Limits and Continuity Test Answer Key with Scoring – AP Calculus AB

Question 1 Evaluate a limit (2 points)

Substitution gives 0/0. Factor: (x − 2)(x + 1)(x − 2)(x + 2) = x + 1x + 2 near 2, which tends to 3/4. Answer (B).

Scoring: 2 points for the correct choice; 1 point if the work shows the correct factorization but the choice is wrong.

Question 2 Reading a graph (3 points)

(a) Both branches approach 3 at x = 2, so the limit is 3.

(b) lim_x→4⁻ k(x) = 1, lim_x→4⁺ k(x) = 3 and k(4) = 1.

(c) Not continuous: the one-sided limits at 4 are different, so the limit does not exist (jump discontinuity).

Scoring: 1 point per part.

Question 3 Limit at infinity (2 points)

Divide numerator and denominator by x³: 7 − 1/x²2/x³ − 5 → 7/(−5). Answer (A).

Scoring: 2 points for the correct choice; 1 point for noting that the degrees are equal.

Question 4 Conjugate method (2 points)

Multiply by the conjugate: (x + 16) − 16x(√(x + 16) + 4) = 1√(x + 16) + 4. At x = 0 this is 1/8. The limit is 1/8.

Scoring: 1 point for using the conjugate, 1 point for the correct simplification and value.

Question 5 Continuity at a point (3 points)

(a) Factor: 2x² − 11x − 6 = (2x + 1)(x − 6), so f(x) = 2x + 1 for x ≠ 6, and the limit is 13.

(b) f(6) = 10 exists and the limit 13 exists, but 10 ≠ 13, so not continuous.

(c) Redefining f(6) = 13 removes the break, so the original is a removable discontinuity.

Scoring: 1 point per part.

Question 6 Discontinuities of a rational function (3 points)

(a) f(x) = (x − 3)(x + 2)(x − 3)(x + 3). At x = 3 the factor cancels and the limit is 5/6: removable. At x = −3 the denominator factor remains: infinite (vertical asymptote).

(b) Near −3 the function behaves like x + 2x + 3 with numerator close to −1. From the left x + 3 < 0, so the limit is +∞; from the right it is −∞.

Scoring: Part (a): 1 point per correct classification with reason. Part (b): 1 point for both correct.

Question 7 Intermediate Value Theorem (2 points)

f is a polynomial, hence continuous on [1, 2]. f(1) = 1 + 1 − 9 = −7 and f(2) = 16 + 2 − 9 = 9. Since 0 is between −7 and 9, the Intermediate Value Theorem gives a c in (1, 2) with f(c) = 0. A solution exists.

Scoring: 1 point for continuity, 1 point for the sign change and conclusion.

Question 8 Asymptotes and one-sided limits (3 points)

(a) For x slightly above 5: numerator 7 > 0, x − 5 small positive, x + 3 = 8 > 0, so +∞; from the left, −∞.

(b) Numerator near −1. From the left of −3: x + 3 < 0, x − 5 < 0 (product of those two is positive small) so the quotient is negative: −∞; from the right: +∞.

(c) The degree of the denominator is greater, so the limit at ±∞ is 0 and the horizontal asymptote is y = 0.

Scoring: 1 point per part.