What this is for. Use this diagnostic to find which topics of Unit 1, Limits and Continuity, to revisit. There is one question per topic, and two for Topics 1.6 and 1.11. These are original Marlbridge practice questions, not past exam questions. They are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Answer everything before opening any answer. c⁻ means from the left, c⁺ from the right. The unit is shared by Calculus AB and Calculus BC.
Question 1 (multiple choice · 1.1)
The average rate of change of f over [1, 1 + h] is 6 + 2h for every h ≠ 0. What is the rate of change of f at x = 1?
- (A) 6
- (B) 8
- (C) Undefined, because h = 0 gives 0/0
- (D) It needs a formula for f.
Answer and explanation
Answer: (A). The rate at an instant is the limit of the averages: lim (h → 0) (6 + 2h) = 6.
- (B) puts h = 1: the average over [1, 2].
- (C) The quotient is 0/0 at h = 0, but a limit never uses h = 0.
- (D) The simplified averages are enough.
If you missed this: Topic 1.1 study guide.
Question 2 (multiple choice · 1.2)
Which statement means the same as lim (x → −2) f(x) = 5?
- (A) f(−2) = 5
- (B) f(x) can be made as close to 5 as you like by taking x close enough to −2, with x ≠ −2.
- (C) As x approaches 5, f(x) approaches −2.
- (D) f(x) < 5 for every x near −2.
Answer and explanation
Answer: (B). A limit describes values near −2, not at −2.
- (A) confuses the limit with f(−2). (C) swaps input and output.
- (D) Values may approach 5 from either side.
If you missed this: Topic 1.2 study guide.
Question 3 (multiple choice · 1.3)
Use Figure 1. Which statement is true?
- (A) lim (x → −1) f(x) = 1
- (B) lim (x → 1) f(x) = 4
- (C) lim (x → 1) f(x) = 2
- (D) lim (x → −1⁻) f(x) = 1
Answer and explanation
Answer: (C). From both sides the graph heads to the open circle at height 2. The dot (1, 4) is f(1), which the limit ignores.
- (A) At −1 the sides head to 3 and 1: no two-sided limit.
- (B) reads f(1). (D) From the left the graph heads to 3.
If you missed this: Topic 1.3 study guide.
Question 4 (multiple choice · 1.4)
Values of k(x) = (2ˣ − 1)/x, to 4 decimal places (k(0) is undefined):
| x | −0.1 | −0.01 | −0.001 | 0.001 | 0.01 | 0.1 |
|---|---|---|---|---|---|---|
| k(x) | 0.6697 | 0.6908 | 0.6929 | 0.6934 | 0.6956 | 0.7177 |
What is the best estimate of lim (x → 0) k(x)?
- (A) 0.67
- (B) 0.69
- (C) 0.72
- (D) It does not exist, since k(0) is undefined.
Answer and explanation
Answer: (B). The values nearest 0 (0.6929 and 0.6934) agree to 0.69, from both sides.
- (A) and (C) use the values furthest from 0.
- (D) A limit does not need k(0).
If you missed this: Topic 1.4 study guide.
Question 5 (multiple choice · 1.5)
You are told that lim (x → 2) f(x) = 4. What is lim (x → 2) [x·f(x) − 3] / √(f(x) + 5)?
- (A) 5/3
- (B) 5/9
- (C) 1/3
- (D) It needs a formula for f.
Answer and explanation
Answer: (A). By the limit properties, (2 × 4 − 3)/√(4 + 5) = 5/3. The bottom limit, 3, is not 0, so the quotient property applies.
- (B) forgets the square root.
- (C) uses f(x) − 3 on top.
- (D) The properties need only the limit of f.
If you missed this: Topic 1.5 study guide.
Question 6 (multiple choice · 1.6)
What is lim (x → −1) (x³ + 1)/(x² − 1)?
- (A) −3/2
- (B) 0
- (C) 3/2
- (D) It does not exist: the bottom is 0 at x = −1.
Answer and explanation
Answer: (A). Substitution gives 0/0. Cancel the common factor (x + 1): for x ≠ −1 the expression is (x² − x + 1)/(x − 1), which gives 3/(−2) = −3/2.
- (B) treats 0/0 as 0. (C) drops the sign of x − 1 = −2.
- (D) 0/0 means “rewrite”, not “no limit”.
If you missed this: Topic 1.6 study guide.
Question 7 (short answer · 1.6)
Find lim (x → 5) (x − 5)/(√(2x − 1) − 3). Show your method.
Worked answer
Substitution gives 0/0, with a square root on the bottom. Multiply top and bottom by the conjugate √(2x − 1) + 3. The bottom becomes (2x − 1) − 9 = 2(x − 5), so for x ≠ 5 the expression is (√(2x − 1) + 3)/2. The limit is (3 + 3)/2 = 3.
If you missed this: Worked example 2 in the Topic 1.6 study guide.
Question 8 (multiple choice · 1.7)
For which limit is factoring and cancelling a common factor the right procedure?
- (A) lim (x → 2) (x² + 4)/(x − 2)
- (B) lim (x → 2) (x² − 3x + 2)/(x² − 4)
- (C) lim (x → 2) (√(3x + 3) − 3)/(x − 2)
- (D) lim (x → 2) (x² − 4)/(x + 2)
Answer and explanation
Answer: (B). Two polynomials give 0/0, so both contain (x − 2): (x − 1)/(x + 2) → 1/4.
- (A) gives 8/0: check signs for unbounded behaviour.
- (C) gives 0/0 from a square root: use the conjugate.
- (D) Substitution gives 0 directly.
If you missed this: Topic 1.7 study guide.
Question 9 (multiple choice · 1.8)
Which fact lets you conclude, by the squeeze theorem, that lim (x → 1) f(x) = 2?
- (A) 1 ≤ f(x) ≤ 3 for all x
- (B) 4x − x² − 1 ≤ f(x) ≤ x² + 1 for all x
- (C) f(x) ≤ x² + 1 for all x
- (D) The bounds in (B), but for x > 1 only
Answer and explanation
Answer: (B). Both bounds tend to 2 as x → 1 (4 − 1 − 1 = 2 and 1 + 1 = 2), and they trap f on both sides of 1.
- (A) The bounds differ. (C) One bound traps nothing.
- (D) gives only the right-hand limit.
If you missed this: Topic 1.8 study guide.
Question 10 (multiple choice · 1.9)
Both one-sided limits of p at x = 2 equal 5, and p(2) = 1. Which formula fits?
- (A) p(x) = x² + 1 for x ≠ 2, and p(2) = 1
- (B) p(x) = x² + 1 for x < 2, and p(x) = 1 for x ≥ 2
- (C) p(x) = 5 + 1/(x − 2) for x ≠ 2, and p(2) = 1
- (D) p(x) = x² + 1 for all x
Answer and explanation
Answer: (A). Near 2 the values approach 2² + 1 = 5 from both sides; p(2) = 1 does not affect the limit.
- (B) The right-hand limit is 1. (C) is unbounded near 2.
- (D) gives p(2) = 5.
If you missed this: Topic 1.9 study guide.
Question 11 (multiple choice · 1.10)
Use Figure 1. How are the discontinuities at x = −1 and x = 1 classified?
- (A) Both are removable.
- (B) Jump at −1, removable at 1
- (C) Removable at −1, jump at 1
- (D) Jump at −1; continuous at 1, because f(1) is defined
Answer and explanation
Answer: (B). At −1 the one-sided limits (3 and 1) differ: a jump. At 1 the limit is 2 but f(1) = 4: removable.
- (A) and (C) A jump has different one-sided limits.
- (D) Being defined is only one of three conditions.
If you missed this: Topic 1.10 study guide.
Question 12 (multiple choice · 1.11)
g(3) = 4 and lim (x → 3⁻) g(x) = 4. What is the least extra fact you need to conclude that g is continuous at x = 3?
- (A) Nothing more
- (B) That lim (x → 3⁺) g(x) = 4
- (C) That g is defined for x > 3
- (D) That g(x) = 4 for all x near 3
Answer and explanation
Answer: (B). Then the two-sided limit is 4 = g(3).
- (A) One side is not enough (see x = −1 in Figure 1).
- (C) says nothing about the limit. (D) is far more than needed.
If you missed this: Topic 1.11 study guide.
Question 13 (short answer · 1.11)
Let g(x) = x² − 1 for x < 2, g(2) = 3, and g(x) = 7 − 2x for x > 2.
(a) Is g continuous at x = 2? Justify using the definition. (b) If the last piece were 8 − 2x, which condition would fail, and what type of discontinuity results?
Worked answer
(a) g(2) = 3 is defined. The left-hand limit is 2² − 1 = 3 and the right-hand limit is 7 − 4 = 3, so lim (x → 2) g(x) = 3 = g(2). g is continuous at x = 2.
(b) The one-sided limits would be 3 and 8 − 4 = 4, so the limit would not exist (the second condition fails): a jump discontinuity.
If you missed this: Topic 1.11 study guide.
Question 14 (multiple choice · 1.12)
Let f(x) = ln(x)/(x² − 4). On which interval is f continuous?
- (A) [0, 2)
- (B) (0, 2)
- (C) (−2, 2)
- (D) (0, 3)
Answer and explanation
Answer: (B). Log and polynomial functions are continuous on their domains. f needs x > 0 and x ≠ 2.
- (A) includes x = 0. (C) includes x ≤ 0.
- (D) includes x = 2.
If you missed this: Topic 1.12 study guide.
Question 15 (multiple choice · 1.13)
Let f(x) = sin(2x)/(x² + 5x) for x ≠ 0 and x ≠ −5. Which value of f(0) makes f continuous at x = 0?
- (A) 0
- (B) 5/2
- (C) 2/5
- (D) No value: the discontinuity cannot be removed.
Answer and explanation
Answer: (C). sin(2x)/(x(x + 5)) = [sin(2x)/(2x)] × 2/(x + 5) → 1 × 2/5 = 2/5. The limit exists, so f(0) = 2/5 removes the break.
- (A) treats 0/0 as 0. (B) inverts the ratio.
- (D) It can be removed because the limit exists.
If you missed this: Topic 1.13 study guide.
Question 16 (multiple choice · 1.14)
What is lim (x → −3⁺) (x − 1)/(x² + 3x)?
- (A) −∞
- (B) +∞
- (C) 4/3
- (D) 0
Answer and explanation
Answer: (B). The top tends to −4. Just above −3, the bottom x(x + 3) is (negative) × (small positive): small and negative. So the quotient is large and positive.
- (A) is the limit from the left.
- (C) and (D) treat nonzero/0 as finite.
If you missed this: the sign check in the Topic 1.14 study guide.
Question 17 (multiple choice · 1.15)
Find every horizontal asymptote of the graph of f(x) = (2x − 5)/|x + 1|.
- (A) y = 2 only
- (B) y = 2 and y = −2
- (C) y = −5
- (D) None; there is only a vertical asymptote, x = −1
Answer and explanation
Answer: (B). For large positive x, |x + 1| = x + 1, so f(x) → 2. For large negative x, |x + 1| = −(x + 1), so f(x) → −2.
- (A) checks only x → ∞. (C) uses the constant term.
- (D) A vertical asymptote does not rule out horizontal ones.
If you missed this: Topic 1.15 study guide.
Question 18 (short answer · 1.16)
A function f is continuous on [1, 4], with f(1) = 6 and f(4) = −3.
(a) Explain why f(c) = 0 for some c in (1, 4). (b) Show that f(c) = c for some c in (1, 4). (c) Does the theorem guarantee f(c) = 7 for some c in (1, 4)?
Worked answer
(a) f is continuous on [1, 4], and 0 is between f(4) = −3 and f(1) = 6. By the Intermediate Value Theorem, f(c) = 0 for some c in (1, 4).
(b) g(x) = f(x) − x is continuous on [1, 4], with g(1) = 5 and g(4) = −7. By the theorem, g(c) = 0, so f(c) = c, for some c in (1, 4).
(c) No. 7 is not between −3 and 6, so the theorem makes no promise (f might still reach 7).
If you missed this: Topic 1.16 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 1.1 Change at an instant | 1 | Guide 1.1 |
| 1.2 Limit notation | 2 | Guide 1.2 |
| 1.3 Limits from graphs | 3 | Guide 1.3 |
| 1.4 Limits from tables | 4 | Guide 1.4 |
| 1.5 Limit properties | 5 | Guide 1.5 |
| 1.6 Algebraic rewriting | 6, 7 | Guide 1.6 |
| 1.7 Choosing a method | 8 | Guide 1.7 |
| 1.8 Squeeze theorem | 9 | Guide 1.8 |
| 1.9 Representations | 10 | Guide 1.9 |
| 1.10 Discontinuity types | 11 | Guide 1.10 |
| 1.11 Continuity at a point | 12, 13 | Guide 1.11 |
| 1.12 Continuity on intervals | 14 | Guide 1.12 |
| 1.13 Removing discontinuities | 15 | Guide 1.13 |
| 1.14 Vertical asymptotes | 16 | Guide 1.14 |
| 1.15 Horizontal asymptotes | 17 | Guide 1.15 |
| 1.16 IVT | 18 | Guide 1.16 |
How to use your result
- Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
- Fix gaps in Topics 1.5–1.7 and 1.11 first; the rest of the unit relies on them.
- For a gap, read the guide, then do the topic’s practice set.
- Then try the Unit 1 mixed review, where each question combines topics.