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 every point marked on a graph has exact coordinates. Notation: lim (x → a) f(x) means “the limit as x approaches a of f(x)”; x → a⁻ means from the left (values less than a) and x → a⁺ means from the right. This set is for both Calculus AB and Calculus BC students.
Questions 1, 2 and 5 use Figure P1.
Question 1 (multiple choice · foundation)
Use Figure P1. What is lim (x → 2) g(x)?
- (A) 0
- (B) 2
- (C) 3
- (D) The limit does not exist.
Answer and explanation
Answer: (C). Follow the middle line towards x = 2 from the left: the heights approach 3. From the right, the heights also approach 3. Both sides agree, so the limit is 3.
- (A) is g(2), the height of the filled dot. The value at x = 2 does not affect the limit.
- (B) gives the x-value, 2, instead of the height.
- (D) assumes a hole means no limit. A hole affects the value, not the limit: both sides still head to the same height.
Question 2 (multiple choice · core)
Use Figure P1. What is the value of lim (x → 0⁻) g(x) + lim (x → 4) g(x)?
- (A) 3
- (B) 5
- (C) 6
- (D) It does not exist, because lim (x → 0) g(x) does not exist.
Answer and explanation
Answer: (A). From the left of 0, the falling line heads to the open circle at (0, −1), so lim (x → 0⁻) g(x) = −1. At x = 4 the two lines meet at (4, 4) with no gap, so both sides approach 4 and lim (x → 4) g(x) = 4. The sum is −1 + 4 = 3.
- (B) misreads the left-hand limit as +1: 1 + 4 = 5. The open circle is below the x-axis.
- (C) uses the right-hand side at 0 (the filled dot at height 2) instead of the left: 2 + 4 = 6.
- (D) confuses the one-sided limit with the two-sided one. The two-sided limit at 0 does not exist, but the question asks only for the left-hand limit, which does exist.
Question 3 (multiple choice · core)
Which of the following limits exists as a real number?
- (A) lim (x → 3) 1/(x − 3)²
- (B) lim (x → 0) cos(1/x)
- (C) lim (x → −1) |x + 1|/(x + 1)
- (D) lim (x → 1) (x − 1)²/|x − 1|
Answer and explanation
Answer: (D). For x ≠ 1, (x − 1)² = |x − 1|², so the expression equals |x − 1|. Its graph is a V shape with a hole at (1, 0). Both sides approach height 0, so the limit is 0.
- (A) is unbounded. As x → 3 from either side, (x − 3)² is a tiny positive number, so 1/(x − 3)² grows without bound. You may write = ∞, but that describes why the limit does not exist.
- (B) oscillates. cos(1/x) = 1 at x = 1/(2π), 1/(4π), 1/(6π), … and cos(1/x) = −1 at x = 1/π, 1/(3π), 1/(5π), … Both lists get as close to 0 as you like, so no single height is approached.
- (C) jumps. For x > −1 the expression is 1; for x < −1 it is −1. The left-hand limit (−1) and right-hand limit (1) are different.
Question 4 (multiple choice · core)
A student graphs y = x + 0.0001/(x − 4)² on a window with −10 ≤ x ≤ 10 and −10 ≤ y ≤ 10. The screen shows what looks like the straight line y = x. Which conclusion about lim (x → 4) of this function is correct?
- (A) The limit is 4, because the graph looks like y = x.
- (B) The limit does not exist, because the values grow without bound as x → 4 from either side.
- (C) The limit is 4.0001, because the extra term adds 0.0001.
- (D) The limit does not exist, because the left-hand and right-hand limits are different.
Answer and explanation
Answer: (B). Close to 4, (x − 4)² is tiny, so the extra term is huge. At x = 3.99 or 4.01, (x − 4)² = 0.0001 and the extra term is 1, giving y ≈ 5. At x = 3.999 or 4.001, the extra term is 100, giving y ≈ 104. The values grow without bound from both sides. The spike is too narrow for the screen to show.
- (A) trusts the picture. The window’s scale hides a very narrow vertical asymptote.
- (C) adds 0.0001 as if the denominator were 1. The denominator (x − 4)² approaches 0, so the term does not stay small.
- (D) gives the wrong reason. Both sides go to +∞, in the same direction. The limit fails because the function is unbounded, not because of a jump.
Question 5 (graph · core)
Use Figure P1.
(a) Find lim (x → 0⁻) g(x), lim (x → 0⁺) g(x) and g(0). (b) Does lim (x → 0) g(x) exist? Justify your answer using part (a). (c) Find lim (x → 2) g(x) and g(2). Explain why it is not a contradiction that these are different. (d) The graph starts at x = −3. Which one-sided limit at x = −3 can you find from the graph, and what is its value?
Worked solution and suggested Marlbridge rubric
(a) lim (x → 0⁻) g(x) = −1 (open circle at (0, −1)). lim (x → 0⁺) g(x) = 2. g(0) = 2 (the filled dot).
(b) No. The left-hand limit is −1 and the right-hand limit is 2. A two-sided limit exists only when both one-sided limits exist and are equal, and −1 ≠ 2.
(c) lim (x → 2) g(x) = 3, because the line approaches height 3 from both sides. g(2) = 0, from the filled dot. A limit uses only x-values near 2, never x = 2 itself, so the separate dot at (2, 0) cannot change it.
(d) Only the right-hand limit can be found, because g is not defined to the left of −3. lim (x → −3⁺) g(x) = 2.
| Point | What earns it |
|---|---|
| 1 | All three values in (a) correct: −1, 2 and 2 |
| 1 | (b) “does not exist”, justified by the unequal one-sided limits (not by “there is a jump” alone without values) |
| 1 | (c) limit 3 and value 0, with the reason that a limit does not depend on the value at x = 2 |
| 1 | (d) right-hand limit only, equal to 2, with the reason that there are no x-values to the left of −3 |
Accept any wording that quotes both one-sided limits in (b).
Question 6 (constructed response · core)
Consider three functions:
- r(x) = 1/(x + 2)²
- s(x) = (x + 2)/|x + 2|
- p(x) = sin(π/x)
(a) Describe the graph of r and of s near x = −2. For each, state whether the limit as x → −2 exists. If it does not, say which kind of behaviour causes this. (b) Show that p(2/5) = 1 and p(2/7) = −1. Give two more x-values between 0 and 2/7 where p takes the values 1 and −1. Use these to explain why lim (x → 0) p(x) does not exist. (c) A student writes: “lim (x → −2) r(x) = ∞, so the limit exists and equals infinity.” Explain what is wrong.
Worked solution and suggested Marlbridge rubric
(a) r: (x + 2)² is a small positive number near x = −2, so r(x) is very large and positive on both sides. For example, r(−1.9) = 1/0.01 = 100 and r(−1.99) = 10 000. The graph rises without bound on both sides of a vertical asymptote at x = −2. The limit does not exist because r is unbounded.
s: For x > −2, x + 2 is positive, so s(x) = 1. For x < −2, x + 2 is negative, so s(x) = −1. The graph is two horizontal pieces with a gap at x = −2. The left-hand limit is −1 and the right-hand limit is 1, so the limit does not exist because the one-sided limits differ (a jump).
(b) p(2/5) = sin(5π/2) = 1 and p(2/7) = sin(7π/2) = −1. Closer to 0: p(2/9) = sin(9π/2) = 1 and p(2/11) = sin(11π/2) = −1. However close to 0 you go, p keeps taking both values 1 and −1, so the heights never settle on one number. The limit does not exist because p oscillates.
(c) ∞ is not a real number. Writing “= ∞” says the values grow without bound, which is one way a limit fails to exist. Correct statement: lim (x → −2) r(x) does not exist, because r is unbounded as x → −2.
| Point | What earns it |
|---|---|
| 1 | r: unbounded on both sides of x = −2, limit does not exist |
| 1 | s: left-hand limit −1, right-hand limit 1, limit does not exist because they differ |
| 1 | p: shows the two given values and gives two correct closer values (such as 2/9 and 2/11) |
| 1 | Explains oscillation (both 1 and −1 occur arbitrarily close to 0) and corrects the student in (c) |
Accept any correct closer values in (b): x = 2/(4n + 1) gives 1 and x = 2/(4n + 3) gives −1, for whole numbers n.
Question 7 (constructed response · stretch)
A function f is defined for every real number except x = 5. It satisfies all of these:
- lim (x → 2⁻) f(x) = 3
- lim (x → 2⁺) f(x) = −1
- f(2) = 1
- lim (x → 5) f(x) = 4
(a) Describe a graph that meets all four conditions, or give a piecewise formula for one. Explain how each condition is shown. (b) Does lim (x → 2) f(x) exist? Explain. (c) A classmate says: “f(5) is undefined, so lim (x → 5) f(x) cannot be 4.” Respond. (d) Explain why a graphing calculator picture of your function might not show anything unusual at x = 5.
Worked solution and suggested Marlbridge rubric
(a) Many answers are possible. One example:
- f(x) = x + 1 for x < 2
- f(2) = 1
- f(x) = (5/3)(x − 2) − 1 for x > 2, x ≠ 5
On a graph: the line y = x + 1 ends at an open circle at (2, 3), so the left-hand limit is 3. A filled dot at (2, 1) gives f(2) = 1. A second line starts with an open circle at (2, −1), so the right-hand limit is −1. It rises with gradient 5/3 and passes through height (5/3)(3) − 1 = 4 at x = 5, where there is an open circle at (5, 4) and no filled dot, so f(5) is undefined but the limit there is 4.
(b) No. The left-hand limit is 3 and the right-hand limit is −1. They are different, so the two-sided limit does not exist. The value f(2) = 1 does not change this.
(c) The classmate is wrong. A limit depends on the heights approached near x = 5, not on the value at 5. Both sides head to height 4, so the limit is 4 even with a hole there.
(d) Most graphing tools do not draw open circles, and a single missing point leaves no visible gap at normal scale. The hole exists but is invisible.
| Point | What earns it |
|---|---|
| 1 | A graph or formula with the correct one-sided limits at x = 2 (3 from the left, −1 from the right) |
| 1 | f(2) = 1 shown as a separate filled dot, and a hole at (5, 4) with no value at x = 5 |
| 1 | (b) “does not exist” because 3 ≠ −1 and (c) limit depends on nearby values, not the value at 5 |
| 1 | (d) explains that a single missing point (hole) is too small to show at the screen’s scale |
A clear sketch with open and filled circles is as good as a formula.
How did you do?
- Q1 or Q5(c) wrong: reread “What a graph tells you about a limit” and Worked example 1(a) in the study guide. The value at a point does not decide the limit.
- Q2 or Q5(a)–(b) wrong: redo Worked example 1(b) and reread “One-sided and two-sided limits”.
- Q3 or Q6 wrong: see “Three ways a limit can fail to exist” and Figure 2 in the guide.
- Q4 or Q7(d) wrong: reread “When the picture can mislead” and Worked example 2.
Then tick off the topic checklist.