AP Calculus AB Quiz: Connecting Multiple Representations Of Limits

What this quiz covers

This quiz focuses on Connecting Multiple Representations Of Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The function is continuous at $x=1$ because the limit exists.
2. The function has a jump discontinuity at $x=1$.
3. The function has a removable discontinuity at $x=1$. (correct answer)
4. The function has an infinite discontinuity at $x=1$.

Explanation: A removable discontinuity occurs at $x=c$ when ${\lim_{x \to c} f(x)}$ exists but is not equal to $f(c)$. Here, the limit as $x$ approaches 1 exists and is 4, but the function value $f(1)$ is 6. This perfectly matches the definition of a removable discontinuity, which would appear as a 'hole' in the graph at $(1, 4)$, with a point at $(1, 6)$. A is incorrect because for continuity, the limit must equal the function value. B is incorrect because a jump discontinuity requires the left and right-sided limits to be different. D is incorrect because an infinite discontinuity requires the limit to be infinite.

Question 2

1. $\lim_{x \to -3^+} f(x) = 5$
2. $\lim_{x \to -3^-} f(x) = 5$ (correct answer)
3. $\lim_{x \to 5} f(x) = -3$
4. $\lim_{x \to -3} f(x) = 5$

Explanation: The given $x$ values ($-3.1, -3.01, -3.001$) are all less than -3 and are approaching -3. This constitutes an approach from the left. The corresponding $f(x)$ values are approaching 5. Therefore, the data suggests that the left-sided limit is 5, which is written as ${\lim_{x \to -3^-} f(x) = 5}$. A is incorrect as the data is for a left-sided approach. C incorrectly swaps the roles of $x$ and $f(x)$. D is a two-sided limit, and we only have evidence for the left side.

Question 3

1. The graph of $g$ has a horizontal asymptote at $y=1$.
2. The graph of $g$ has a jump discontinuity at $x=1$.
3. The graph of $g$ has a vertical asymptote at $x=1$. (correct answer)
4. The graph of $g$ is continuous but not differentiable at $x=1$.

Explanation: When the one-sided limit of a function as $x$ approaches a finite number $c$ is either positive or negative infinity, it signifies the presence of a vertical asymptote at $x=c$. In this case, as $x$ approaches 1, the function's values are unbounded, indicating a vertical asymptote at $x=1$. The fact that the limits are different from each side describes the behavior on either side of the asymptote. A describes end behavior, not behavior at a finite point. B describes a discontinuity where both one-sided limits are finite but different. D is incorrect because the function is not continuous where it has an infinite limit.

Question 4

1. $\lim_{x \to 2^-} f(x) = 4$ and $\lim_{x \to 2^+} f(x) = 4$ (correct answer)
2. $\lim_{x \to 2^-} f(x) = 2$ and $\lim_{x \to 2^+} f(x) = 2$
3. $\lim_{x \to 2^-} f(x) = 4$ and $\lim_{x \to 2^+} f(x) = 2$
4. $\lim_{x \to 2^-} f(x) = 2$ and $\lim_{x \to 2^+} f(x) = 4$

Explanation: To find the left-sided limit ($x \to 2^-$), we use the piece of the function defined for $x < 2$, which is $f(x) = x^2$. So, ${\lim_{x \to 2^-} x^2 = 2^2 = 4}$. To find the right-sided limit ($x \to 2^+$), we use the piece of the function defined for $x > 2$, which is $f(x) = 3x - 2$. So, ${\lim_{x \to 2^+} (3x-2) = 3(2) - 2 = 4}$. Both one-sided limits are equal to 4. The other choices contain incorrect calculations for one or both of the one-sided limits.

Question 5

1. The function is continuous at $x=0$.
2. The function has a jump discontinuity at $x=0$ and is continuous from the right at $x=0$. (correct answer)
3. The function has a removable discontinuity at $x=0$.
4. The function has a jump discontinuity at $x=0$ and is continuous from the left at $x=0$.

Explanation: The left-sided limit (-1) and the right-sided limit (1) are not equal, so the function has a jump discontinuity at $x=0$. The definition of continuity from the right at a point $c$ is ${\lim_{x \to c^+} f(x) = f(c)}$. In this case, ${\lim_{x \to 0^+} f(x) = 1}$ and $f(0) = 1$, so the function is continuous from the right. The function is not continuous from the left because ${\lim_{x \to 0^-} f(x) = -1 \neq f(0)}$. Thus, B is the most complete and accurate description. A is false due to the jump. C is false because the two-sided limit doesn't exist. D is false because it is not continuous from the left.

Question 6

1. $\lim_{x \to 5^-} f(x) = \infty$
2. $\lim_{x \to 5^+} f(x) = \infty$ (correct answer)
3. $\lim_{x \to \infty} f(x) = 5$
4. $\lim_{x \to 5} f(x) = \infty$

Explanation: The description 'as $x$ approaches 5 from the right' corresponds to a right-sided limit, denoted by $x \to 5^+$. The description 'the graph of $f(x)$ increases without bound' means the function's values approach positive infinity. Therefore, the correct notation is ${\lim_{x \to 5^+} f(x) = \infty}$. A represents the behavior as $x$ approaches 5 from the left. C describes end behavior, which corresponds to a horizontal asymptote, not a vertical one. D represents a two-sided limit, but the description only provides information about the right-sided behavior.

Question 7

1. $\lim_{x \to 4} g(x) = -\infty$
2. $\lim_{x \to -\infty} g(x) = 4$ (correct answer)
3. $\lim_{x \to 4^-} g(x) = \infty$
4. $\lim_{x \to \infty} g(x) = -4$

Explanation: The phrase 'as $x$ becomes large in the negative direction' means $x$ is approaching negative infinity ($x \to -\infty$). The phrase 'its values approach 4' means the function's output approaches 4. Therefore, the correct notation is ${\lim_{x \to -\infty} g(x) = 4}$. A and C describe behavior near the finite value $x=4$. D describes behavior as $x$ approaches positive infinity and incorrectly states the limit value is -4.

Question 8

1. $\lim_{x \to 1} f(x) = 5$ (correct answer)
2. $\lim_{x \to 1} f(x) = \infty$
3. $\lim_{x \to 5} f(x) = 1$
4. The limit does not exist because $f(1)$ is not given.

Explanation: The table of values shows that as $x$ approaches 1 from the left (values like 0.9, 0.99, 0.999), the values of $f(x)$ get closer to 5. As $x$ approaches 1 from the right (values like 1.1, 1.01, 1.001), the values of $f(x)$ also get closer to 5. Since the function approaches the same value from both sides, the limit is 5. B is incorrect because the values are approaching a finite number. C incorrectly reverses the roles of the input and the limit value. D is incorrect because the existence of a limit at a point does not depend on the function's value at that point.

Question 9

1. $g(-1) = 7$
2. $\lim_{x \to 7} g(x) = -1$
3. $\lim_{x \to -1} g(x) = 7$ (correct answer)
4. $\lim_{x \to -1^+} g(x) = 7$ and $\lim_{x \to -1^-} g(x) \neq 7$

Explanation: The verbal description indicates that as the input $x$ approaches -1, the output $g(x)$ approaches 7. This is the definition of a two-sided limit. The notation ${\lim_{x \to -1} g(x) = 7}$ correctly represents this statement. A represents the function value at a point, not the limit. B reverses the roles of the input and the limit value. D describes a situation where the right-sided limit is 7 but the left-sided limit is not, so the two-sided limit would not exist.

Question 10

1. The value of the function at $x=4$ is exactly 2.
2. As $x$ approaches 4, the values of $f(x)$ can be made arbitrarily close to 2. (correct answer)
3. The value of $f(x)$ approaches 4 as $x$ approaches 2.
4. For every value of $x$ near 4, the corresponding value of $f(x)$ is 2.

Explanation: The statement ${\lim_{x \to 4} f(x) = 2}$ describes the behavior of the function $f(x)$ as $x$ gets close to 4. It means the function's values approach 2. The value of $f(4)$ itself is not relevant to the limit. Choice B correctly states this concept. A is incorrect because the limit of a function as $x$ approaches a point does not depend on the value of the function at that point. C incorrectly swaps the roles of $x$ and $f(x)$. D is too strong; the values only need to approach 2, not be exactly 2.

Question 11

1. The graph has a vertical asymptote at $x = -3$.
2. The graph has a hole at the point where the y-coordinate is -3.
3. The graph has a horizontal asymptote at $y = -3$. (correct answer)
4. The graph intersects the y-axis at $y = -3$.

Explanation: A limit at infinity describes the end behavior of a function. If the limit of $f(x)$ as $x$ approaches infinity is a finite number $L$, this means the graph of the function gets arbitrarily close to the horizontal line $y=L$. Therefore, the graph has a horizontal asymptote at $y = -3$. A describes the behavior ${\lim_{x \to -3} f(x) = \pm\infty}$. B describes a removable discontinuity, which is related to a limit at a finite point. D describes the y-intercept, which is $f(0)$, not a limit at infinity.

Question 12

1. $f(2) = 5$
2. $f(x)$ is continuous at $x=2$.
3. $\lim_{x \to 2} f(x) = 5$ (correct answer)
4. The graph of $f(x)$ has a vertical asymptote at $x=2$.

Explanation: For a two-sided limit to exist, the left-sided limit and the right-sided limit must exist and be equal. Since both the left-sided and right-sided limits as $x$ approaches 2 are equal to 5, the two-sided limit ${\lim_{x \to 2} f(x)}$ must exist and be equal to 5. A is not necessarily true; the value of the function at the point is independent of the limit. B is not necessarily true; for continuity, we would also need to know that $f(2) = 5$. D is incorrect; asymptotes involve infinite limits, not finite ones.

Question 13

1. As $x$ approaches -2 from the left, the values of $f(x)$ decrease without bound. (correct answer)
2. As $x$ approaches -2 from the right, the values of $f(x)$ decrease without bound.
3. The values of $f(x)$ can be made arbitrarily close to -2 by taking $x$ sufficiently large and negative.
4. As $x$ approaches negative infinity, the values of $f(x)$ approach -2.

Explanation: The notation $x \to -2^-$ means 'as x approaches -2 from the left' (from values less than -2). The notation $f(x) \to -\infty$ means 'the values of f(x) decrease without bound.' Choice A correctly combines these two interpretations. B describes the right-sided limit, ${\lim_{x \to -2^+} f(x) = -\infty}$. C and D both describe end behavior (limits at infinity), not the behavior near a finite point.

Question 14

1. $\lim_{x \to 3} h(x) = -2$ and $\lim_{x \to -1} h(x) = -2$
2. $\lim_{x \to -2} h(x) = 3$ and $\lim_{x \to -2} h(x) = -1$
3. $\lim_{x \to -2^-} h(x) = 3$ and $\lim_{x \to -2^+} h(x) = -1$ (correct answer)
4. $h(-2) = 3$ and the limit does not exist.

Explanation: The description 'as $x$ gets closer to -2 from the left, the $y$-values get closer to 3' corresponds to the left-sided limit ${\lim_{x \to -2^-} h(x) = 3}$. The description 'as $x$ gets closer to -2 from the right, the $y$-values get closer to -1' corresponds to the right-sided limit ${\lim_{x \to -2^+} h(x) = -1}$. Choice C correctly pairs these statements. A incorrectly swaps the inputs and outputs. B is logically impossible, as a limit cannot equal two different values. D makes a conclusion about the function value $h(-2)$, which is not given in the description of the limits.

Question 15

1. A removable discontinuity (a hole)
2. A vertical asymptote
3. A sharp corner or cusp
4. A jump discontinuity (correct answer)

Explanation: The numerical data indicates that the left-sided limit (${\lim_{x \to c^-} f(x) = 5}$) and the right-sided limit (${\lim_{x \to c^+} f(x) = -5}$) both exist but are not equal. This is the definition of a jump discontinuity. The graph 'jumps' from one finite value to another at $x=c$. A is incorrect because a removable discontinuity has a two-sided limit that exists. B is incorrect because a vertical asymptote involves an infinite limit. C is incorrect because a corner or cusp is a point where the function is continuous but not differentiable; the limit would exist.

Question 16

1. $p(2) = 0$ and $q(2) = 0$
2. $p(2) \neq 0$ and $q(2) = 0$ (correct answer)
3. $p(2) = 0$ and $q(2) \neq 0$
4. $p(2) \neq 0$ and $q(2) \neq 0$

Explanation: An infinite limit for a rational function typically occurs at a vertical asymptote. A vertical asymptote at $x=c$ happens when the denominator is zero and the numerator is non-zero at $x=c$. Therefore, for the limit to be infinite at $x=2$, we would expect $q(2)=0$ and $p(2) \neq 0$. A would lead to an indeterminate form $0/0$, which usually corresponds to a removable discontinuity (a hole). C would result in a limit of 0. D would result in a finite, non-zero limit, and the function would be continuous at $x=2$.

Question 17

1. The graph of $f(x)$ must pass through the point $(c, L)$.
2. For any horizontal strip around $y=L$, there is a vertical strip around $x=c$ such that the graph within the vertical strip stays within the horizontal strip. (correct answer)
3. The graph of $f(x)$ must be a straight line with slope $L$ near the point $x=c$.
4. For any vertical strip around $x=L$, there is a horizontal strip around $y=c$ such that the graph within the horizontal strip stays within the vertical strip.

Explanation: This question describes the epsilon-delta definition of a limit in graphical terms. 'Making $f(x)$ as close as we want to $L$' corresponds to defining a small horizontal strip (an epsilon-neighborhood) around the line $y=L$. 'By making $x$ sufficiently close to $c$' corresponds to finding a vertical strip (a delta-neighborhood) around the line $x=c$ such that the portion of the graph inside this vertical strip (excluding possibly at $x=c$) is contained entirely within the horizontal strip. Choice B accurately describes this relationship. A is incorrect; the function value at $c$ is irrelevant. C confuses the limit value $L$ with the slope. D incorrectly swaps the roles of the horizontal and vertical strips.

Question 18

1. As $x$ approaches 0, the values of $g(x)$ alternate between 1 and -1 but do not settle on a single value. (correct answer)
2. As $x$ approaches 0 from the left, $g(x)$ approaches 2, and from the right, $g(x)$ approaches -2.
3. As $x$ approaches 0, the values of $g(x)$ grow infinitely large.
4. As $x$ approaches 0, the values of $g(x)$ get closer and closer to $g(0)$.

Explanation: A limit fails to exist due to oscillation when the function's values do not approach a single number but instead fluctuate between two or more values as $x$ gets closer to the limit point. The classic example is $g(x) = \sin(1/x)$ near $x=0$. Choice A describes this behavior. B describes a jump discontinuity, a different reason for a limit not to exist. C describes an infinite limit, another reason for a limit not to exist, but it's not oscillation. D describes a function that is continuous at $x=0$, so the limit exists.

Question 19

1. The graph of $f$ has a hole at $x=-1$.
2. The graph of $f$ is continuous at $x=-1$.
3. The graph of $f$ has a jump discontinuity at $x=-1$. (correct answer)
4. The graph of $f$ has a sharp corner at $x=-1$.

Explanation: If the limit ${\lim_{x \to -1} f(x)}$ exists, it means that ${\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x)}$. A jump discontinuity is defined by the left- and right-sided limits existing but being unequal. Therefore, if the two-sided limit exists, the graph cannot have a jump discontinuity. A, B, and D are all situations where the two-sided limit exists. A hole (removable discontinuity) has a limit that exists but is not equal to the function value. A continuous point has a limit that exists and equals the function value. A sharp corner has a limit that exists and equals the function value (it's continuous but not differentiable).