1/2.
| Point | What earns it |
|---|---|
| 1 | Both one-sided limits correct in terms of k: 3 + k and 2k − 3 |
| 1 | Sets the one-sided limits equal, with the reason (the two-sided limit needs them equal) |
| 1 | k = 6 and limit 9 |
| 1 | Limit 1/2, naming the difference, sum and quotient properties and checking the bottom limit is nonzero |
Acceptable alternative for (c): substitute q(x) = 2k − x = 12 − x for x > 3 and x² − 2x + 6 for x < 3, and find each one-sided limit of the quotient separately; both give 1/2. Reading the answer from a table of values earns no mark for (c), because it does not justify the value.
Question 7 (constructed response · stretch)
You are told that lim (x → a) f(x) = 4 and lim (x → a) [f(x) + g(x)] = 1. You are also told that lim (x → a) h(x) does not exist.
(a) Find lim (x → a) g(x). Justify your answer. (b) Find lim (x → a) f(x)g(x) / [2 + g(x)]. (c) Explain why lim (x → a) [f(x) + h(x)] cannot exist. (d) A student says: “If neither lim (x → a) u(x) nor lim (x → a) v(x) exists, then lim (x → a) [u(x) + v(x)] cannot exist.” Give an example that shows the student is wrong.
Worked solution and suggested Marlbridge rubric
(a) Write g(x) = [f(x) + g(x)] − f(x). Both limits on the right exist, so by the difference property, lim (x → a) g(x) = 1 − 4 = −3.
(b) Top: 4 × (−3) = −12 (product). Bottom: 2 + (−3) = −1 (sum). The bottom limit is not 0, so the quotient property gives −12/(−1) = 12.
(c) Suppose lim [f(x) + h(x)] did exist. Then h(x) = [f(x) + h(x)] − f(x) would be a difference of two functions whose limits exist, so lim h(x) would exist by the difference property. That contradicts what you are told. So lim [f(x) + h(x)] cannot exist.
(d) Take u(x) = |x − a|/(x − a) and v(x) = −u(x). Near a, u is −1 on the left and 1 on the right, so lim u does not exist; the same is true for v. But u(x) + v(x) = 0 for all x ≠ a, so lim [u(x) + v(x)] = 0. Any pair of jumps that cancel works, as in Question 5(b).
| Point | What earns it |
|---|---|
| 1 | lim g = −3, justified by writing g as a difference of functions with known limits |
| 1 | 12, with the bottom limit −1 checked as nonzero |
| 1 | Argument for (c): if the sum had a limit, the difference property would give a limit for h, a contradiction |
| 1 | A valid counterexample in (d), with both one-sided limits shown for one function and the sum’s limit stated |
Acceptable alternative for (d): a graphical example, such as the f and g of Question 5 at x = 2, with one-sided values stated.
How did you do?
- Q1, Q2 or Q7(b) wrong: revisit the properties table and Worked example 1 in the study guide.
- Q3 or Q5(d) wrong: reread “Limits of composite functions”, especially the side of approach.
- Q4 or Q7(c) wrong: check the condition for each property, especially a nonzero bottom limit.
- Q5 or Q6 wrong: redo Worked examples 2 and 3: work one side at a time.
Then tick off the topic checklist.