Limits and Continuity Worksheet Answer Key – AP Calculus AB

Problem 1 Limits from a graph

Read the open circles. Approaching from the left the branch heads to height 3; from the right it heads to 4.

1. 3
2. 4
3. The one-sided limits differ, so the limit is does not exist.
4. The filled point is at height 5, so h(1) = 5.
5. Near x = 3 the branch is 5 − x, which approaches 2.

Problem 2 Estimating from a table

As x gets closer to 2 from either side the outputs approach 12 (11.4100 at 1.9, 11.9401 at 1.99, 11.9940 at 1.999, and 12.0060 at 2.001). The estimate is 12.

Problem 3 Direct substitution

The function is a polynomial, so substitute: 3(4) + 8 + 1 = 12 + 8 + 1 = 21.

Problem 4 Factoring a 0/0 form

Substitution gives 0/0. Factor the numerator: (x − 4)(x + 3). Cancel x − 4 to get x + 3, then substitute 4: 7.

Problem 5 Using a conjugate

Multiply top and bottom by √(x + 25) + 5. The numerator becomes (x + 25) − 25 = x, so the fraction is xx(√(x + 25) + 5) = 1√(x + 25) + 5. At x = 0 this is 1/10: 1/10.

Problem 6 One-sided limits with absolute value

For x > 5, |2x − 10| = 2(x − 5), so g = 2. For x < 5, |2x − 10| = −2(x − 5), so g = −2. The limits are −2 from the left and 2 from the right, so the two-sided limit does not exist.

Problem 7 Limit at infinity of a rational function

Divide every term by x²: 6 − 1/x2 + 5/x². As x → ∞, the small terms go to 0, leaving 6/2. The limit is 3, so the horizontal asymptote is y = 3.

Problem 8 Limits at infinity with a root

For large |x|, √(x² + 9) behaves like |x|. As x → ∞, |x| = x, so the ratio tends to 4/1 = 4. As x → −∞, |x| = −x, so the ratio tends to 4/(−1) = −4. The limits are 4 and −4.

Problem 9 Infinite one-sided limits

Neither factor of the denominator cancels with the numerator, so the asymptotes are x = 7 and x = −3. Near 7 the numerator is about 8 and x + 3 is about 10, both positive. For x > 7 the factor x − 7 is a small positive number, so lim_x→7⁺ f(x) = +∞. For x < 7 it is a small negative number, so lim_x→7⁻ f(x) = −∞.

Problem 10 Squeeze Theorem

Since −1 ≤ cos(7/x) ≤ 1, multiplying by x² (which is nonnegative) gives −x² ≤ x²cos(7/x) ≤ x². Both bounds approach 0 as x → 0, so the limit is 0.

Problem 11 Making a function continuous

Factor: x² − 2x − 35 = (x − 7)(x + 5), so f(x) = x + 5 for x ≠ 7 and the limit at 7 is 12. Continuity requires f(7) = limit, so k = 12.

Problem 12 Classifying discontinuities

The function is undefined at −5 and at 2. At x = −5 the factor cancels, leaving x − 6x − 2, whose value at −5 is −11/−7 = 11/7. That is a removable discontinuity with limit 11/7. At x = 2 the factor x − 2 remains and the numerator tends to −4, so there is an infinite discontinuity (vertical asymptote): −∞ from the right, +∞ from the left.

Problem 13 A piecewise function and one parameter

The right limit is 7(2) − 9 = 5. The left limit equals f(2) = 4 + 2a. Set 4 + 2a = 5, so a = 1/2.

Problem 14 Intermediate Value Theorem for a polynomial

f is a polynomial, so it is continuous on [1, 2]. f(1) = 1 + 3 − 11 = −7 and f(2) = 8 + 6 − 11 = 3. Since 0 is between −7 and 3, the Intermediate Value Theorem guarantees a number c in (1, 2) with f(c) = 0. A zero exists in (1, 2).

Problem 15 Counting guaranteed zeros

(a) The sign changes between 0 and 2 (−3 to 4), between 2 and 4 (4 to −1), and between 4 and 6 (−1 to 5). Each interval has at least one zero, so at least 3 zeros.

(b) On [4, 6], 4.5 lies between −1 and 5, so the theorem guarantees a solution. On [0, 2], 4.5 is not between −3 and 4, so no solution is guaranteed.

Problem 16 Parking garage fees

(a) The cost just below 4 hours is $9 and just above is $12, so the limits are 9 and 12.

(b) At t = 3 the cost is constant at 9 nearby, so C is continuous at 3. At t = 4 the one-sided limits differ, so C has a jump discontinuity at 4.

(c) 4.5 is in (4, 6], so the cost is $12.

Problem 17 Medicine level in the blood

C(0) = 20/10 = 2 mg/L. As t → ∞ the ratio of leading coefficients is 45/3 = 15, so the long-run level is 15 mg/L. Solve 45t + 20 = 14(3t + 10) = 42t + 140, so 3t = 120 and t = 40 hours. The level approaches 15 but never reaches it.

Problem 18 Repairing a cubic quotient

Factor: (x − 3)(x² + 3x + 9)(x − 3)(x + 3). At x = 3 the factor cancels, and the limit is 9 + 9 + 96 = 9/2 — a removable discontinuity; define f(3) = 9/2. At x = −3 the numerator tends to 9 while the denominator tends to 0, so it is an infinite discontinuity with a vertical asymptote: the limit is −∞ from the left and +∞ from the right.

Problem 19 Horizontal and vertical asymptotes with a root

For large |x|, √(9x² + 4x) ≈ 3|x|. As x → ∞ the ratio tends to 3; as x → −∞, |x| = −x, so it tends to −3. The horizontal asymptotes are y = 3 and y = −3. At x = −1 the radicand is 9 − 4 = 5, so the numerator is √5 > 0 while x + 1 → 0. Hence lim_x→−1⁺ f(x) = +∞ and lim_x→−1⁻ f(x) = −∞: a vertical asymptote at x = −1.

Problem 20 Continuity of a three-piece function

At x = 2 the right limit is 4 + 1 = 5 and the middle piece gives 2b + 3. So 2b + 3 = 5 and b = 1. At x = −1 the middle piece gives −b + 3 = 2, and the left piece gives −2 + a. So −2 + a = 2 and a = 4. The values are a = 4 and b = 1.