AP Calculus AB Quiz: Defining Continuity At A Point

What this quiz covers

This quiz focuses on Defining Continuity At A Point, 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. Yes; $\lim_{x\to2}P(x)=0$ and $P(2)=0$. (correct answer)
2. No; $\lim_{x\to2}P(x)=2$ but $P(2)=0$.
3. No; $P(2)$ is undefined.
4. No; $\lim_{x\to2}P(x)$ does not exist because the function is linear.
5. Yes; the limit exists, so $P$ is continuous even if $P(2)$ differed.

Explanation: At x=2, P(2)=0, both limits 0 from x-2, continuous. Pieces agree at limit. Common: confusing with value. Continuous. Checklist: (1) Defined? (2) Limit? (3) Equals?

Question 2

1. Yes; $\lim_{x\to2^-}w(x)=8$, $\lim_{x\to2^+}w(x)=8$, and $w(2)=8$. (correct answer)
2. No; $\lim_{x\to2}w(x)=8$ but $w(2)$ is undefined.
3. No; $\lim_{x\to2^-}w(x)=8$ and $\lim_{x\to2^+}w(x)=0$, so the limit does not exist.
4. No; $\lim_{x\to2}w(x)=2$ but $w(2)=8$.
5. Yes; $w(2)=8$ so continuity is automatic.

Explanation: Continuity at x=2 needs w(2) defined, limit, and equality. w(2)=8, left limit from x³ is 8, right from constant 8 is 8, so continuous. The pieces connect smoothly. Omission: not checking right limit separately. Continuous here. Strategy: (1) f(a)? (2) Limit via sides? (3) Match?

Question 3

1. Yes; $\lim_{x\to0}e(x)=2$ and $e(0)=2$.
2. No; $\lim_{x\to0}e(x)=1$ but $e(0)=2$. (correct answer)
3. No; $\lim_{x\to0}e(x)$ does not exist because cosine oscillates.
4. No; $e(0)$ is undefined.
5. Yes; cosine is continuous, so any value at $0$ makes it continuous.

Explanation: Continuity at a point means the limit matches the function value, with both defined and limit existing. For e(x) at x = 0, e(0) = 2, but lim cos x = 1 ≠ 2, so discontinuous. A common mistake is assuming trigonometric functions are continuous everywhere without checking redefinitions. Cosine is continuous, but the point change disrupts it. This illustrates point discontinuities in otherwise continuous functions. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 4

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

Explanation: For continuity at x=0, H(0)=1, but left limit -1, right 1, DNE, not continuous. Sign function jump. Common: thinking point fixes limit issue. Jump discontinuity. Checklist: (1) Defined? (2) Limit? (3) Match?

Question 5

1. No; $\lim_{x\to-2}L(x)$ does not exist because the function is constant.
2. Yes; $\lim_{x\to-2}L(x)=2$ and $L(-2)=2$. (correct answer)
3. No; $L(-2)$ is undefined.
4. No; $\lim_{x\to-2}L(x)=0$ but $L(-2)=2$.
5. Yes; $L(-2)=2$ so the limit is unnecessary.

Explanation: Continuity at x=-2: L(-2)=2, limit 2, continuous. Constant function. Omission: unnecessary worry over notation. Continuous. Use: (1) f(a)? (2) Limit? (3) Match?

Question 6

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

Explanation: Continuity requires the function value, existing limit, and their match at the point. For c(x) at x = 1, c(1) = 2, but left limit = 1 and right = 2, so limit does not exist, discontinuous. A typical omission is not checking if one-sided limits agree. This is a step function discontinuity. Constants on intervals need to match for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 7

1. Yes; $\lim_{x\to2}Z(x)=0$ and $Z(2)=0$.
2. No; $\lim_{x\to2}Z(x)=0$ but $Z(2)=1$. (correct answer)
3. No; $\lim_{x\to2}Z(x)$ does not exist because it is quadratic.
4. No; $Z(2)$ is undefined.
5. Yes; $Z(2)=1$ so the limit must be $1$.

Explanation: A function is continuous at x = a if all three conditions hold: f(a) defined, limit exists, and they equal. For Z(x) at x = 2, Z(2) = 1, but the limit of (x-2)² is 0, which does not equal 1, so discontinuous. Students commonly omit comparing the limit to the redefined value, thinking the limit alone suffices. Here, the quadratic approaches 0, but the point is set to 1, creating a discontinuity. Redefining doesn't ensure continuity unless it matches the limit. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 8

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

Explanation: Continuity at a point: f(a) defined, limit exists, equal. For o(x) at x = 0, o(0) = -1, but left x²-1 → -1, right 1-x² → 1, DNE, discontinuous. Common omission: not verifying one-sided limits match. Parabolic pieces disagree. Symmetry matters for limit existence. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 9

1. Yes; $p(0)=0$ and both one-sided limits equal $0$. (correct answer)
2. No; $\lim_{x\to0}p(x)$ does not exist because $\sqrt{x}$ is undefined for $x<0$.
3. No; $\lim_{x\to0}p(x)=1$ but $p(0)=0$.
4. No; $p(0)$ is undefined.
5. Yes; only the right-hand limit matters at $0$.

Explanation: For continuity, check definition, limit existence, and equality. At x = 0 for p(x), p(0) = 0 (from √x), left limit 0 (constant), right √x → 0, so limit = 0 matches, continuous. People often forget that for domains like square root, left limit is from the extension. Here, the extension to 0 for x<0 makes left continuous. This shows how to extend functions for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 10

1. No; $\lim_{x\to0}\sin x/x$ does not exist because $\sin 0=0$.
2. Yes; $f(0)=1$ and $\lim_{x\to0}f(x)=1$, so $\lim_{x\to0}f(x)=f(0)$. (correct answer)
3. No; $f(0)$ is undefined since $\sin 0/0$ is undefined.
4. No; $\lim_{x\to0}f(x)=0$ but $f(0)=1$.
5. Yes; the limit exists, so continuity holds even if $f(0)$ were different.

Explanation: Continuity at x=0 requires f(0) defined, limit existing, and equality. Here, f(0)=1, and the known limit of sin(x)/x is 1, so all conditions hold, making f continuous. The piecewise definition fills the hole at 0 where sin(x)/x is undefined. A frequent omission is assuming the function is undefined at 0 without checking the piecewise rule. This extends the sinc function continuously. Use this transferable checklist: (1) Check if f(a) is defined, (2) Confirm the limit exists, (3) Verify limit equals f(a).

Question 11

1. Yes; $\lim_{x\to3}p(x)=6$ and $p(3)=6$.
2. No; $\lim_{x\to3}p(x)=6$ but $p(3)$ is undefined, so continuity fails. (correct answer)
3. No; $\lim_{x\to3}p(x)$ does not exist because the numerator is zero.
4. Yes; $p(3)$ exists because cancellation always makes rational functions defined.
5. No; $\lim_{x\to3}p(x)=0$ so it cannot match $p(3)$.

Explanation: For continuity at x=3, p(3) must be defined, the limit must exist, and they must equal. However, p(3) is undefined due to division by zero, even though the limit is 6 after simplifying to x+3. Without p(3) defined, continuity fails despite the existing limit. Learners often forget that the function must be defined at the point, focusing only on the limit. This rational function has a hole at x=3, illustrating a removable discontinuity. To verify continuity, apply this checklist: (1) Is f(a) defined? (2) Does lim_{x→a} f(x) exist? (3) Is lim_{x→a} f(x) = f(a)?

Question 12

1. No; $u(1)$ is undefined because the fraction has $x-1$ in the denominator.
2. Yes; $\lim_{x\to1}u(x)=2$ exists and equals $u(1)=2$. (correct answer)
3. No; $\lim_{x\to1}u(x)$ does not exist since the expression is $0/0$ at $x=1$.
4. No; $\lim_{x\to1}u(x)=1$ exists, but $u(1)=2$.
5. Yes; $u$ is continuous because it is a rational function.

Explanation: For continuity at x = 1, we need u(1) defined, lim[x→1] u(x) to exist, and these values to match. The function defines u(1) = 2, satisfying condition 1. To find the limit, we simplify the rational expression: lim[x→1] (x² - 1)/(x - 1) = lim[x→1] (x + 1)(x - 1)/(x - 1) = lim[x→1] (x + 1) = 2, so the limit exists and equals 2. Since lim[x→1] u(x) = 2 = u(1), all continuity conditions are satisfied. This shows how removable discontinuities can be "removed" by defining the function value to equal the limit. Continuity checklist: (1) u(1) = 2 ✓, (2) Limit = 2 after factoring ✓, (3) They match ✓.

Question 13

1. Yes; $t(1)=0$ and $\lim_{x\to1^+}t(x)=0$, so the limit equals the value on its domain. (correct answer)
2. No; $\lim_{x\to1}t(x)$ does not exist because there is no left-hand limit.
3. No; $t(1)$ is undefined since $\sqrt{0}$ is undefined.
4. No; $\lim_{x\to1^+}t(x)=1$ but $t(1)=0$.
5. Yes; the limit exists, so continuity holds even if $t(1)$ were not defined.

Explanation: Continuity at endpoint x=1 for domain x≥1 requires f(1) defined, the one-sided limit from the right exists, and they equal. t(1)=0, and right limit is 0, so it is continuous (using one-sided definition for endpoints). No left limit exists due to domain, but that's acceptable. A common omission is insisting on two-sided limits even at boundaries. This square root function is continuous on its domain. Checklist: (1) f(a) defined? (2) Appropriate limit exists? (3) Equals f(a)?

Question 14

1. Yes; $\lim_{x\to-1}F(x)=0$ and $F(-1)=0$. (correct answer)
2. No; $\lim_{x\to-1}F(x)=0$ but $F(-1)=1$.
3. No; $\lim_{x\to-1}F(x)$ does not exist because the function is linear.
4. No; $F(-1)$ is undefined.
5. Yes; $F(-1)=0$ even though the limit is $-2$.

Explanation: To check continuity at x=-1, ensure F(-1) defined, limit exists, equals. F(-1)=0, limit of x+1 is 0, matches, continuous. Piecewise overrides linear. Common: assuming original expression defines point. Removable if not redefined. Checklist: (1) Defined? (2) Limit? (3) Equal?

Question 15

1. Yes; $v(0)=0$ and $\lim_{x\to0}1/x=0$.
2. No; $v(0)$ is undefined.
3. No; $\lim_{x\to0}v(x)$ does not exist, so it cannot equal $v(0)=0$. (correct answer)
4. Yes; the left-hand limit exists, so continuity holds.
5. No; $\lim_{x\to0}v(x)=1$ but $v(0)=0$.

Explanation: For continuity at x=0, v(0) must be defined, limit exist, and equal. v(0)=0, but limit of 1/x doesn't exist (diverges to ±∞), so not continuous. The point value can't overcome the limit failure. Common mistake: thinking a defined value implies continuity. Infinite discontinuity here. Checklist: (1) Defined? (2) Limit exists? (3) Equals?

Question 16

1. No; $\lim_{x\to0}G(x)=0$ but $G(0)=1$.
2. Yes; $\lim_{x\to0^-}G(x)=0$ and $\lim_{x\to0^+}G(x)=1$, so the limit exists.
3. No; $\lim_{x\to0^-}G(x)=0$ and $\lim_{x\to0^+}G(x)=1$, so $\lim_{x\to0}G(x)$ does not exist. (correct answer)
4. Yes; $G(0)=0$ and both one-sided limits equal $0$.
5. No; $G(0)$ is undefined because there are two rules.

Explanation: Continuity at x=0 requires G(0) defined, limit, equality. G(0)=0 from x² (≤0), but left limit 0, right 1 from x+1, limit DNE, not continuous. Jump due to mismatch. Omission: not computing sides. Jump discontinuity. Use: (1) f(a)? (2) Limit exists? (3) Equals?

Question 17

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

Explanation: For continuity, the limit must exist and equal f(a), with f(a) defined. At x = 2 for k(x), k(2) = 4, but left limit 4 and right 5 differ, so limit DNE, discontinuous. Often, students forget to check one-sided agreement in piecewise constants. This is a jump discontinuity. Matching pieces are crucial. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 18

1. No; $\lim_{x\to1}K(x)=2$ but $K(1)=3$. (correct answer)
2. Yes; $\lim_{x\to1}K(x)=3$ and $K(1)=3$.
3. No; $\lim_{x\to1}K(x)$ does not exist because of division by zero.
4. No; $K(1)$ is undefined.
5. Yes; cancellation shows $K(x)=x+1$, so it is continuous even if $K(1)$ differed.

Explanation: At x=1, K(1)=3, limit 2 from x+1, mismatch, not continuous. Simplification key. Common: not simplifying. Removable if redefined. Checklist: (1) Defined? (2) Limit? (3) Equals?

Question 19

1. Yes; $\lim_{x\to\pi}f(x)=1$ and $f(\pi)=1$. (correct answer)
2. No; $f(\pi)$ is undefined because of division by zero.
3. No; $\lim_{x\to\pi}f(x)=0$ but $f(\pi)=1$.
4. No; $\lim_{x\to\pi}f(x)$ does not exist since sine oscillates.
5. Yes; the limit exists, so continuity holds even if $f(\pi)$ differed.

Explanation: A function is continuous at x = a if f(a) is defined, lim exists, and equals f(a). For f(x) at x = π, f(π) = 1, and lim sin(u)/u = 1 where u = x-π, so continuous. People often omit recognizing the sinc function limit as 1. This fixes a removable discontinuity in sin(x-π)/(x-π). Special limits like this are key for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 20

1. No; $\lim_{x\to5}d(x)$ does not exist because the denominator is $0$.
2. Yes; $\lim_{x\to5}d(x)=10$ and $d(5)=10$. (correct answer)
3. No; $\lim_{x\to5}d(x)=5$ but $d(5)=10$.
4. No; $d(5)$ is undefined.
5. Yes; $d(5)=10$ so the limit must be $10$.

Explanation: For continuity, verify f(a) defined, limit exists, and equals f(a). At x = 5 for d(x), d(5) = 10, and simplified limit (x+5) → 10, so continuous. Students often forget to simplify rational expressions before taking limits. The removable discontinuity is fixed by redefining. This shows how holes can be filled for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).