AP Calc AB Limits and Continuity — Questions and Answers
Question 1: What is lim(x→2) (x² − 4)/(x − 2)?
- 0
- 2
- 4 (Correct answer)
- The limit does not exist
Correct answer: 4
Factor the numerator as (x − 2)(x + 2) and cancel the common factor. That leaves lim(x→2) (x + 2) = 4.
Direct substitution gives the indeterminate form 0/0, which means the expression needs algebraic simplification. The numerator is a difference of squares, so x² − 4 = (x − 2)(x + 2). For x ≠ 2 the factor (x − 2) cancels, and the function equals x + 2. A limit depends only on values near x = 2, not at it, so the limit is 2 + 2 = 4. The original function has a hole at (2, 4).
Question 2: What is lim(x→∞) (3x² + 5x)/(2x² − 1)?
Correct answer: 3/2
The numerator and denominator have the same degree, so the limit is the ratio of the leading coefficients, 3/2.
Divide the numerator and denominator by x², the highest power present. This gives (3 + 5/x)/(2 − 1/x²). As x → ∞, the terms 5/x and 1/x² go to 0, leaving 3/2. When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. When the numerator's degree is smaller, the limit is 0. When it is larger, the limit is unbounded.
Question 3: Let f(x) = kx + 1 for x ≤ 3 and f(x) = x² − 5 for x > 3. For what value of k is f continuous at x = 3?
- 4/3
- 3
- 1/3
- 1 (Correct answer)
Correct answer: 1
The right-hand limit at 3 is 3² − 5 = 4. The left piece must give 3k + 1 = 4, so k = 1.
Continuity at x = 3 requires that the left-hand limit, the right-hand limit, and f(3) all agree. From the left, the limit and f(3) both equal 3k + 1. From the right, the limit is 9 − 5 = 4. Setting 3k + 1 = 4 gives 3k = 3, so k = 1. Solving 3k = 4 by mistake gives the distractor 4/3.
Question 4: What is lim(x→0) sin(5x)/(2x)?
- 5/2 (Correct answer)
- 2/5
- 5
- 1
Correct answer: 5/2
Rewrite the expression as (5/2)·[sin(5x)/(5x)]. Since sin(u)/u → 1 as u → 0, the limit is 5/2.
The special limit lim(u→0) sin(u)/u = 1 needs the same expression inside the sine and in the denominator. Multiply and divide by 5 to get (5/2)·sin(5x)/(5x). As x → 0, u = 5x → 0, so the fraction sin(5x)/(5x) approaches 1. The limit is then (5/2)(1) = 5/2. Forgetting the 2 in the denominator gives 5, and flipping the ratio gives 2/5.
Question 5: The function f is continuous on the closed interval [1, 4], with f(1) = −3 and f(4) = 6. Which statement must be true?
- f(x) > 0 for every x in (1, 4)
- f′(c) = 0 for some c in (1, 4)
- f has a maximum value of 6 on [1, 4]
- f(c) = 0 for some c in (1, 4) (Correct answer)
Correct answer: f(c) = 0 for some c in (1, 4)
By the Intermediate Value Theorem, a continuous f takes every value between −3 and 6 on (1, 4). Since 0 lies between them, f(c) = 0 for some c in (1, 4).
The Intermediate Value Theorem says that if f is continuous on [a, b], then f takes every value between f(a) and f(b). Here 0 lies between −3 and 6, so a zero must exist in the open interval (1, 4). The other statements are not guaranteed. f may stay negative for a while before crossing zero, and its maximum may exceed 6. Nothing says f is differentiable, so no horizontal tangent is guaranteed.
Question 6: What is lim(x→3⁻) 2/(x − 3)?
- −∞ (Correct answer)
- +∞
- 0
- 2/3
Correct answer: −∞
As x approaches 3 from the left, x − 3 is a small negative number. Dividing 2 by it gives a large negative value, so the function decreases without bound.
A nonzero numerator over a denominator approaching 0 produces an unbounded result. The sign depends on the side of approach. For x slightly less than 3, such as 2.99, x − 3 = −0.01, so 2/(x − 3) = −200. The values become more negative as x gets closer to 3. The left-hand limit is −∞, while the right-hand limit is +∞, so the line x = 3 is a vertical asymptote.
Question 7: Let f(x) = (x² − x − 6)/((x − 3)(x + 1)). At which value of x does f have a removable discontinuity?
Correct answer: 3
The numerator factors as (x − 3)(x + 2), so the factor (x − 3) cancels and leaves a hole at x = 3. At x = −1 the denominator still vanishes, which gives a vertical asymptote.
Factoring gives f(x) = (x − 3)(x + 2)/((x − 3)(x + 1)). A factor that cancels corresponds to a removable discontinuity, because the limit exists there. After canceling, the limit as x → 3 is (3 + 2)/(3 + 1) = 5/4. The factor (x + 1) does not cancel, so x = −1 is a nonremovable infinite discontinuity. The value x = −2 is a zero of f, not a discontinuity.
Question 8: What is lim(x→0) (√(x + 9) − 3)/x?
- 1/3
- 0
- 1/6 (Correct answer)
- 3
Correct answer: 1/6
Multiply the numerator and denominator by the conjugate √(x + 9) + 3. The expression becomes 1/(√(x + 9) + 3), which approaches 1/6.
Direct substitution gives 0/0. Multiplying by (√(x + 9) + 3)/(√(x + 9) + 3) turns the numerator into (x + 9) − 9 = x. The x in the numerator cancels the x in the denominator, leaving 1/(√(x + 9) + 3). As x → 0 this becomes 1/(3 + 3) = 1/6. This result equals the derivative of √x at x = 9.
Question 9: Let f(x) = x² for x ≠ 1 and f(1) = 3. Which statement is true?
- lim(x→1) f(x) = 1, but f is not continuous at x = 1 (Correct answer)
- f is continuous at x = 1 because f(1) is defined
- lim(x→1) f(x) = 3
- lim(x→1) f(x) does not exist
Correct answer: lim(x→1) f(x) = 1, but f is not continuous at x = 1
Limits ignore the value at the point, so lim(x→1) f(x) = lim(x→1) x² = 1. Because this differs from f(1) = 3, f is not continuous at x = 1.
Continuity at a requires three things: f(a) is defined, the limit exists, and the limit equals f(a). Near x = 1 (but not at it), f(x) = x², so the limit is 1. The function value is f(1) = 3, which does not match. This is a removable discontinuity, and redefining f(1) = 1 would make f continuous. Having f(1) defined is not enough for continuity.
Question 10: What is lim(x→−∞) (2x + 1)/√(x² + 4)?
Correct answer: −2
For large negative x, √(x² + 4) behaves like |x| = −x. The ratio then approaches 2x/(−x) = −2.
Divide the numerator and denominator by |x|, which equals −x as x → −∞. The denominator becomes √(1 + 4/x²), which approaches 1. The numerator becomes (2x + 1)/(−x) = −2 − 1/x, which approaches −2. So the limit is −2. As x → +∞, the same expression approaches +2, so the function has two different horizontal asymptotes.
