AP Calculus BC Quiz: Exploring Types Of Discontinuities

What this quiz covers

This quiz focuses on Exploring Types Of Discontinuities, 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. Removable discontinuity
2. Oscillating discontinuity
3. Infinite discontinuity
4. No discontinuity
5. Jump discontinuity (correct answer)

Explanation: Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For the piecewise r(x), at x = 1, the left-hand limit is 1 from x², the right-hand limit is 3 from x² + 2, and r(1) = 1. The one-sided limits are finite but unequal, so the limit does not exist, indicating a jump discontinuity. The graph jumps from 1 to 3 across x = 1. A tempting distractor might be 'no discontinuity' since it's defined at x = 1, but it fails because the right limit differs. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.

Question 2

1. Jump discontinuity
2. Removable discontinuity (correct answer)
3. Infinite discontinuity
4. Oscillating discontinuity
5. No discontinuity

Explanation: This problem tests your ability to classify discontinuities by analyzing function behavior at a specific point. The function f(x) = (x² - 9)/(x - 3) can be simplified by factoring the numerator as (x - 3)(x + 3)/(x - 3) = x + 3 for x ≠ 3. This means the limit as x approaches 3 exists and equals 6, but f(3) is defined as 5, which differs from this limit value. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. A jump discontinuity would require different left and right limits, which doesn't occur here since both approach 6. To classify discontinuities, always check: (1) if the limit exists, (2) if the function is defined at that point, and (3) if they're equal.

Question 3

1. Jump discontinuity
2. Removable discontinuity
3. Oscillating discontinuity (correct answer)
4. Infinite discontinuity
5. No discontinuity

Explanation: Classifying types of discontinuities is essential for analyzing function continuity in AP Calculus BC. For p(x) = sin(1/x) with p(0) = 0, as x approaches 0, 1/x becomes arbitrarily large, causing sin(1/x) to oscillate rapidly between -1 and 1. The limit does not exist because the function values do not approach a single number, instead oscillating indefinitely. This type of behavior is classified as an oscillating discontinuity. A tempting distractor is infinite discontinuity, but that fails because the function remains bounded and does not approach infinity. A transferable classification strategy is to observe if the function oscillates without converging to a limit, distinguishing it from jump or infinite types.

Question 4

1. Jump discontinuity
2. No discontinuity
3. Infinite discontinuity
4. Removable discontinuity (correct answer)
5. Oscillating discontinuity

Explanation: Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For u(x) = (x - 4)/(√x - 2) for x ≥ 0 and u(4) = 7, simplifying by rationalizing gives u(x) = √x + 2 for x ≠ 4, so the limit as x approaches 4 is 4. However, u(4) = 7 ≠ 4, creating a discontinuity where the limit exists finitely. This is removable since redefining u(4) to 4 would ensure continuity. A tempting distractor might be 'infinite discontinuity' before simplifying, but it fails because the indeterminate form resolves to a finite limit. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.

Question 5

1. Jump discontinuity
2. Removable discontinuity (correct answer)
3. Infinite discontinuity
4. Oscillating discontinuity
5. No discontinuity

Explanation: Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For the function f(x) = (x² - 9)/(x - 3) when x ≠ 3 and f(3) = 5, simplifying the expression gives f(x) = x + 3 for x ≠ 3, so the limit as x approaches 3 is 6. However, since f(3) = 5, which does not equal the limit, the function is discontinuous at x = 3, but the limit exists. This makes it a removable discontinuity because redefining f(3) to 6 would make the function continuous at that point. A tempting distractor might be 'no discontinuity' since the function is defined at x = 3, but it fails because the limit does not match the function value. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.

Question 6

1. Infinite discontinuity
2. Jump discontinuity (correct answer)
3. No discontinuity
4. Removable discontinuity
5. Oscillating discontinuity

Explanation: Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For s(x) = |x|/x when x ≠ 0 and s(0) = 0, the left-hand limit is -1 and the right-hand limit is 1, while s(0) = 0. Since the one-sided limits exist but are not equal, the overall limit does not exist, characterizing a jump discontinuity. The function jumps from -1 to 1 across x = 0, with the defined value in between not affecting the classification. A tempting distractor might be 'removable discontinuity' because it's defined at 0, but it fails as no single redefinition can match both sides. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.

Question 7

1. No discontinuity
2. Infinite discontinuity
3. Removable discontinuity (correct answer)
4. Jump discontinuity
5. Oscillating discontinuity

Explanation: Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For p(x) = sin(x)/x when x ≠ 0 and p(0) = 2, the limit as x approaches 0 is 1, which is well-known from calculus. However, p(0) = 2 does not equal this limit, so there is a discontinuity at x = 0, but the limit exists finitely. This is a removable discontinuity since redefining p(0) to 1 would make it continuous. A tempting distractor might be 'no discontinuity' because the function is defined everywhere, but it fails since the limit and function value mismatch. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.

Question 8

1. Removable discontinuity
2. No discontinuity
3. Infinite discontinuity (correct answer)
4. Jump discontinuity
5. Oscillating discontinuity

Explanation: This problem asks about the discontinuity of u(x) = 2x/(x² - 16) at x = 4. First, factor the denominator: x² - 16 = (x - 4)(x + 4), so u(x) = 2x/[(x - 4)(x + 4)]. As x approaches 4, the numerator approaches 8 while the denominator approaches 0, with (x - 4) → 0 and (x + 4) → 8. This causes the function to approach ±∞, creating an infinite discontinuity. The sign depends on the direction of approach: from the left it's -∞, from the right it's +∞. Unlike removable discontinuities, there's no common factor to cancel. Infinite discontinuities occur at vertical asymptotes where denominators approach zero without cancellation.

Question 9

1. Oscillating discontinuity
2. Removable discontinuity
3. Infinite discontinuity
4. Jump discontinuity (correct answer)
5. No discontinuity

Explanation: This problem requires classifying the discontinuity of the piecewise function t(x) at x = 1. From the left: lim[x→1⁻] t(x) = lim[x→1⁻] x² = 1² = 1, and t(1) = 1² = 1. From the right: lim[x→1⁺] t(x) = lim[x→1⁺] (x + 2) = 1 + 2 = 3. Since the left-hand limit (1) differs from the right-hand limit (3), and both are finite, this is a jump discontinuity. The function "jumps" from y = 1 to y = 3 as x passes through 1. For piecewise functions, always calculate both one-sided limits to detect jumps.

Question 10

1. Infinite discontinuity (correct answer)
2. Jump discontinuity
3. Removable discontinuity
4. Oscillating discontinuity
5. No discontinuity

Explanation: This problem asks us to identify the discontinuity type for h(x) = 1/(x - 2)² at x = 2. As x approaches 2 from either side, the denominator (x - 2)² approaches 0 while remaining positive, causing the function to approach +∞. Since lim[x→2] h(x) = +∞, this is an infinite discontinuity. The function cannot be made continuous by redefining h(2) because the limit doesn't exist as a finite value. It's not removable because we can't "fill the hole" with any finite value. To identify infinite discontinuities, look for vertical asymptotes where the function approaches ±∞.

Question 11

1. Oscillating discontinuity
2. Infinite discontinuity
3. Jump discontinuity (correct answer)
4. Removable discontinuity
5. No discontinuity

Explanation: This problem requires identifying the discontinuity type for q(x) = |x|/x at x = 0. For x > 0, |x| = x, so q(x) = x/x = 1, giving lim[x→0⁺] q(x) = 1. For x < 0, |x| = -x, so q(x) = -x/x = -1, giving lim[x→0⁻] q(x) = -1. Since the left and right limits exist but are different (and finite), this is a jump discontinuity. The fact that q(0) = 0 is defined doesn't change the discontinuity type. To identify jump discontinuities, check if both one-sided limits exist as finite values but are unequal.

Question 12

1. Jump discontinuity
2. Infinite discontinuity
3. Removable discontinuity (correct answer)
4. Oscillating discontinuity
5. No discontinuity

Explanation: This problem asks us to classify the type of discontinuity at x = 3. The function f(x) = (x² - 9)/(x - 3) can be simplified by factoring the numerator: (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 for x ≠ 3. The limit as x approaches 3 is lim[x→3] (x + 3) = 6, but f(3) = 10, which differs from this limit value. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. It's not a jump discontinuity because there's only one limit value (not different left and right limits). To identify removable discontinuities, check if the limit exists but differs from the function value at that point.

Question 13

1. Removable discontinuity
2. Infinite discontinuity (correct answer)
3. Jump discontinuity
4. No discontinuity
5. Oscillating discontinuity

Explanation: This question tests recognition of infinite discontinuities in rational functions. The function $h(x)=\frac{1}{(x+2)^2}$ has a denominator that equals zero when $x=-2$, while the numerator remains 1. As $x$ approaches $-2$ from either side, $(x+2)^2$ approaches 0 through positive values, making $h(x)$ approach $+\infty$. This behavior defines an infinite discontinuity. A removable discontinuity would require a common factor in numerator and denominator, which isn't present here. To identify infinite discontinuities, look for points where the denominator approaches zero while the numerator doesn't.

Question 14

1. No discontinuity
2. Jump discontinuity
3. Removable discontinuity
4. Infinite discontinuity (correct answer)
5. Oscillating discontinuity

Explanation: This question tests recognition of infinite discontinuities in logarithmic functions. The function $s(x)=\ln(x-2)$ is only defined for $x>2$ since logarithms require positive arguments. As $x$ approaches 2 from the right, $(x-2)$ approaches 0 through positive values, making $\ln(x-2)$ approach $-\infty$. The function isn't defined for $x\leq 2$, so there's no left-sided approach. This one-sided infinite behavior characterizes an infinite discontinuity. A removable discontinuity would require a finite limit, which doesn't exist here. For logarithmic functions, discontinuities at domain boundaries are typically infinite.

Question 15

1. Oscillating discontinuity
2. Jump discontinuity
3. Infinite discontinuity (correct answer)
4. Removable discontinuity
5. No discontinuity

Explanation: This question involves classifying discontinuities in trigonometric functions. The function $t(x)=\tan\left(\frac{\pi x}{2}\right)$ has vertical asymptotes where $\frac{\pi x}{2}=\frac{\pi}{2}+n\pi$ for integer $n$. When $x=1$, we get $\frac{\pi}{2}$, where tangent has a vertical asymptote. As $x$ approaches 1 from the left, $\tan\left(\frac{\pi x}{2}\right)$ approaches $+\infty$, and from the right, it approaches $-\infty$. This behavior defines an infinite discontinuity. An oscillating discontinuity would involve rapid oscillation, not divergence to infinity. To classify discontinuities in periodic functions, identify where vertical asymptotes occur.

Question 16

1. Jump discontinuity
2. No discontinuity
3. Infinite discontinuity
4. Removable discontinuity (correct answer)
5. Oscillating discontinuity

Explanation: This question involves factoring to identify discontinuity types. The function u(x) = (x² - 4)/(x + 2) can be factored as (x + 2)(x - 2)/(x + 2), which simplifies to x - 2 for all x ≠ -2. Since lim[x→-2] u(x) = lim[x→-2] (x - 2) = -4 exists and is finite, but u(-2) is undefined, this creates a removable discontinuity. The discontinuity is not infinite (option C) because the limit is finite. To identify removable discontinuities in rational functions, factor and simplify to check if the problematic factor cancels.

Question 17

1. Infinite discontinuity
2. No discontinuity
3. Jump discontinuity (correct answer)
4. Removable discontinuity
5. Oscillating discontinuity

Explanation: This question involves analyzing discontinuities in functions with absolute values. For s(x) = |x|/x, we find lim[x→0⁻] s(x) = lim[x→0⁻] (-x)/x = -1 and lim[x→0⁺] s(x) = lim[x→0⁺] x/x = 1. Since the left and right limits exist but are unequal (-1 ≠ 1), and s(0) is undefined, this creates a jump discontinuity at x = 0. This cannot be an infinite discontinuity (option A) because both one-sided limits are finite. When one-sided limits exist finitely but differ, classify the discontinuity as a jump.

Question 18

1. No discontinuity (correct answer)
2. Infinite discontinuity
3. Removable discontinuity
4. Jump discontinuity
5. Oscillating discontinuity

Explanation: This question tests whether you can identify when a piecewise function has no discontinuity. For x ≠ 1, t(x) = (x² - 1)/(x - 1) = (x - 1)(x + 1)/(x - 1) = x + 1, so lim(x→1) t(x) = 1 + 1 = 2. The function is defined at x = 1 with t(1) = 2, which equals the limit. Since the limit exists, the function is defined at that point, and these values are equal, the function is continuous at x = 1—there is no discontinuity. A removable discontinuity would occur if t(1) were defined differently from the limit value. To verify continuity, always check three conditions: (1) the limit exists, (2) the function is defined, and (3) the limit equals the function value.

Question 19

1. Infinite discontinuity (correct answer)
2. Jump discontinuity
3. Removable discontinuity
4. Oscillating discontinuity
5. No discontinuity

Explanation: This problem involves classifying the discontinuity of h(x) = 1/(x + 2)² at x = -2. As x approaches -2 from either direction, the denominator (x + 2)² approaches 0 while remaining positive (since it's squared), causing the function to approach positive infinity. This behavior—where the function grows without bound as x approaches the discontinuity from both sides—defines an infinite discontinuity. A jump discontinuity would require finite but different one-sided limits, which doesn't occur here. To identify infinite discontinuities, look for denominators that approach zero without cancellation, particularly when the function approaches ±∞ from at least one side.

Question 20

1. Removable discontinuity
2. Jump discontinuity
3. Infinite discontinuity
4. Oscillating discontinuity (correct answer)
5. No discontinuity

Explanation: This question tests recognition of oscillating discontinuities, which occur when a function oscillates infinitely as it approaches a point. For p(x) = sin(1/x), as x approaches 0, the argument 1/x grows without bound, causing sin(1/x) to oscillate rapidly between -1 and 1. This means the limit as x approaches 0 does not exist—not because the function approaches infinity, but because it oscillates without settling on any value. Even though p(0) is defined as 0, the non-existence of the limit due to oscillation creates an oscillating discontinuity. A removable discontinuity would require the limit to exist, which it doesn't here. To identify oscillating discontinuities, look for compositions involving periodic functions with arguments that approach infinity.