AP Calculus AB Flashcards: Exploring Types Of Discontinuities

What this deck covers

This deck focuses on Exploring Types Of Discontinuities, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify the discontinuity: $f(x) = \tan(x)$ at $x = \frac{\pi}{2}$.

Answer: Infinite discontinuity. Tangent has vertical asymptotes where cosine equals zero.

Flashcard 2: Identify the discontinuity in $f(x) = \frac{x^2 - 1}{x+1}$ at $x = -1$.

Answer: Removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x+1}$, creates removable hole at $x=-1$.

Flashcard 3: What is the condition for a function to have a removable discontinuity at $x = c$?

Answer: The limit exists, but the function value is either undefined or different. The limit approaches a finite value, but function is undefined or has wrong value.

Flashcard 4: Determine the discontinuity for $f(x) = \frac{x^2 + 4x + 4}{x + 2}$ at $x = -2$.

Answer: Removable discontinuity. Factor: $\frac{(x+2)^2}{x+2}$, simplifies to $(x+2)$ with hole at $x=-2$.

Flashcard 5: What condition indicates a function is discontinuous at a point?

Answer: The limit does not equal the function value or does not exist. Either the limit doesn't exist or doesn't equal the function value.

Flashcard 6: Identify the type of discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.

Answer: Removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, undefined at $x=1$ but limit exists.

Flashcard 7: Which type of discontinuity occurs when $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$?

Answer: Jump discontinuity. When one-sided limits exist but are unequal, the function "jumps" between values.

Flashcard 8: What is a removable discontinuity?

Answer: A discontinuity that can be removed by redefining the function. The function has a hole that can be filled by defining the value at that point.

Flashcard 9: Identify the discontinuity in $f(x) = \frac{x+3}{x^2-9}$ at $x = -3$.

Answer: Removable discontinuity. Factor: $\frac{x+3}{(x+3)(x-3)}$, simplifies with hole at $x=-3$.

Flashcard 10: What type of discontinuity does $f(x) = \lfloor x \rfloor$ have at integer points?

Answer: Jump discontinuity. Floor function creates jumps of size 1 at every integer value.

Flashcard 11: What is required for a function to be continuous at a point?

Answer: The limit and function value must be equal at that point. The function value must exist and equal the limit at that point.

Flashcard 12: Identify the discontinuity in $g(x) = \frac{1}{(x-2)^2}$ at $x = 2$.

Answer: Infinite discontinuity. Squared denominator ensures function approaches $+\infty$ from both sides.

Flashcard 13: Define a point of discontinuity.

Answer: A point where a function is not continuous. Any location where the function fails to meet continuity requirements.

Flashcard 14: Which type of discontinuity occurs when $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$?

Answer: Jump discontinuity. When one-sided limits exist but are unequal, the function "jumps" between values.

Flashcard 15: Determine discontinuity type in $f(x) = \frac{x^2 - 9}{x - 3}$ at $x = 3$.

Answer: Removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3}$, creates removable hole at $x=3$.

Flashcard 16: What type of discontinuity is present in $f(x) = \frac{x^2 - x - 2}{x - 2}$ at $x = 2$?

Answer: Removable discontinuity. Factor: $\frac{(x-2)(x+1)}{x-2}$, creates hole at $x=2$.

Flashcard 17: What is the definition of a function being discontinuous at a point?

Answer: The function is not continuous at that point. The function violates at least one of the three continuity conditions.

Flashcard 18: Identify the discontinuity in $f(x) = \frac{x^2 - 1}{x+1}$ at $x = -1$.

Answer: Removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x+1}$, creates removable hole at $x=-1$.

Flashcard 19: Determine discontinuity type in $f(x) = \frac{1}{x^2 - 1}$ at $x = 1$.

Answer: Infinite discontinuity. Factor $x^2-1=(x+1)(x-1)$, so denominator is zero at $x=1$.

Flashcard 20: Identify the discontinuity in $f(x) = \lfloor x \rfloor$ at $x = 1$.

Answer: Jump discontinuity. Left limit is 0, right limit is 1, creating a jump of size 1.

Flashcard 21: Find the discontinuity type for $g(x) = \frac{x+2}{x^2-4}$ at $x = -2$.

Answer: Removable discontinuity. Factor: $\frac{x+2}{(x+2)(x-2)}$, cancels to $\frac{1}{x-2}$ with hole at $x=-2$.

Flashcard 22: What is the difference between removable and non-removable discontinuities?

Answer: Removable can be fixed by redefining; non-removable cannot. Removable means "fixable"; non-removable means permanent breaks or asymptotes.

Flashcard 23: Determine the discontinuity for $f(x) = \frac{1}{x-3}$ at $x = 3$.

Answer: Infinite discontinuity. Denominator equals zero creating vertical asymptote where function approaches infinity.

Flashcard 24: What is an essential discontinuity?

Answer: A discontinuity that is neither removable nor a jump. Includes oscillatory discontinuities where limits don't exist in any form.

Flashcard 25: Identify the discontinuity for $f(x) = \frac{x}{x^2 - 4}$ at $x = 2$.

Answer: Infinite discontinuity. Factor denominator: $x^2-4=(x+2)(x-2)$, zero at $x=2$.

Flashcard 26: Identify the discontinuity in $f(x) = \frac{1}{x^2}$ at $x = 0$.

Answer: Infinite discontinuity. Positive denominator creates vertical asymptote as function approaches $+\infty$.

Flashcard 27: What is the definition of continuity on an interval?

Answer: A function is continuous if it is continuous at every point on the interval. No breaks, holes, or jumps exist anywhere within the given interval.

Flashcard 28: What type of discontinuity occurs if $\lim_{x \to c} f(x)$ does not exist?

Answer: Non-removable discontinuity. When the limit fails to exist, the discontinuity cannot be removed.

Flashcard 29: Identify the discontinuity in $f(x) = \frac{1}{x}$ at $x = 0$.

Answer: Infinite discontinuity. Division by zero creates a vertical asymptote as $f(x) \to \pm\infty$.

Flashcard 30: What type of discontinuity does $f(x) = \frac{1}{x(x-1)}$ have at $x = 0$?

Answer: Infinite discontinuity. Both factors in denominator create vertical asymptotes at $x=0$ and $x=1$.

Flashcard 31: What type of discontinuity is in $f(x) = \frac{1}{x(x-3)}$ at $x = 3$?

Answer: Infinite discontinuity. Denominator has factors $x$ and $(x-3)$, both creating vertical asymptotes.

Flashcard 32: What type of discontinuity is present in $f(x) = \frac{x^2 - x - 2}{x - 2}$ at $x = 2$?

Answer: Removable discontinuity. Factor: $\frac{(x-2)(x+1)}{x-2}$, creates hole at $x=2$.

Flashcard 33: Determine the discontinuity for $f(x) = \frac{x^2 + 4x + 4}{x + 2}$ at $x = -2$.

Answer: Removable discontinuity. Factor: $\frac{(x+2)^2}{x+2}$, simplifies to $(x+2)$ with hole at $x=-2$.

Flashcard 34: What is an infinite discontinuity?

Answer: A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote where values grow without bound.

Flashcard 35: What type of discontinuity does $f(x) = \lfloor x \rfloor$ have at integer points?

Answer: Jump discontinuity. Floor function creates jumps of size 1 at every integer value.

Flashcard 36: What does it mean if a function is continuous at a point?

Answer: The function's limit, value, and limit from both sides are equal. All three continuity conditions: function exists, limit exists, and they're equal.

Flashcard 37: Identify the discontinuity in $g(x) = \frac{1}{(x-2)^2}$ at $x = 2$.

Answer: Infinite discontinuity. Squared denominator ensures function approaches $+\infty$ from both sides.

Flashcard 38: What condition indicates a function is discontinuous at a point?

Answer: The limit does not equal the function value or does not exist. Either the limit doesn't exist or doesn't equal the function value.

Flashcard 39: What is the condition for a function to have a removable discontinuity at $x = c$?

Answer: The limit exists, but the function value is either undefined or different. The limit approaches a finite value, but function is undefined or has wrong value.

Flashcard 40: Identify the discontinuity: $f(x) = \tan(x)$ at $x = \frac{\pi}{2}$.

Answer: Infinite discontinuity. Tangent has vertical asymptotes where cosine equals zero.

Flashcard 41: What type of discontinuity does $f(x) = \frac{1}{x(x-1)}$ have at $x = 0$?

Answer: Infinite discontinuity. Both factors in denominator create vertical asymptotes at $x=0$ and $x=1$.

Flashcard 42: Determine the discontinuity for $f(x) = \frac{1}{x-3}$ at $x = 3$.

Answer: Infinite discontinuity. Denominator equals zero creating vertical asymptote where function approaches infinity.

Flashcard 43: What is a removable discontinuity?

Answer: A discontinuity that can be removed by redefining the function. The function has a hole that can be filled by defining the value at that point.

Flashcard 44: Identify the discontinuity in $f(x) = \lfloor x \rfloor$ at $x = 1$.

Answer: Jump discontinuity. Left limit is 0, right limit is 1, creating a jump of size 1.

Flashcard 45: What condition creates a jump discontinuity at $x = c$?

Answer: The left-hand and right-hand limits at $c$ exist but differ. The graph has a finite "jump" between the left and right limit values.

Flashcard 46: What condition creates an infinite discontinuity?

Answer: The function approaches infinity or negative infinity at the point. Denominator approaches zero while numerator approaches nonzero value.

Flashcard 47: Determine if $h(x) = |x|$ has any discontinuities.

Answer: No discontinuities. Absolute value function is continuous everywhere with no breaks or jumps.

Flashcard 48: Determine discontinuity type in $f(x) = \frac{1}{x^2 - 1}$ at $x = 1$.

Answer: Infinite discontinuity. Factor $x^2-1=(x+1)(x-1)$, so denominator is zero at $x=1$.

Flashcard 49: What is the difference between removable and non-removable discontinuities?

Answer: Removable can be fixed by redefining; non-removable cannot. Removable means "fixable"; non-removable means permanent breaks or asymptotes.

Flashcard 50: What is the definition of a function being discontinuous at a point?

Answer: The function is not continuous at that point. The function violates at least one of the three continuity conditions.

Flashcard 51: Determine if $h(x) = |x|$ has any discontinuities.

Answer: No discontinuities. Absolute value function is continuous everywhere with no breaks or jumps.

Flashcard 52: What does it mean if a function is continuous at a point?

Answer: The function's limit, value, and limit from both sides are equal. All three continuity conditions: function exists, limit exists, and they're equal.

Flashcard 53: Identify the discontinuity in $f(x) = \frac{x+3}{x^2-9}$ at $x = -3$.

Answer: Removable discontinuity. Factor: $\frac{x+3}{(x+3)(x-3)}$, simplifies with hole at $x=-3$.

Flashcard 54: Identify the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

Answer: Removable discontinuity. Factor: $\frac{(x+2)(x-2)}{x-2}$, simplifies with removable hole at $x=2$.

Flashcard 55: What condition creates a jump discontinuity at $x = c$?

Answer: The left-hand and right-hand limits at $c$ exist but differ. The graph has a finite "jump" between the left and right limit values.

Flashcard 56: What is an infinite discontinuity?

Answer: A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote where values grow without bound.

Flashcard 57: What is the discontinuity in $f(x) = \frac{x^2+2x+1}{x+1}$ at $x = -1$?

Answer: Removable discontinuity. Factor: $\frac{(x+1)^2}{x+1}$, simplifies to $(x+1)$ with hole at $x=-1$.

Flashcard 58: Identify the discontinuity for $f(x) = \frac{x}{x^2 - 4}$ at $x = 2$.

Answer: Infinite discontinuity. Factor denominator: $x^2-4=(x+2)(x-2)$, zero at $x=2$.

Flashcard 59: What type of discontinuity is present in $f(x) = \frac{x^2 - 9}{x-3}$ at $x = 3$?

Answer: Removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3}$, creates removable hole at $x=3$.

Flashcard 60: What is a jump discontinuity?

Answer: A discontinuity where the left and right limits exist but are not equal. The function has different left and right limits creating a "jump" in the graph.

Flashcard 61: Find the discontinuity type for $g(x) = \frac{x+2}{x^2-4}$ at $x = -2$.

Answer: Removable discontinuity. Factor: $\frac{x+2}{(x+2)(x-2)}$, cancels to $\frac{1}{x-2}$ with hole at $x=-2$.

Flashcard 62: What is an essential discontinuity?

Answer: A discontinuity that is neither removable nor a jump. Includes oscillatory discontinuities where limits don't exist in any form.

Flashcard 63: What condition creates an infinite discontinuity?

Answer: The function approaches infinity or negative infinity at the point. Denominator approaches zero while numerator approaches nonzero value.

Flashcard 64: Identify the discontinuity in $f(x) = \frac{1}{x^2}$ at $x = 0$.

Answer: Infinite discontinuity. Positive denominator creates vertical asymptote as function approaches $+\infty$.

Flashcard 65: What type of discontinuity is in $f(x) = \frac{1}{x(x-3)}$ at $x = 3$?

Answer: Infinite discontinuity. Denominator has factors $x$ and $(x-3)$, both creating vertical asymptotes.

Flashcard 66: What is the definition of continuity on an interval?

Answer: A function is continuous if it is continuous at every point on the interval. No breaks, holes, or jumps exist anywhere within the given interval.

Flashcard 67: Identify the discontinuity in $f(x) = \frac{1}{x}$ at $x = 0$.

Answer: Infinite discontinuity. Division by zero creates a vertical asymptote as $f(x) \to \pm\infty$.

Flashcard 68: Define a point of discontinuity.

Answer: A point where a function is not continuous. Any location where the function fails to meet continuity requirements.

Flashcard 69: Identify the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

Answer: Removable discontinuity. Factor: $\frac{(x+2)(x-2)}{x-2}$, simplifies with removable hole at $x=2$.

Flashcard 70: What type of discontinuity occurs if $\lim_{x \to c} f(x)$ does not exist?

Answer: Non-removable discontinuity. When the limit fails to exist, the discontinuity cannot be removed.

Flashcard 71: What is the discontinuity in $f(x) = \frac{x^2+2x+1}{x+1}$ at $x = -1$?

Answer: Removable discontinuity. Factor: $\frac{(x+1)^2}{x+1}$, simplifies to $(x+1)$ with hole at $x=-1$.

Flashcard 72: What type of discontinuity is present in $f(x) = \frac{x^2 - 9}{x-3}$ at $x = 3$?

Answer: Removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3}$, creates removable hole at $x=3$.

Flashcard 73: What is required for a function to be continuous at a point?

Answer: The limit and function value must be equal at that point. The function value must exist and equal the limit at that point.

Flashcard 74: What is a jump discontinuity?

Answer: A discontinuity where the left and right limits exist but are not equal. The function has different left and right limits creating a "jump" in the graph.

Flashcard 75: Identify the type of discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.

Answer: Removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, undefined at $x=1$ but limit exists.

Flashcard 76: Determine discontinuity type in $f(x) = \frac{x^2 - 9}{x - 3}$ at $x = 3$.

Answer: Removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3}$, creates removable hole at $x=3$.