AP Calc BC Limits and Continuity — Questions and Answers
Question 1: What is lim(x→3) (x² − 9)/(x − 3)?
- 0
- 6 (Correct answer)
- 3
- The limit does not exist
Correct answer: 6
Factor the numerator as (x − 3)(x + 3) and cancel (x − 3). The limit is then lim(x→3) (x + 3) = 6.
Direct substitution gives 0/0, which is an indeterminate form, so the limit has to be found another way. Factoring x² − 9 into (x − 3)(x + 3) lets the common factor cancel for all x ≠ 3. A limit only depends on values near 3, not at 3, so the cancellation is valid. The remaining expression x + 3 is continuous, so the limit is 3 + 3 = 6.
Question 2: What is lim(x→0) sin(5x)/(2x)?
- 2/5
- 1
- 5/2 (Correct answer)
- 5
Correct answer: 5/2
Rewrite as (5/2)·[sin(5x)/(5x)]. Since sin(u)/u → 1 as u → 0, the limit is 5/2.
The basic trig limit is lim(u→0) sin(u)/u = 1. Here the argument of sine is 5x, so write sin(5x)/(2x) = (5/2)·sin(5x)/(5x). As x → 0, 5x → 0, so the fraction sin(5x)/(5x) approaches 1. The limit is therefore 5/2.
Question 3: Let f(x) = kx + 2 for x ≤ 2 and f(x) = x² − k for x > 2. For what value of k is f continuous at x = 2?
- 2
- 1
- 3/2
- 2/3 (Correct answer)
Correct answer: 2/3
Continuity requires the left and right limits to equal f(2). So 2k + 2 = 4 − k, which gives k = 2/3.
A function is continuous at a point when the limit exists and equals the function value there. For a piecewise function, this means both one-sided limits must match f(2). The left piece gives 2k + 2, which is also f(2). The right piece gives 2² − k = 4 − k. Setting 2k + 2 = 4 − k gives 3k = 2, so k = 2/3.
Question 4: What is lim(x→∞) (3x² − 5x + 1)/(2x² + 7)?
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 every term by x², the highest power in the denominator. This gives (3 − 5/x + 1/x²)/(2 + 7/x²). As x → ∞, the terms 5/x, 1/x² and 7/x² all go to 0. The limit is 3/2, so y = 3/2 is a horizontal asymptote.
Question 5: What is lim(x→2⁻) 1/(x − 2)?
- +∞
- −∞ (Correct answer)
- 0
- 1/2
Correct answer: −∞
For x slightly less than 2, x − 2 is a small negative number, so 1/(x − 2) is a large negative number. The limit is −∞.
The notation x → 2⁻ means x approaches 2 from values smaller than 2. In that case the denominator x − 2 is negative and its magnitude shrinks toward 0. A fixed positive numerator divided by a tiny negative number is a very large negative number. The function decreases without bound, so the one-sided limit is −∞ and x = 2 is a vertical asymptote.
Question 6: A function f is continuous on [1, 4] with f(1) = −3 and f(4) = 5. Which statement must be true?
- f is increasing on [1, 4]
- f(x) = 0 has exactly one solution in (1, 4)
- f(c) = 0 for at least one c in (1, 4) (Correct answer)
- f(c) = 6 for some c in (1, 4)
Correct answer: f(c) = 0 for at least one c in (1, 4)
0 lies between f(1) = −3 and f(4) = 5, so the Intermediate Value Theorem guarantees at least one c in (1, 4) with f(c) = 0.
The Intermediate Value Theorem says that if f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for at least one c in (a, b). Here 0 is between −3 and 5, so a root is guaranteed. The theorem says nothing about how many such values exist or whether f is monotonic. The value 6 is outside the interval from −3 to 5, so it is not guaranteed.
Question 7: What type of discontinuity does f(x) = (x² − x − 6)/(x − 3) have at x = 3?
- Jump discontinuity
- Infinite discontinuity
- Oscillating discontinuity
- Removable discontinuity (Correct answer)
Correct answer: Removable discontinuity
The numerator factors as (x − 3)(x + 2), so f(x) = x + 2 for x ≠ 3. The limit is 5 even though f(3) is undefined, which makes the discontinuity removable.
At x = 3 the function is undefined, so it is not continuous there. Factoring gives (x − 3)(x + 2)/(x − 3), which simplifies to x + 2 when x ≠ 3. The limit as x → 3 exists and equals 5. A discontinuity where the limit exists but the function value is missing or different is called removable. Defining f(3) = 5 would make the function continuous.
Question 8: What is lim(x→0) (1 − cos x)/x²?
Correct answer: 1/2
Multiply by (1 + cos x)/(1 + cos x) to get sin²x / [x²(1 + cos x)]. This approaches 1·(1/2) = 1/2.
Direct substitution gives 0/0. Multiplying by the conjugate turns the numerator into 1 − cos²x = sin²x. The expression becomes (sin x / x)² · 1/(1 + cos x). As x → 0, sin x / x → 1 and 1/(1 + cos x) → 1/2. The product is 1/2.
Question 9: What is lim(x→2) |x − 2|/(x − 2)?
- 1
- Does not exist (Correct answer)
- −1
- 0
Correct answer: Does not exist
From the right the expression equals 1, and from the left it equals −1. The one-sided limits differ, so the two-sided limit does not exist.
A two-sided limit exists only when both one-sided limits exist and are equal. If x > 2, then |x − 2| = x − 2, so the quotient is 1. If x < 2, then |x − 2| = −(x − 2), so the quotient is −1. Because 1 ≠ −1, the function has a jump at x = 2 and the limit does not exist.
Question 10: What are the horizontal asymptotes of f(x) = 2x/√(x² + 1)?
- y = 2 only
- y = −2 only
- y = 2 and y = −2 (Correct answer)
- There are no horizontal asymptotes
Correct answer: y = 2 and y = −2
As x → ∞, √(x² + 1) ≈ x, so f → 2. As x → −∞, √(x² + 1) ≈ |x| = −x, so f → −2.
For large |x|, the denominator √(x² + 1) behaves like |x|. When x is positive, |x| = x and the ratio 2x/x approaches 2. When x is negative, |x| = −x and the ratio 2x/(−x) approaches −2. The two end behaviors give different limits, so the graph has two horizontal asymptotes, y = 2 and y = −2.
