Practice Set B: Limits and Continuity | AP Calculus AB

Question 1

Which of the following is true about the limit lim_{x→2} (x^2 - 4)?

Solution: The limit lim_{x→2} (x^2 - 4) = 0 because the function approaches 0 as x approaches 2, regardless of the value of f(2).

Question 2

For a function f(x) with a hole at x = -1 and the function approaching -3 from both sides, what is lim_{x→-1} f(x)?

Solution: The limit as x approaches -1 is the value the function approaches, which is -3, regardless of the hole at x = -1.

Question 3

What is lim_{x→-3} (4x + 7)?

Solution: Using direct substitution, lim_{x→-3} (4x + 7) = 4(-3) + 7 = -12 + 7 = -5.

Question 4

Which type of discontinuity does f(x) = 1 / (x^2) have at x = 0?

Solution: The function f(x) = 1 / (x^2) has a vertical asymptote at x = 0, indicating an infinite discontinuity.

Question 5

For the function f(x) = (x^2 - 4) / (x - 2), what is lim_{x→2} f(x)?

Solution: The function simplifies to f(x) = x + 2 for x ≠ 2. Thus, lim_{x→2} f(x) = 2 + 2 = 4.

Question 6

Given the table of values for f(x) near x = 2, what is lim_{x→2} f(x)? | x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 | |---|-----|-----|------|------|-----|-----| | f(x) | 3.9 | 3.99 | 3.999 | 4.001 | 4.01 | 4.1 |

Solution: As x approaches 2 from both sides, f(x) approaches 4, so lim_{x→2} f(x) = 4.

Question 7

If -x^2 ≤ f(x) ≤ x^2 for all x ≠ 0, what is lim_{x→0} f(x)?

Solution: By the Squeeze Theorem, since lim_{x→0} (-x^2) = lim_{x→0} (x^2) = 0, lim_{x→0} f(x) = 0.

Question 8

For f(x) to be continuous at x = a, which of the following must be true?

Solution: Continuity at a point requires three conditions: f(a) is defined, lim_{x→a} f(x) exists, and lim_{x→a} f(x) = f(a).

Question 9

What is lim_{x→∞} (4x^2 + 3x - 1) / (2x^2 - 5)?

Solution: Divide numerator and denominator by x^2: lim_{x→∞} (4 + 3/x - 1/x^2) / (2 - 5/x^2) = 4/2 = 2.

Question 10

Which of the following functions f(x) has a limit as x approaches 0, but the function is not defined at x = 0?

Solution: The limit as x approaches 0 of sin(x)/x is 1, but the function is undefined at x=0 because of division by zero. Option A has different one-sided limits, Option C has an infinite limit, and Option D has different one-sided limits.

Question 11

What is the horizontal asymptote of a function f(x) = (2 - 5x^3) / (4x^3 + x^2 - 1)?

Solution: For a rational function where the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients, which is -5/4.

Question 12

Let f(x) be a continuous function on the closed interval [0, 4] such that f(0) = 2 and f(4) = 8. Which of the following statements must be true?

Solution: By the Intermediate Value Theorem, since f is continuous, it must take on every value between 2 and 8 on the interval [0, 4]. 5 is in this range, so there must be an x such that f(x) = 5.

Question 13

Which of the following functions has a jump discontinuity at x = 1?

Solution: For a jump discontinuity the one-sided limits must both exist but be unequal. For C, lim_{x->1^-} = 1 + 2 = 3 and lim_{x->1^+} = 1 + 4 = 5, a finite jump. A is an infinite discontinuity, B is removable, and D oscillates (no one-sided limits).

Question 14

Find the value of lim_{x→3^+} (2x) / (x - 3).

Solution: As x approaches 3 from the right, (x - 3) is a very small positive number. (2x) is approximately 6. A positive number divided by a small positive number approaches positive infinity.

Question 15

A function f is defined by f(x) = { kx^2 for x ≤ 2; x + k for x > 2 }. For what value of k is f continuous at x = 2?

Solution: For continuity, the limits from both sides must equal the function value: k(2)^2 = 2 + k. So, 4k = 2 + k, which means 3k = 2, or k = 2/3.

Question 16

Which of the following best describes the instantaneous rate of change of f(x) at x = a?

Solution: The instantaneous rate of change at a specific point is represented by the slope of the tangent line at that point.

Question 17

Evaluate lim_{x→∞} (2x^2 + 1) / (x^2 - 3x).

Solution: Divide both numerator and denominator by x^2: (2 + 1/x^2) / (1 - 3/x). As x approaches infinity, this becomes (2 + 0) / (1 - 0) = 2.

Question 18

On which of the following intervals is f(x) = 1 / (x - 2) continuous?

Solution: The function has a vertical asymptote at x = 2. It is continuous everywhere except x = 2. Interval (2, 5] does not include 2, so it is continuous there.

Question 19

Evaluate lim_{x→1} (x - 1) / (sqrt(x) - 1).

Solution: Factor the numerator as a difference of squares: (sqrt(x)-1)(sqrt(x)+1) / (sqrt(x)-1). Canceling the terms gives sqrt(x)+1, and substituting 1 gives 1 + 1 = 2.

Question 20

A function f has a vertical asymptote at x = 3. Which of the following is most likely true?

Solution: A vertical asymptote at x = c implies that the function grows without bound (to infinity or negative infinity) as x approaches c from at least one side.