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 in $f(x) = \frac{1}{x^2}$ at $x = 0$.

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

Flashcard 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: Identify the discontinuity: $f(x) = \tan(x)$ at $x = \frac{\pi}{2}$.

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

Flashcard 9: 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 10: 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 11: 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 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: 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 14: 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 15: 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 16: 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 17: Determine if $h(x) = |x|$ has any discontinuities.

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

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 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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.