AP Calculus AB 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 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. Jump discontinuity
2. Removable discontinuity (correct answer)
3. Infinite discontinuity
4. No discontinuity (continuous)
5. Corner discontinuity

Explanation: This function has a removable discontinuity at x = 2. The expression η(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2) simplifies to η(x) = x+2 for all x ≠ 2, but η(2) is undefined since we get 0/0. The limit as x approaches 2 is lim(x→2) (x+2) = 4, but the function is undefined at x = 2. This creates a removable discontinuity because we could define η(2) = 4 to make the function continuous. Students often recognize this as a standard difference of squares factorization. To identify removable discontinuities: look for factorable expressions where common factors in numerator and denominator cancel to yield finite limits.

Question 2

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

Explanation: This function has a removable discontinuity at x = 3. When x ≠ 3, we can factor the numerator as (x-3)(x+3), so f(x) = (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) = 2. Since the limit exists but doesn't equal the function value, this creates a removable discontinuity. Students often confuse this with an infinite discontinuity because of the rational form, but the key is that the factor (x-3) cancels from numerator and denominator. To classify discontinuities: check if the limit exists and compare it to the function value at that point.

Question 3

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

Explanation: This function has an infinite discontinuity at x = 0. The expression β(x) = (x²-9)/(x²-3x) = (x²-9)/(x(x-3)) has a denominator that equals zero when x = 0, while the numerator equals -9 ≠ 0. As x approaches 0, the function approaches ±∞, creating a vertical asymptote. This defines an infinite discontinuity. Students might confuse this with the behavior at x = 3, but each zero of the denominator must be analyzed separately. To identify infinite discontinuities: check points where denominators approach zero while numerators approach non-zero values.

Question 4

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

Explanation: This function has a removable discontinuity at x = 3. The expression T(x) = (x-3)(x+1)/(x-3) simplifies to T(x) = x+1 for all x ≠ 3, but T(3) is undefined since we get 0/0. The limit as x approaches 3 is lim(x→3) (x+1) = 4, but the function is undefined at x = 3. This creates a removable discontinuity because we could define T(3) = 4 to make the function continuous. Students often see the rational form and assume infinite discontinuity, but factor cancellation is key. To identify removable discontinuities: look for common factors in numerator and denominator that create 0/0 forms.

Question 5

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

Explanation: This function has a removable discontinuity at x = -4. The expression r(x) = (x+4)/(x+4) simplifies to r(x) = 1 for all x ≠ -4, but r(-4) is undefined since we get 0/0. The limit as x approaches -4 is lim(x→-4) 1 = 1, but the function is undefined at x = -4. This creates a removable discontinuity because we could define r(-4) = 1 to make the function continuous. Students often think this is an infinite discontinuity because of the rational form, but the key is that both numerator and denominator approach zero simultaneously. To identify removable discontinuities: look for 0/0 indeterminate forms that can be simplified to find a finite limit.

Question 6

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

Explanation: This function has a jump discontinuity at x = 0. The left-hand limit is lim(x→0⁻) (x+1) = 1, while the right-hand limit is lim(x→0⁺) 3 = 3, and g(0) = 3. Since the left and right limits exist but are unequal (1 ≠ 3), this creates a jump discontinuity. The function literally 'jumps' from approaching 1 on the left to the value 3 on the right. Students might think this is continuous because g(0) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: check that both one-sided limits exist but are different values.

Question 7

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

Explanation: This function has a removable discontinuity at x = 3. The expression V(x) = (x²-6x+9)/(x-3) = (x-3)²/(x-3) simplifies to V(x) = x-3 for all x ≠ 3, but V(3) is undefined since we get 0/0. The limit as x approaches 3 is lim(x→3) (x-3) = 0, but the function is undefined at x = 3. This creates a removable discontinuity because we could define V(3) = 0 to make the function continuous. Students might not recognize the perfect square factorization, but the key is that both numerator and denominator have the factor (x-3). To identify removable discontinuities: factor completely and look for common factors that cancel.

Question 8

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

Explanation: This function has a removable discontinuity at x = 5. The expression d(x) = (x-5)/((x-5)(x+1)) simplifies to d(x) = 1/(x+1) for all x ≠ 5, but d(5) is undefined since we get 0/0. The limit as x approaches 5 is lim(x→5) 1/(x+1) = 1/6, but the function is undefined at x = 5. This creates a removable discontinuity because we could define d(5) = 1/6 to make the function continuous. Students often miss the factor cancellation, thinking this is an infinite discontinuity. To identify removable discontinuities: look for common factors in numerator and denominator that create 0/0 forms.

Question 9

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

Explanation: This function has an infinite discontinuity at x = 1. The expression ι(x) = 1/(x²-1) = 1/((x-1)(x+1)) has a denominator that approaches 0 as x approaches 1, while the numerator remains 1. This causes the function to approach ±∞, creating a vertical asymptote at x = 1. The sign of the denominator determines whether the function approaches +∞ or -∞ from each side. This defines an infinite discontinuity. Students might confuse this with the behavior at x = -1, but both zeros of the denominator create infinite discontinuities. To identify infinite discontinuities: find values where denominators approach zero while numerators approach non-zero values.

Question 10

1. Jump discontinuity
2. No discontinuity (continuous at $x=0$)
3. Infinite discontinuity
4. Removable discontinuity (correct answer)
5. Oscillating discontinuity

Explanation: The function p(x) = (x²-4x)/x = x(x-4)/x simplifies to x-4 for x≠0. Therefore, lim(x→0) p(x) = 0-4 = -4. Since p(0) is defined as 7, which differs from the limit value of -4, we have a removable discontinuity at x=0. The discontinuity is called "removable" because we could "remove" it by redefining p(0) = -4 to make the function continuous. Students often assume any function undefined at a point must have an infinite discontinuity, but when simplification yields a finite limit, it's removable. To identify removable discontinuities: simplify the expression, find the limit, and check if it's finite but different from the function value.

Question 11

1. Jump discontinuity
2. Removable discontinuity
3. Infinite discontinuity (correct answer)
4. No discontinuity (continuous at $x=-4$)
5. Oscillating discontinuity

Explanation: The function h(x) = 1/(x+4)² has a denominator that equals zero when x=-4, making the function undefined at this point. As x approaches -4 from either direction, (x+4)² approaches 0 through positive values, causing 1/(x+4)² to approach positive infinity. Since the function approaches infinity (not a finite value) as x approaches -4, this is an infinite discontinuity. Students sometimes confuse this with a removable discontinuity, but removable discontinuities require finite limits. The classification strategy: when a rational function's denominator has a zero that isn't canceled by the numerator, check the limit—if it's infinite, it's an infinite discontinuity.

Question 12

1. Infinite discontinuity
2. Removable discontinuity
3. Jump discontinuity (correct answer)
4. No discontinuity; $v$ is continuous at $x=4$
5. Hole with limit $0$

Explanation: The function $v(x)=\frac{x-4}{|x-4|}$ has a jump discontinuity at $x=4$. For $x<4$, we have $x-4<0$, so $|x-4|=-(x-4)$, giving $v(x)=\frac{x-4}{-(x-4)}=-1$. For $x>4$, we have $x-4>0$, so $|x-4|=x-4$, giving $v(x)=\frac{x-4}{x-4}=1$. Therefore, $\lim_{x\to 4^-}v(x)=-1$ and $\lim_{x\to 4^+}v(x)=1$. The function is undefined at $x=4$ due to division by zero. Since the one-sided limits exist but differ (-1 vs 1), this is a jump discontinuity. This function is similar to $\frac{|x|}{x}$ but shifted to $x=4$. Jump discontinuities are characterized by finite but unequal one-sided limits, creating a "jump" in the function's graph.

Question 13

1. No discontinuity; $r$ is continuous at $x=1$
2. Infinite discontinuity
3. Jump discontinuity
4. Removable discontinuity (correct answer)
5. Vertical asymptote at $x=1$

Explanation: The function $r(x)=\frac{x^2-1}{x-1}$ has a removable discontinuity at $x=1$. We can factor the numerator as $(x-1)(x+1)$, so for $x\neq 1$, $r(x)=\frac{(x-1)(x+1)}{x-1}=x+1$. This means $\lim_{x\to 1}r(x)=2$. However, we're told that $r(1)=5$, which doesn't equal the limit. Since the limit exists and is finite (2) but doesn't equal the function value (5), this is a removable discontinuity. The discontinuity could be "removed" by redefining $r(1)=2$. Students often overlook that a removable discontinuity can occur when the function is defined but has the "wrong" value, not just when it's undefined. To identify removable discontinuities: check if the limit exists and is finite, then verify if it matches the function value.

Question 14

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

Explanation: This function has a removable discontinuity at $x = 1$. When $x \neq 1$, we can factor the numerator as $(x-1)(x+1)$, so $p(x) = \frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. The limit as $x$ approaches 1 is $\lim_{x \to 1} (x+1) = 2$, but $p(1) = 5$. Since the limit exists (equals 2) but doesn't equal the function value (5), this creates a removable discontinuity. The discontinuity could be 'removed' by redefining $p(1) = 2$. Students often miss that the rational expression simplifies, thinking it's an infinite discontinuity. To classify discontinuities: always simplify rational functions first, then check if limits exist and match function values.

Question 15

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

Explanation: This function has a jump discontinuity at x = 1. The left-hand limit is lim(x→1⁻) x³ = 1, the function value is W(1) = 2, and the right-hand limit is lim(x→1⁺) (x³+1) = 2. Since the left-hand limit (1) differs from both the function value and right-hand limit (both 2), this creates a jump discontinuity. The function jumps from approaching 1 on the left to the value 2 at and to the right of x = 1. Students might think this is continuous because W(1) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that the one-sided limits exist but are different.

Question 16

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

Explanation: This function has an infinite discontinuity at $x = 0$. The left-hand limit is $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$, while the right-hand limit is $\lim_{x \to 0^{+}} 1 = 1$, and $\delta(0) = 1$. Since the left-hand limit is infinite, this creates an infinite discontinuity. The function exhibits vertical asymptote behavior as x approaches 0 from the left. Students might focus on the fact that the right limit and function value exist, but an infinite discontinuity occurs when at least one one-sided limit is infinite. To identify infinite discontinuities: check if either one-sided limit approaches $\pm \infty$, regardless of the other limit's behavior.

Question 17

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

Explanation: This function has no discontinuity at x = 0. The limit of sin(x)/x as x approaches 0 is a fundamental limit equal to 1, and t(0) is defined as 1. Since lim(x→0) sin(x)/x = 1 = t(0), the function is continuous at x = 0. This is a classic example where a rational expression that's undefined at a point can be made continuous by defining the function value to equal the limit. Students often think this is a removable discontinuity because sin(x)/x is undefined at x = 0, but since t(0) = 1 matches the limit, there's no discontinuity. To determine continuity: check that the limit exists and equals the function value at the point.

Question 18

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

Explanation: This function has a jump discontinuity at x = 3. The left-hand limit is lim(x→3⁻) 0 = 0, while the right-hand limit is lim(x→3⁺) 1 = 1, and c(3) = 1. Since both one-sided limits exist but are unequal (0 ≠ 1), this creates a jump discontinuity. The function jumps from 0 on the left to 1 on the right at x = 3. Students might think this is continuous because c(3) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

Question 19

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

Explanation: This function has a jump discontinuity at x = -1. The left-hand limit is lim(x→-1⁻) (x+5) = 4, while the right-hand limit is lim(x→-1⁺) x² = 1, and θ(-1) = (-1)² = 1. Since both one-sided limits exist but are unequal (4 ≠ 1), this creates a jump discontinuity. The function jumps from approaching 4 on the left to the value 1 at and to the right of x = -1. Students might think this is continuous because θ(-1) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

Question 20

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

Explanation: This function has a jump discontinuity at x = 2. The left-hand limit is lim(x→2⁻) (x-2) = 0, the function value is S(2) = 0, and the right-hand limit is lim(x→2⁺) (x+1) = 3. Since the left and right limits exist but are unequal (0 ≠ 3), this creates a jump discontinuity. The function jumps from 0 on the left to 3 on the right at x = 2. Students might think this is continuous because S(2) equals the left-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.