AP Calculus BC Flashcards: Removing Discontinuities

What this deck covers

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

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

Answer: Removable discontinuity. The factor $(x-2)$ cancels, leaving a hole at $x=2$.

Flashcard 2: How do you remove a removable discontinuity?

Answer: Redefine the function at the discontinuity point. Set the function value equal to the limit at that point.

Flashcard 3: What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$?

Answer: Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

Flashcard 4: What is the behavior of $f(x)$ at a jump discontinuity?

Answer: Differing left and right-hand limits. Left and right limits exist but are unequal.

Flashcard 5: What must be true for $f(x)$ to be continuous at $x = a$?

Answer: $\lim_{x \to a} f(x) = f(a)$. All three continuity conditions must hold at the point.

Flashcard 6: What condition creates a removable discontinuity?

Answer: A factor cancels in the numerator and denominator. Common factors in numerator and denominator create holes.

Flashcard 7: Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$.

Answer: Redefine $f(3) = 6$. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.

Flashcard 8: Identify the discontinuity type: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

Answer: Removable discontinuity. The factor $(x-2)$ cancels, creating a hole at $x=2$.

Flashcard 9: Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$.

Answer: $x = 3$. The denominator equals zero when $x=3$, creating a hole.

Flashcard 10: What is the necessary condition for a function to be continuous?

Answer: The function is defined and limits match the function value. Function exists, limit exists, and both values are equal.

Flashcard 11: Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$.

Answer: $x = 0$. The factor $x$ in denominator creates a hole when $x=0$.

Flashcard 12: What is the condition for a removable discontinuity?

Answer: A cancelable factor in the expression. A common factor exists in both numerator and denominator.

Flashcard 13: What happens to $f(x)$ at an essential discontinuity?

Answer: The limits don't exist or are undefined. The function oscillates wildly or has no pattern near the point.

Flashcard 14: Determine the discontinuity type: $f(x) = \frac{1}{x}$ at $x = 0$.

Answer: Infinite discontinuity. Division by zero with no cancellation creates a vertical asymptote.

Flashcard 15: State the condition for a function to have no discontinuities.

Answer: The function must be continuous everywhere in its domain. No holes, jumps, or vertical asymptotes anywhere.

Flashcard 16: What is the limit definition of continuity at $x = a$?

Answer: $\lim_{x \to a} f(x) = f(a)$. The limit as $x$ approaches $a$ equals the function value.

Flashcard 17: What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$?

Answer:

1. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, so limit is $1+1=2$.

Flashcard 18: What is an infinite discontinuity?

Answer: A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote at that point.

Flashcard 19: What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$?

Answer: Infinite discontinuity at $x = a$. The function approaches infinity, creating a vertical asymptote.

Flashcard 20: Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.

Answer: Redefine $f(2) = 4$. The limit at $x=2$ is 4, so define $f(2)=4$ for continuity.

Flashcard 21: Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

Answer:

1. Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $2+2=4$.

Flashcard 22: How is a jump discontinuity characterized?

Answer: Differing left and right-hand limits. The function has different values approaching from each side.

Flashcard 23: Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$.

Answer: $x = 5$. The denominator becomes zero when $x=5$, creating a hole.

Flashcard 24: Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$?

Answer: Removable. The factor $(x-1)$ cancels, leaving a hole at $x=1$.

Flashcard 25: What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$?

Answer: Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

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

Answer: Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

Flashcard 27: What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$?

Answer:

1. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.

Flashcard 28: Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.

Answer: Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.

Flashcard 29: Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.

Answer: Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.

Flashcard 30: What is the condition for a removable discontinuity?

Answer: A cancelable factor in the expression. A common factor exists in both numerator and denominator.