AP Calculus AB Quiz: Confirming Continuity Over An Interval

What this quiz covers

This quiz focuses on Confirming Continuity Over An Interval, 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. No; $\lim_{x\to1}Q(x)=1$ but $Q(1)=0$.
2. Yes; $\lim_{x\to1}Q(x)=0$ exists and equals $Q(1)=0$. (correct answer)
3. No; $\lim_{x\to1^-}Q(x)=0$ and $\lim_{x\to1^+}Q(x)=1$, so the limit does not exist.
4. No; $Q(1)$ is undefined.
5. No; $\lim_{x\to1}Q(x)=-1$ but $Q(1)=0$.

Explanation: Q at 1: Q(1)=0, both limits 0 matches. Continuous. Omission: assuming symmetry causes issue. Checklist: f(a), sides, match.

Question 2

1. Yes; $\lim_{x\to0}V(x)=4$ exists and equals $V(0)=4$. (correct answer)
2. No; $\lim_{x\to0^-}V(x)=4$ and $\lim_{x\to0^+}V(x)=3$, so the limit does not exist.
3. No; $V(0)$ is undefined.
4. No; $\lim_{x\to0}V(x)=0$ but $V(0)=4$.
5. No; $\lim_{x\to0}V(x)=5$ but $V(0)=4$.

Explanation: V at 0: V(0)=4, left 4, right 4 matches. Continuous. Omission: miscalculating right limit as 3. Checklist: f(a), sides, match.

Question 3

1. No; $\lim_{x\to5}E(x)=10$ but $E(5)=5$.
2. Yes; $\lim_{x\to5}E(x)=10$ exists and equals $E(5)=10$. (correct answer)
3. No; $\lim_{x\to5}E(x)$ does not exist because the denominator is zero.
4. No; $E(5)$ is undefined.
5. No; $\lim_{x\to5}E(x)=0$ but $E(5)=10$.

Explanation: Continuity of E at x = 5 demands the limit exists, E(5) is defined, and they are equal. E(5) = 10 is defined. Simplifying ($x^2$ - 25)/(x - 5) = x + 5 for x ≠ 5, the limit as x approaches 5 is 10. Since the limit equals E(5), it is continuous. A common omission is failing to simplify removable discontinuities before evaluating the limit. Another is assuming undefined at the point means no continuity without checking the limit. Checklist: confirm f(a) defined, compute the limit (simplifying if needed), ensure limit equals f(a).

Question 4

1. Yes; $\lim_{x\to0}P(x)=-4$ exists and equals $P(0)=-4$. (correct answer)
2. No; $P(0)$ is undefined.
3. No; $\lim_{x\to0}P(x)=0$ but $P(0)=-4$.
4. No; $\lim_{x\to0}P(x)$ does not exist because one-sided limits differ.
5. No; $\lim_{x\to0}P(x)=4$ but $P(0)=-4$.

Explanation: P at 0: P(0)=-4, simplified limit x-4=-4 matches. Continuous. Omission: simplifying. Checklist: f(a), limit, equality.

Question 5

1. Yes; $\lim_{x\to1}M(x)=2$ equals $M(1)=2$.
2. No; $\lim_{x\to1^-}M(x)=1$ and $\lim_{x\to1^+}M(x)=2$, so the limit does not exist. (correct answer)
3. No; $M(1)$ is undefined.
4. No; $\lim_{x\to1}M(x)=1$ but $M(1)=2$.
5. No; $\lim_{x\to1}M(x)=3$ but $M(1)=2$.

Explanation: M at 1: M(1)=2, left limit 1, right 2, no limit. Not continuous. Omission: boundary inclusion. Checklist: f(a), sides agree, match.

Question 6

1. Yes; $\lim_{x\to2}J(x)=4$ exists and equals $J(2)=4$.
2. No; $\lim_{x\to2^-}J(x)=5$ and $\lim_{x\to2^+}J(x)=4$, so the limit does not exist. (correct answer)
3. No; $\lim_{x\to2}J(x)=5$ but $J(2)=4$.
4. No; $J(2)$ is undefined.
5. No; $\lim_{x\to2}J(x)=3$ but $J(2)=4$.

Explanation: J at 2: J(2)=4 defined, left limit 4+1=5, right 4, differ so no limit. Not continuous. Omission: not checking sides. Checklist: f(a), left/right agree, match.

Question 7

1. Yes; $\lim_{x\to-1}X(x)=1$ exists and equals $X(-1)=1$. (correct answer)
2. No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=5$.
3. No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=1$ is undefined.
4. No; $\lim_{x\to-1}X(x)$ does not exist because one-sided limits differ.
5. No; $\lim_{x\to-1}X(x)=0$ but $X(-1)=1$.

Explanation: X at -1: X(-1)=1, limit 2x+3=1 matches. Continuous. Omission: thinking middle differs. Checklist: f(a), limit, equality.

Question 8

1. Yes; $\lim_{x\to0}s(x)=1=s(0)$.
2. No; $\lim_{x\to0}s(x)=0\ne s(0)=1$. (correct answer)
3. No; $s(0)$ is undefined.
4. No; $\lim_{x\to0}s(x)$ does not exist because $\sin x$ oscillates.
5. Yes; $\sin x$ is continuous, so redefining $s(0)$ cannot affect continuity.

Explanation: To determine continuity at x = 0, we verify: (1) s(0) exists, (2) lim[x→0] s(x) exists, and (3) they are equal. The function defines s(0) = 1, so condition (1) is met. For the limit, since s(x) = sin x for x ≠ 0, we have lim[x→0] s(x) = lim[x→0] sin x = sin(0) = 0. We have s(0) = 1 but lim[x→0] s(x) = 0, so 1 ≠ 0, violating condition (3). This is a classic example where redefining a continuous function at a single point creates a discontinuity—the natural value of sin(0) = 0 has been changed to 1. Remember the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?

Question 9

1. Yes; $r(1)$ exists, so $r$ is continuous at $1$.
2. No; $\lim_{x\to1}r(x)=2$ but $r(1)=0$. (correct answer)
3. No; $r(1)$ is undefined.
4. Yes; $\lim_{x\to1}r(x)=0$ equals $r(1)$.
5. No; $\lim_{x\to1}r(x)$ does not exist because the denominator is $0$ at $x=1$.

Explanation: For r(x) = (x² - 1)/(x - 1) when x ≠ 1, we can factor and simplify: r(x) = (x + 1)(x - 1)/(x - 1) = x + 1 for x ≠ 1. Therefore, lim[x→1] r(x) = lim[x→1] (x + 1) = 2. We're told r(1) = 0, which is defined. However, since the limit (2) doesn't equal r(1) = 0, the function is not continuous at x = 1. This is a removable discontinuity that hasn't been properly removed—to make r continuous, we would need to define r(1) = 2, not 0. Students often assume that defining the function at a point automatically makes it continuous there, but the definition must match the limit. Continuity checklist: ✓ r(1) defined, ✓ Limit exists (= 2), ✗ Limit ≠ r(1) (2 ≠ 0).

Question 10

1. No; $\lim_{x\to1^-}s(x)=2$ and $\lim_{x\to1^+}s(x)=2$, but $s(1)=1$.
2. Yes; $\lim_{x\to1}s(x)$ exists and equals $2=s(1)$. (correct answer)
3. No; $s(1)$ is undefined.
4. No; $\lim_{x\to1}s(x)$ does not exist because the rule changes at $x=1$.
5. Yes; $s$ is continuous at $1$ because both pieces are linear.

Explanation: For this piecewise function, we need to check continuity at the transition point x = 1. From the definition, s(1) = 2(1) = 2 using the first piece (x ≤ 1). The left-hand limit: lim[x→1⁻] s(x) = lim[x→1⁻] 2x = 2. The right-hand limit: lim[x→1⁺] s(x) = lim[x→1⁺] 2 = 2. Since both one-sided limits equal 2, the two-sided limit exists and equals 2. With s(1) = 2 matching this limit, all continuity conditions are satisfied. The function smoothly transitions from the line y = 2x to the constant y = 2 at x = 1. Students sometimes incorrectly assume piecewise functions are automatically discontinuous at transition points. Continuity checklist for piecewise: ✓ s(1) defined (= 2), ✓ Left and right limits equal (both 2), ✓ Common limit equals s(1).

Question 11

1. Yes; $\lim_{x\to2}T(x)=6$ exists and equals $T(2)=6$.
2. No; $\lim_{x\to2^-}T(x)=6$ and $\lim_{x\to2^+}T(x)=7$, so the limit does not exist. (correct answer)
3. No; $\lim_{x\to2}T(x)=7$ but $T(2)=6$.
4. No; $T(2)$ is undefined.
5. No; $\lim_{x\to2}T(x)=5$ but $T(2)=6$.

Explanation: T at 2: T(2)=6, left 6, right 7, no limit. Not continuous. Omission: close values. Checklist: f(a), sides agree, match.

Question 12

1. Yes; $\lim_{x\to-2}G(x)=7$ exists and equals $G(-2)=7$.
2. No; $\lim_{x\to-2^-}G(x)=7$ and $\lim_{x\to-2^+}G(x)=3$, so the limit does not exist. (correct answer)
3. No; $\lim_{x\to-2}G(x)=3$ but $G(-2)=7$.
4. No; $G(-2)$ is undefined.
5. No; $\lim_{x\to-2}G(x)=5$ but $G(-2)=7$.

Explanation: For G continuous at x = -2, the limit must exist, G(-2) defined, and equal. G(-2) = 7 is defined, but left limit is (-2)^2 + 3 = 7 and right is -2 + 5 = 3, so limit does not exist. Hence, not continuous. Common omission: not computing one-sided limits for piecewise functions. Often, people check only the function value. Checklist: define f(a), check left/right limits agree, ensure match with f(a).

Question 13

1. No; $\lim_{x\to2}L(x)=2$ but $L(2)=0$.
2. Yes; $\lim_{x\to2}L(x)=0$ exists and equals $L(2)=0$. (correct answer)
3. No; $L(2)$ is undefined.
4. No; $\lim_{x\to2}L(x)$ does not exist because of a corner.
5. No; $\lim_{x\to2}L(x)=1$ but $L(2)=0$.

Explanation: L at 2: L(2)=0, limit of |x-2|=0 matches. Continuous. Omission: thinking absolute value causes jump. Checklist: f(a), limit, equality.

Question 14

1. Yes; $\lim_{x\to0}I(x)=1$ equals $I(0)=1$.
2. No; $\lim_{x\to0}I(x)$ does not exist (unbounded), though $I(0)=1$. (correct answer)
3. No; $I(0)$ is undefined.
4. No; $\lim_{x\to0}I(x)=0$ but $I(0)=1$.
5. No; $\lim_{x\to0}I(x)=-1$ but $I(0)=1$.

Explanation: I continuous at 0: limit exists, I(0) defined, equal. I(0)=1 defined, but 1/x limit unbounded, does not exist. Not continuous. Omission: assuming defined point fixes discontinuity. Checklist: f(a) defined, finite limit exists, equals f(a).

Question 15

1. Yes; $\lim_{x\to-3}K(x)=0$ exists and equals $K(-3)=0$.
2. No; $\lim_{x\to-3}K(x)=-1$ exists but $K(-3)=0$. (correct answer)
3. No; $K(-3)$ is undefined.
4. No; $\lim_{x\to-3}K(x)$ does not exist because one-sided limits differ.
5. No; $\lim_{x\to-3}K(x)=1$ but $K(-3)=0$.

Explanation: K at -3: K(-3)=0 defined, limit of x+2 = -1 ≠0. Not continuous. Omission: assuming redefinition matches limit. Checklist: f(a), limit, equality.

Question 16

1. No; $\lim_{x\to1^-}D(x)=0$ and $\lim_{x\to1^+}D(x)=1$, so the limit does not exist. (correct answer)
2. Yes; $\lim_{x\to1}D(x)=0$ exists and equals $D(1)=0$.
3. No; $D(1)$ is undefined.
4. No; $\lim_{x\to1}D(x)=1$ but $D(1)=0$.
5. No; $\lim_{x\to1}D(x)=-1$ but $D(1)=0$.

Explanation: For D to be continuous at x = 1, the limit must exist, D(1) must be defined, and they must match. D(1) = 0 is defined, but the left limit is 1^2 - 1 = 0 and the right limit is 2 - 1 = 1, so the limit does not exist due to differing one-sided limits. Therefore, D is not continuous at x = 1. A common omission is not checking both one-sided limits separately when the function is piecewise. People often assume the function value dictates continuity without limit agreement. Use this checklist for continuity: verify f(a) exists, ensure left and right limits exist and equal each other, and confirm that value matches f(a).

Question 17

1. Yes; $\lim_{x\to\pi/2}F(x)=0$ equals $F(\pi/2)=0$.
2. No; $\lim_{x\to\pi/2}F(x)$ does not exist (vertical asymptote), though $F(\pi/2)=0$. (correct answer)
3. No; $F(\pi/2)$ is undefined.
4. No; $\lim_{x\to\pi/2}F(x)=1$ but $F(\pi/2)=0$.
5. No; $\lim_{x\to\pi/2}F(x)=-1$ but $F(\pi/2)=0$.

Explanation: To check if F is continuous at x = π/2, verify the limit exists, F(π/2) is defined, and they match. F(π/2) = 0 is defined, but tan x has a vertical asymptote at π/2, so the limit does not exist (approaches ±∞). Thus, F is not continuous. A common omission is not recognizing that infinite limits prevent existence for continuity. Explanations may forget to assess behavior near asymptotes. For continuity checks: ensure f(a) defined, verify limit exists finitely, confirm equality with f(a).

Question 18

1. Yes; $\lim_{x\to2}O(x)=2$ exists and equals $O(2)=2$.
2. No; $\lim_{x\to2^-}O(x)=0$ and $\lim_{x\to2^+}O(x)=2$, so the limit does not exist. (correct answer)
3. No; $O(2)$ is undefined.
4. No; $\lim_{x\to2}O(x)=0$ but $O(2)=2$.
5. No; $\lim_{x\to2}O(x)=4$ but $O(2)=0$.

Explanation: O at 2: O(2)=0? Wait, x≤2 is x²-2x, at 2=0, but right limit 2, left 0, no limit. Not continuous. Omission: evaluating at point. Checklist: f(a), sides, match.

Question 19

1. No; $q(0)$ is undefined since $\frac{\sin 0}{0}$ is undefined.
2. No; $\lim_{x\to0}q(x)=0$ but $q(0)=1$.
3. Yes; $\lim_{x\to0}q(x)$ exists and equals $1=q(0)$. (correct answer)
4. No; $\lim_{x\to0}q(x)$ does not exist because $\sin x$ oscillates.
5. Yes; $q$ is continuous because it is a quotient of continuous functions.

Explanation: This is the classic sinc function, important in calculus. To check continuity at x = 0, we need to evaluate lim[x→0] (sin x)/x, which is a fundamental limit equal to 1. We're given q(0) = 1, so the function is defined at x = 0. Since lim[x→0] (sin x)/x = 1 and q(0) = 1, all three continuity conditions are satisfied: the function is defined, the limit exists, and they're equal. This careful definition removes what would otherwise be a removable discontinuity. Students sometimes incorrectly think the limit doesn't exist because of the 0/0 form, but L'Hôpital's rule or the squeeze theorem confirms the limit is 1. Continuity checklist: ✓ q(0) defined, ✓ Limit exists (= 1), ✓ Limit equals q(0).

Question 20

1. Yes; $\lim_{x\to-2}v(x)=5=v(-2)$.
2. No; $v(-2)$ is undefined.
3. No; $\lim_{x\to-2}v(x)=4\ne v(-2)=5$. (correct answer)
4. No; $\lim_{x\to-2}v(x)$ does not exist because $x^2$ is not continuous.
5. Yes; since $v(-2)$ exists and $x^2$ has a limit, $v$ is continuous regardless of their values.

Explanation: For continuity at x = -2, we must check: (1) v(-2) exists, (2) lim[x→-2] v(x) exists, and (3) they are equal. The function defines v(-2) = 5, so condition (1) is satisfied. For x ≠ -2, v(x) = x², so lim[x→-2] v(x) = lim[x→-2] x² = (-2)² = 4. We have v(-2) = 5 but lim[x→-2] v(x) = 4, so 5 ≠ 4, violating condition (3). This illustrates that arbitrarily redefining a continuous function at a single point creates a jump discontinuity—the natural value of (-2)² = 4 has been changed to 5. Remember the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?