AP Calculus BC 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 BC.

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; $s(3)$ is undefined, so continuity fails.
2. Yes; $s(3)$ exists, $\lim_{x\to3}s(x)$ exists, and they are equal. (correct answer)
3. No; $\lim_{x\to3}s(x)$ does not exist, so continuity fails.
4. No; $\lim_{x\to3}s(x)\ne s(3)$, so continuity fails.
5. No; $s(3)$ exists but $\lim_{x\to3}s(x)=\infty$, so continuity fails.

Explanation: This problem involves a rational function with a removable discontinuity that has been "filled in" at x = 3. For continuity at x = 3, we need s(3) to exist, lim[x→3] s(x) to exist, and these to be equal. We're given s(3) = 6, and for x ≠ 3, we can simplify: s(x) = (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3. Therefore, lim[x→3] s(x) = lim[x→3] (x + 3) = 3 + 3 = 6. Since s(3) = 6 = lim[x→3] s(x), all three continuity conditions are satisfied. The key insight is recognizing that defining the function value to match the limit of the simplified expression creates continuity. For removable discontinuities: factor and simplify first, find the limit, then verify the defined value matches.

Question 2

1. No; $f(1)$ is undefined, so continuity fails.
2. No; $\lim_{x\to1}f(x)$ does not exist, so continuity fails.
3. No; $\lim_{x\to1}f(x)\ne f(1)$, so continuity fails. (correct answer)
4. Yes; $f(1)$ exists, $\lim_{x\to1}f(x)$ exists, and they are equal.
5. No; $f(1)$ exists but $\lim_{x\to1}f(x)=\infty$, so continuity fails.

Explanation: This question tests whether a function is continuous at a point by checking if all three conditions are satisfied. For continuity at x = 1, we need: (1) f(1) to exist, (2) lim[x→1] f(x) to exist, and (3) these two values to be equal. Here, f(1) = 3 (given by the piecewise definition), and lim[x→1] f(x) = lim[x→1] (x² - 1) = 1² - 1 = 0. Since f(1) = 3 ≠ 0 = lim[x→1] f(x), the function fails the third condition for continuity. A common mistake is assuming that having a defined value at a point guarantees continuity, but all three conditions must be met. To check continuity: verify the function value exists, calculate the limit, then compare—if any step fails or they're unequal, the function is discontinuous.

Question 3

1. Yes; $v(-2)$ exists and equals $\lim_{x\to-2}v(x)$.
2. No; $\lim_{x\to-2}v(x)$ exists but is not equal to $v(-2)$. (correct answer)
3. No; $\lim_{x\to-2}v(x)$ does not exist because the one-sided limits are unequal.
4. No; $v(-2)$ is undefined, so continuity fails.
5. No; $v(-2)$ equals the limit, but $v$ is not differentiable at $x=-2$.

Explanation: This question evaluates continuity at a point for a modified quadratic function. Continuity at x=-2 requires v(-2) to be defined, the limit as x approaches -2 to exist, and equality. v(-2)=0 is defined, the limit is (-2)^2=4 from both sides, but 4 ≠ 0, so it fails. This is a removable discontinuity. A wrong choice might say the limit does not exist, but it does exist as 4. To check continuity at a point, use this checklist: verify f(a) is defined, confirm the limit exists by checking one-sided limits match, and ensure they equal f(a).

Question 4

1. No; $v(\pi)$ is not defined.
2. Yes; $v(\pi)$ is defined, $\lim_{x\to\pi}v(x)$ exists, and equals $v(\pi)$. (correct answer)
3. No; $\lim_{x\to\pi}v(x)$ exists but does not equal $v(\pi)$.
4. No; $\lim_{x\to\pi}v(x)$ does not exist.
5. No; $v(\pi)$ is defined but $\lim_{x\to\pi}v(x)$ does not exist.

Explanation: This question tests whether redefining a continuous function at one point maintains continuity. The function $v(x) = \cos x$ for $x \neq \pi$ normally has $\cos(\pi) = -1$. Since $\lim_{x \to \pi} v(x) = \lim_{x \to \pi} \cos x = \cos(\pi) = -1$, and we're told $v(\pi) = -1$, all three continuity conditions are satisfied: $v(\pi)$ is defined, the limit exists and equals -1, and this equals $v(\pi)$. A student might think that treating $x = \pi$ as a special case breaks continuity, but since the redefined value matches what the function would naturally be, continuity is preserved. When a piecewise definition matches the original function's value, continuity is maintained.

Question 5

1. No; $\lim_{x\to1}u(x)$ exists but is not equal to $u(1)$.
2. Yes; $\lim_{x\to1}u(x)$ exists and equals $u(1)$. (correct answer)
3. No; $u(1)$ is undefined, so continuity fails at $x=1$.
4. No; $\lim_{x\to1}u(x)$ does not exist because one-sided limits are unequal.
5. No; $u(1)$ exists but $\lim_{x\to1}u(x)$ does not exist.

Explanation: This question involves a square root function at the edge of its domain, testing continuity where the function begins. For continuity at x = 1, we need u(1) defined, lim[x→1] u(x) to exist, and these to be equal. Since u(x) = √(x - 1) is defined for x ≥ 1, we have u(1) = √(1 - 1) = √0 = 0. For the limit, we only consider the right-hand limit since the function isn't defined for x < 1: lim[x→1⁺] u(x) = lim[x→1⁺] √(x - 1) = √0 = 0. Since this one-sided limit equals u(1), the function is continuous at x = 1. Choice A incorrectly suggests the limit doesn't equal u(1), but both equal 0. For functions with restricted domains, continuity at boundary points only requires checking the limit from within the domain.

Question 6

1. No; $r(2)$ is not defined. (correct answer)
2. Yes; $r(2)$ is defined, $\lim_{x\to2}r(x)$ exists, and equals $r(2)$.
3. No; $\lim_{x\to2}r(x)$ exists but does not equal $r(2)$.
4. No; $\lim_{x\to2}r(x)$ does not exist.
5. No; $r(2)$ is defined but $\lim_{x\to2}r(x)$ does not exist.

Explanation: This question tests recognition of when a function cannot be made continuous at a point. The function $r(x) = \frac{1}{x-2}$ for $x \neq 2$ has a vertical asymptote at $x = 2$. As $x$ approaches 2 from the right, $r(x) \to +\infty$, and as $x$ approaches 2 from the left, $r(x) \to -\infty$. Most importantly, $r(2)$ is not defined in the problem statement. Students might think they could define $r(2)$ to make it continuous, but no finite value would work since the limit doesn't exist. For rational functions with non-removable discontinuities: check if the denominator's zero persists after simplification - if yes, continuity is impossible.

Question 7

1. Yes; $\lim_{x\to1}f(x)$ exists and equals $f(1)$.
2. No; $f(1)$ is undefined, so continuity fails at $x=1$.
3. No; $\lim_{x\to1}f(x)$ does not exist because one-sided limits are unequal.
4. No; $\lim_{x\to1}f(x)$ exists but is not equal to $f(1)$. (correct answer)
5. No; $f(1)$ exists but $\lim_{x\to1}f(x)$ does not exist.

Explanation: This question tests whether a function is continuous at a point by checking the three conditions for continuity. For continuity at x = 1, we need: (1) f(1) to be defined, (2) lim[x→1] f(x) to exist, and (3) these two values to be equal. Here, f(1) = 3 (defined), and for x ≠ 1, f(x) = x² - 1, so lim[x→1] f(x) = lim[x→1] (x² - 1) = 1² - 1 = 0. Since lim[x→1] f(x) = 0 ≠ 3 = f(1), the function fails the third condition for continuity. Choice A incorrectly claims the limit equals f(1), when actually 0 ≠ 3. To check continuity at a point: verify the function is defined there, calculate the limit, then compare these two values—all three conditions must be satisfied.

Question 8

1. No; $\lim_{x\to2}q(x)$ does not exist, so continuity fails. (correct answer)
2. No; $q(2)$ is undefined, so continuity fails.
3. Yes; $q(2)$ exists, $\lim_{x\to2}q(x)$ exists, and $\lim_{x\to2}q(x)=q(2)$.
4. Yes; $q(2)$ exists, so $q$ is continuous at $2$.
5. No; $\lim_{x\to2}q(x)\ne q(2)$, so continuity fails.

Explanation: This question examines continuity at a point where a piecewise function changes its formula. For continuity at x = 2, we need q(2) to exist, lim[x→2] q(x) to exist, and these to be equal. Since 2 satisfies x ≤ 2, we have q(2) = 2 + 3 = 5. For the limit, we check: lim[x→2⁻] q(x) = lim[x→2⁻] (x + 3) = 5 and lim[x→2⁺] q(x) = lim[x→2⁺] 7 = 7. Since 5 ≠ 7, the one-sided limits differ, so lim[x→2] q(x) doesn't exist. Choice E might seem plausible if you incorrectly think the limit exists, but when one-sided limits differ, the two-sided limit doesn't exist. At piecewise transition points, always verify both one-sided limits match before concluding the limit exists.

Question 9

1. Yes; $g(2)$ exists, $\lim_{x\to2}g(x)$ exists, and $\lim_{x\to2}g(x)=g(2)$.
2. No; $g(2)$ is undefined, so continuity fails. (correct answer)
3. No; $\lim_{x\to2}g(x)$ does not exist, so continuity fails.
4. No; $\lim_{x\to2}g(x)\ne g(2)$, so continuity fails.
5. Yes; $\lim_{x\to2}g(x)$ exists, so $g$ is continuous at $2$.

Explanation: This question asks about continuity at a point where the function has a removable discontinuity. For continuity at x = 2, we need g(2) to exist, lim[x→2] g(x) to exist, and these values to be equal. The function g(x) = (x² - 4)/(x - 2) is undefined at x = 2 because we get 0/0, so g(2) doesn't exist. Even though we could simplify to g(x) = (x + 2)(x - 2)/(x - 2) = x + 2 for x ≠ 2, giving lim[x→2] g(x) = 4, the function value g(2) remains undefined. Choice C might seem correct since the limit does exist, but the first failure point is that g(2) is undefined. Check conditions in order: if the function value doesn't exist, continuity immediately fails.

Question 10

1. Yes; $q(0)$ is defined, $\lim_{x\to0}q(x)$ exists, and equals $q(0)$. (correct answer)
2. No; $q(0)$ is not defined.
3. No; $\lim_{x\to0}q(x)$ exists but does not equal $q(0)$.
4. No; $\lim_{x\to0}q(x)$ does not exist.
5. No; $q(0)$ is defined but $\lim_{x\to0}q(x)$ does not exist.

Explanation: This question involves a function with a removable discontinuity that has been properly addressed. For $q(x) = \frac{x^2}{x}$ when $x \neq 0$, we can simplify to $q(x) = x$ for $x \neq 0$. Therefore, $\lim_{x \to 0} q(x) = \lim_{x \to 0} x = 0$. We're given that $q(0) = 0$, so all three continuity conditions are satisfied: $q(0)$ is defined, $\lim_{x \to 0} q(x)$ exists and equals 0, and this equals $q(0)$. A common error would be to think the original fraction form prevents continuity at $x = 0$, but the simplified form shows the limit exists. When dealing with rational functions: simplify first, then evaluate limits using the simplified form.

Question 11

1. No; $p(0)$ is not defined.
2. No; $\lim_{x\to0}p(x)$ exists but does not equal $p(0)$. (correct answer)
3. No; $\lim_{x\to0}p(x)$ does not exist.
4. Yes; $p(0)$ is defined, $\lim_{x\to0}p(x)$ exists, and equals $p(0)$.
5. No; $p(0)$ is defined but $\lim_{x\to0}p(x)$ does not exist.

Explanation: This question tests understanding of continuity when a function value is redefined at a point. The function $p(x) = \sin x$ for $x \neq 0$ has $\lim_{x \to 0} p(x) = \lim_{x \to 0} \sin x = \sin(0) = 0$. However, we're told that $p(0) = 1$. Since $p(0)$ is defined as 1, the limit exists and equals 0, but $\lim_{x \to 0} p(x) = 0 \neq 1 = p(0)$, the function is not continuous at $x = 0$. Students might assume that since sine is normally continuous everywhere, this function must be continuous, but redefining the value at a single point can break continuity. To verify continuity: always check that the limit equals the actual function value, not what it "should" be.

Question 12

1. No; $h(0)$ is not defined.
2. No; $\lim_{x\to0}h(x)$ exists but does not equal $h(0)$.
3. Yes; $h(0)$ is defined, $\lim_{x\to0}h(x)$ exists, and equals $h(0)$.
4. No; $\lim_{x\to0}h(x)$ does not exist. (correct answer)
5. No; $h(0)$ is defined but $\lim_{x\to0}h(x)$ does not exist.

Explanation: This question examines continuity at a point where the function has different behaviors from each side. The function $h(x) = \frac{|x|}{x}$ equals 1 when $x > 0$ and -1 when $x < 0$. We have $h(0) = 0$ by definition. For the limit: $\lim_{x \to 0^+} h(x) = 1$ and $\lim_{x \to 0^-} h(x) = -1$. Since these one-sided limits are different, $\lim_{x \to 0} h(x)$ does not exist. A student might think that defining $h(0) = 0$ makes the function continuous, but the limit must exist independently of how we define the function at the point. When checking continuity involving absolute values or sign functions: always evaluate both one-sided limits separately.

Question 13

1. Yes; $p(0)$ exists and equals the limit at $0$.
2. No; $p(0)$ is undefined, so continuity fails.
3. No; the one-sided limits at $0$ are not equal, so the limit does not exist. (correct answer)
4. No; the limit at $0$ exists but is not equal to $p(0)$.
5. Yes; the limit at $0$ exists regardless of the one-sided limits.

Explanation: This question tests the definition of continuity at a point. For p to be continuous at x=0, p(0) must be defined, the limit as x approaches 0 must exist, and they must be equal. Here, p(0)=1 is defined, but the left-hand limit is -1 and the right-hand limit is 1, so the two-sided limit does not exist. Therefore, p is not continuous at x=0. A tempting distractor is choice A, which claims the limit equals p(0), but the differing one-sided limits prevent the limit from existing. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

Question 14

1. No; the one-sided limits at $x=1$ are not equal, so the limit does not exist. (correct answer)
2. Yes; $g(1)$ exists and both one-sided limits equal $g(1)$.
3. No; $g(1)$ is undefined, so continuity fails.
4. No; the limit at $x=1$ exists but is not equal to $g(1)$.
5. Yes; the limit at $x=1$ exists, regardless of $g(1)$.

Explanation: This question tests the definition of continuity at a point. For g to be continuous at x=1, g(1) must be defined, the limit as x approaches 1 must exist, and they must be equal. Here, g(1)=4 is defined, but the left-hand limit is 3 and the right-hand limit is 4, so the two-sided limit does not exist. Therefore, g is not continuous at x=1. A tempting distractor is choice B, which assumes both one-sided limits equal g(1), but the left limit differs. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

Question 15

1. Yes; $s(-1)$ exists and equals the limit at $-1$.
2. No; $s(-1)$ is undefined, so continuity fails.
3. No; the limit at $-1$ exists but is not equal to $s(-1)$. (correct answer)
4. No; the one-sided limits at $-1$ are not equal, so the limit does not exist.
5. Yes; the limit at $-1$ exists, so $s$ is continuous there.

Explanation: This question tests the definition of continuity at a point. For s to be continuous at x=-1, s(-1) must be defined, the limit as x approaches -1 must exist, and they must be equal. Here, s(-1)=3 is defined, and the limit is 1 (from x²), which exists but does not equal 3. Therefore, s is not continuous at x=-1. A tempting distractor is choice A, which claims continuity because the limit equals s(-1), but 1≠3. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

Question 16

1. No; $h(3)$ is undefined, so continuity fails.
2. No; $\lim_{x\to3}h(x)$ does not exist because the one-sided limits are unequal.
3. Yes; $h(3)$ exists and equals $\lim_{x\to3}h(x)$. (correct answer)
4. No; $\lim_{x\to3}h(x)$ exists but is not equal to $h(3)$.
5. No; $h(3)$ exists and equals the limit, but $h$ has no derivative at $x=3$.

Explanation: This question evaluates continuity at an endpoint of the domain for a square root function. For h to be continuous at x=3, where the domain starts, h(3) must be defined and equal the right-hand limit as x approaches 3 from above, since the left-hand limit is not applicable. Here, h(3)=0, and the right-hand limit is also 0, satisfying the conditions for continuity at the endpoint. In calculus, functions like square root are considered continuous on their closed domain intervals at the endpoints. A tempting distractor might argue the two-sided limit does not exist due to the domain, but continuity at endpoints uses the appropriate one-sided limit. To check continuity at a point, use this checklist: verify f(a) is defined, confirm the limit exists by checking one-sided limits match, and ensure they equal f(a).

Question 17

1. Yes; $\lim_{x\to0}p(x)$ exists and equals $p(0)$.
2. No; $p(0)$ is undefined, so continuity fails at $x=0$.
3. No; $p(0)$ exists but $\lim_{x\to0}p(x)$ does not exist. (correct answer)
4. No; $\lim_{x\to0}p(x)$ exists but is not equal to $p(0)$.
5. No; $\lim_{x\to0}p(x)$ does not exist because one-sided limits are unequal.

Explanation: This question involves a function with oscillatory behavior near the point of interest. For continuity at x = 0, we need p(0) defined, lim[x→0] p(x) to exist, and these to be equal. We have p(0) = 0 (given), but for x ≠ 0, p(x) = sin(1/x) oscillates rapidly between -1 and 1 as x approaches 0. As x → 0⁺, 1/x → +∞ and sin(1/x) oscillates without approaching any single value; similarly as x → 0⁻, 1/x → -∞ and sin(1/x) oscillates. Therefore, lim[x→0] p(x) does not exist. Choice D incorrectly suggests the limit exists but doesn't equal p(0)—the limit doesn't exist at all due to oscillation. When dealing with compositions involving 1/x, check for unbounded behavior or oscillation that prevents the limit from existing.

Question 18

1. No; $q(0)$ is undefined, so continuity fails at $x=0$.
2. No; $\lim_{x\to0}q(x)$ does not exist because one-sided limits are unequal. (correct answer)
3. No; $\lim_{x\to0}q(x)$ exists but is not equal to $q(0)$.
4. Yes; $\lim_{x\to0}q(x)$ exists and equals $q(0)$.
5. No; $q(0)$ exists but $\lim_{x\to0}q(x)$ does not exist.

Explanation: This question tests continuity of a function involving absolute value at a point where behavior changes. For continuity at x = 0, we need q(0) defined, lim[x→0] q(x) to exist, and these to be equal. We're given q(0) = 0, and for x ≠ 0, q(x) = |x|/x. When x > 0, |x| = x, so q(x) = x/x = 1; when x < 0, |x| = -x, so q(x) = -x/x = -1. Thus lim[x→0⁺] q(x) = 1 and lim[x→0⁻] q(x) = -1. Since the one-sided limits are unequal (1 ≠ -1), lim[x→0] q(x) does not exist. Choice C incorrectly claims the limit exists; the unequal one-sided limits prove otherwise. For functions with absolute value, always check behavior from both sides as the sign of x affects the simplified form.

Question 19

1. No; $\lim_{x\to-1}v(x)$ exists but is not equal to $v(-1)$.
2. No; $\lim_{x\to-1}v(x)$ does not exist because one-sided limits are unequal.
3. Yes; $\lim_{x\to-1}v(x)$ exists and equals $v(-1)$. (correct answer)
4. No; $v(-1)$ is undefined, so continuity fails at $x=-1$.
5. No; $v(-1)$ exists but $\lim_{x\to-1}v(x)$ does not exist.

Explanation: This question presents a piecewise function where the special value at a point matches what the limit would be anyway. For continuity at x = -1, we need v(-1) defined, lim[x→-1] v(x) to exist, and these to be equal. We have v(-1) = 1 (from the second piece), and for x ≠ -1, v(x) = x². Therefore, lim[x→-1] v(x) = lim[x→-1] x² = (-1)² = 1. Since v(-1) = 1 and lim[x→-1] v(x) = 1, all continuity conditions are satisfied. Choice A incorrectly claims the limit doesn't equal v(-1), when both equal 1. When a piecewise function's special value matches the natural limit of the other piece, the function is continuous—this often indicates a removable discontinuity was "filled in" correctly.

Question 20

1. Yes; $r(-1)$ exists, $\lim_{x\to-1}r(x)$ exists, and $\lim_{x\to-1}r(x)=r(-1)$.
2. No; $\lim_{x\to-1}r(x)\ne r(-1)$, so continuity fails. (correct answer)
3. No; $r(-1)$ is undefined, so continuity fails.
4. No; $\lim_{x\to-1}r(x)$ does not exist, so continuity fails.
5. Yes; $\lim_{x\to-1}r(x)$ exists, so $r$ is continuous at $-1$.

Explanation: This question presents a function with a point discontinuity where the function value differs from what the formula suggests. For continuity at x = -1, we need r(-1) to exist, lim[x→-1] r(x) to exist, and these to be equal. From the definition, r(-1) = 0. For x ≠ -1, r(x) = x², so lim[x→-1] r(x) = lim[x→-1] x² = (-1)² = 1. Since r(-1) = 0 and lim[x→-1] r(x) = 1, we have 0 ≠ 1, violating the third continuity condition. Choice D might attract students who think the limit doesn't exist because of the piecewise definition, but the limit of x² as x approaches -1 clearly exists. When a function is redefined at a single point, the limit still follows the original formula.