AP Calculus BC Flashcards: Removing Discontinuities

Study Removing Discontinuities in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Calculus BC

Removing Discontinuities

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QUESTION

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

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ANSWER

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

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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.

All flashcards

Flashcard 1: 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 2: What is a jump discontinuity?

Answer: A discontinuity where a function has differing left and right limits at a point. The function 'jumps' from one value to another at the point.

Flashcard 3: 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 4: 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 5: 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 6: 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 7: 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 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: 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 10: What is the general approach to remove a discontinuity?

Answer: Simplify and redefine the function. Factor out common terms and define the function at the hole.

Flashcard 11: 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 12: What is the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ ?

Answer:

Factor: $\frac{(x^2+x+1)(x-1)}{x-1} = x^2+x+1$ , so limit is 3.

Flashcard 13: 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 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: 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 16: 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 17: 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 18: What condition creates a removable discontinuity?

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

Flashcard 19: What does it mean for a function to be continuous?

Answer: The function is continuous if it's defined and its limit equals the function value. Three conditions: defined, limit exists, and they're equal.

Flashcard 20: What is the limit as $x \to 2$ of $\frac{x^2 - 4}{x - 2}$ ?

Answer:

Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$ .

Flashcard 21: What is a removable discontinuity in a function?

Answer: A point where a function is not defined but can be redefined to make it continuous. The limit exists but doesn't equal the function value.

Flashcard 22: What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$ ?

Answer:

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

Flashcard 23: Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Answer:

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

Flashcard 24: What condition creates a removable discontinuity?

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

Flashcard 25: Find and remove the discontinuity: $f(x) = \frac{x^2 - 4x}{x - 4}$ .

Answer: Redefine $f(4) = 4$ . Factor: $\frac{x(x-4)}{x-4} = x$ , so limit at $x=4$ is 4.

Flashcard 26: 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 27: 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 28: 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 29: 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 30: How is a jump discontinuity characterized?

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

Flashcard 31: 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 32: 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 33: 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 34: 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 35: What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ ?

Answer:

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

Flashcard 36: 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 37: 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 38: 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 39: 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 40: 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 41: What is the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ ?

Answer:

Factor: $\frac{(x^2+x+1)(x-1)}{x-1} = x^2+x+1$ , so limit is 3.

Flashcard 42: 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 43: 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 44: What is the limit as $x \to 2$ of $\frac{x^2 - 4}{x - 2}$ ?

Answer:

Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$ .

Flashcard 45: 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 46: Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Answer:

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

Flashcard 47: When is a discontinuity considered non-removable?

Answer: If limits do not exist or differ at a point. No common factors cancel or limits become infinite.

Flashcard 48: 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 49: 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 50: 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 51: 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 52: What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$ ?

Answer:

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

Flashcard 53: 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 54: When is a discontinuity considered non-removable?

Answer: If limits do not exist or differ at a point. No common factors cancel or limits become infinite.

Flashcard 55: 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 56: 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 57: What is a jump discontinuity?

Answer: A discontinuity where a function has differing left and right limits at a point. The function 'jumps' from one value to another at the point.

Flashcard 58: What is the general approach to remove a discontinuity?

Answer: Simplify and redefine the function. Factor out common terms and define the function at the hole.

Flashcard 59: 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 60: 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 61: 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 62: What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ ?

Answer:

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

Flashcard 63: Find and remove the discontinuity: $f(x) = \frac{x^2 - 4x}{x - 4}$ .

Answer: Redefine $f(4) = 4$ . Factor: $\frac{x(x-4)}{x-4} = x$ , so limit at $x=4$ is 4.

Flashcard 64: 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 65: 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 66: How is a jump discontinuity characterized?

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

Flashcard 67: 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 68: What does it mean for a function to be continuous?

Answer: The function is continuous if it's defined and its limit equals the function value. Three conditions: defined, limit exists, and they're equal.

Flashcard 69: 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 70: What is a removable discontinuity in a function?

Answer: A point where a function is not defined but can be redefined to make it continuous. The limit exists but doesn't equal the function value.

Flashcard 71: 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 72: 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.

AP Calculus BC Flashcards: Removing Discontinuities

Removing Discontinuities

What this deck covers

How to use these flashcards

All flashcards

Flashcard 1: What is the behavior of f(x)f(x)f(x) at a jump discontinuity?

Flashcard 2: What is a jump discontinuity?

Flashcard 3: Which discontinuity is present in f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ at x=1x=1x=1?

Flashcard 4: What kind of discontinuity does f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ have at x=0x = 0x=0?

Flashcard 5: What kind of discontinuity does f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ have at x=0x = 0x=0?

Flashcard 6: Find the removable discontinuity in f(x)=x2−25x−5f(x) = \frac{x^2 - 25}{x - 5}f(x)=x−5x2−25​.

Flashcard 7: Determine the discontinuity type: f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ at x=0x = 0x=0.

Flashcard 8: Identify the discontinuity type: f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​ at x=2x = 2x=2.

Flashcard 9: Remove the discontinuity in f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​.

Flashcard 10: What is the general approach to remove a discontinuity?

Flashcard 11: Identify the discontinuity at x=0x = 0x=0 for f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​.

Flashcard 12: What is the limit lim⁡x→1x3−1x−1\lim_{x \to 1} \frac{x^3 - 1}{x - 1}limx→1​x−1x3−1​?

Flashcard 13: What is the limit definition of continuity at x=ax = ax=a?

Flashcard 14: Determine the discontinuity type: f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ at x=0x = 0x=0.

Flashcard 15: Find the removable discontinuity: f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​.

Flashcard 16: What is the behavior of f(x)f(x)f(x) at a jump discontinuity?

Flashcard 17: State the condition for a function to have no discontinuities.

Flashcard 18: What condition creates a removable discontinuity?

Flashcard 19: What does it mean for a function to be continuous?

Flashcard 20: What is the limit as x→2x \to 2x→2 of x2−4x−2\frac{x^2 - 4}{x - 2}x−2x2−4​?

Flashcard 21: What is a removable discontinuity in a function?

Flashcard 22: What is the limit as x→3x \to 3x→3 of x2−9x−3\frac{x^2 - 9}{x - 3}x−3x2−9​?

Flashcard 23: Find the limit: lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​.

Flashcard 24: What condition creates a removable discontinuity?

Flashcard 25: Find and remove the discontinuity: f(x)=x2−4xx−4f(x) = \frac{x^2 - 4x}{x - 4}f(x)=x−4x2−4x​.

Flashcard 26: What must be true for f(x)f(x)f(x) to be continuous at x=ax = ax=a?

Flashcard 27: Identify the removable discontinuity in f(x)=x2−4xx(x−4)f(x) = \frac{x^2 - 4x}{x(x - 4)}f(x)=x(x−4)x2−4x​.

Flashcard 28: What is the necessary condition for a function to be continuous?

Flashcard 29: How do you remove a removable discontinuity?

Flashcard 30: How is a jump discontinuity characterized?

Flashcard 31: Determine the discontinuity: f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​ at x=2x = 2x=2.

Flashcard 32: State the condition for a function to have no discontinuities.

Flashcard 33: What is the condition for a removable discontinuity?

Flashcard 34: What is an infinite discontinuity?

Flashcard 35: What is the limit of lim⁡x→1x2−1x−1\lim_{x \to 1} \frac{x^2 - 1}{x - 1}limx→1​x−1x2−1​?

Flashcard 36: How do you remove a removable discontinuity?

Flashcard 37: Which discontinuity is present in f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ at x=1x=1x=1?

Flashcard 38: Identify and remove the discontinuity of f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​.

Flashcard 39: Identify the discontinuity at x=0x = 0x=0 for f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​.

Flashcard 40: What happens to f(x)f(x)f(x) at an essential discontinuity?

Flashcard 41: What is the limit lim⁡x→1x3−1x−1\lim_{x \to 1} \frac{x^3 - 1}{x - 1}limx→1​x−1x3−1​?

Flashcard 42: Identify the discontinuity type: f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​ at x=2x = 2x=2.

Flashcard 43: What happens to f(x)f(x)f(x) at an essential discontinuity?

Flashcard 44: What is the limit as x→2x \to 2x→2 of x2−4x−2\frac{x^2 - 4}{x - 2}x−2x2−4​?

Flashcard 45: What is the condition for a removable discontinuity?

Flashcard 46: Find the limit: lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​.

Flashcard 47: When is a discontinuity considered non-removable?

Flashcard 48: What is an infinite discontinuity?

Flashcard 49: What must be true for f(x)f(x)f(x) to be continuous at x=ax = ax=a?

Flashcard 50: Identify the discontinuity type: f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ at x=1x = 1x=1.

Flashcard 51: Identify the removable discontinuity in f(x)=x2−4xx(x−4)f(x) = \frac{x^2 - 4x}{x(x - 4)}f(x)=x(x−4)x2−4x​.

Flashcard 52: What is the limit as x→3x \to 3x→3 of x2−9x−3\frac{x^2 - 9}{x - 3}x−3x2−9​?

Flashcard 53: What type of discontinuity occurs at f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ at x=0x = 0x=0?

Flashcard 54: When is a discontinuity considered non-removable?

Flashcard 55: What does lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \inftylimx→a​f(x)=∞ imply about f(x)f(x)f(x) at x=ax = ax=a?

Flashcard 56: What is the limit definition of continuity at x=ax = ax=a?

Flashcard 57: What is a jump discontinuity?

Flashcard 58: What is the general approach to remove a discontinuity?

Flashcard 59: Remove the discontinuity in f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​.

Flashcard 60: Find the removable discontinuity: f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​.

Flashcard 61: Identify and remove the discontinuity of f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​.

Flashcard 62: What is the limit of lim⁡x→1x2−1x−1\lim_{x \to 1} \frac{x^2 - 1}{x - 1}limx→1​x−1x2−1​?

Flashcard 63: Find and remove the discontinuity: f(x)=x2−4xx−4f(x) = \frac{x^2 - 4x}{x - 4}f(x)=x−4x2−4x​.

Flashcard 64: Identify the discontinuity type: f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ at x=1x = 1x=1.

Flashcard 65: What type of discontinuity occurs at f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ at x=0x = 0x=0?

Flashcard 66: How is a jump discontinuity characterized?

Flashcard 67: Determine the discontinuity: f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​ at x=2x = 2x=2.

Flashcard 68: What does it mean for a function to be continuous?

Flashcard 69: What does lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \inftylimx→a​f(x)=∞ imply about f(x)f(x)f(x) at x=ax = ax=a?

Flashcard 70: What is a removable discontinuity in a function?

Flashcard 71: What is the necessary condition for a function to be continuous?

Flashcard 72: Find the removable discontinuity in f(x)=x2−25x−5f(x) = \frac{x^2 - 25}{x - 5}f(x)=x−5x2−25​.

Flashcard 1: What is the behavior of $f(x)$ at a jump discontinuity?

Flashcard 3: Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$ ?

Flashcard 4: What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$ ?

Flashcard 5: What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$ ?

Flashcard 6: Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$ .

Flashcard 7: Determine the discontinuity type: $f(x) = \frac{1}{x}$ at $x = 0$ .

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

Flashcard 9: Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$ .

Flashcard 11: Identify the discontinuity at $x = 0$ for $f(x) = \frac{1}{x^2}$ .

Flashcard 12: What is the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ ?

Flashcard 13: What is the limit definition of continuity at $x = a$ ?

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

Flashcard 15: Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$ .

Flashcard 16: What is the behavior of $f(x)$ at a jump discontinuity?

Flashcard 20: What is the limit as $x \to 2$ of $\frac{x^2 - 4}{x - 2}$ ?

Flashcard 22: What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$ ?

Flashcard 23: Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Flashcard 25: Find and remove the discontinuity: $f(x) = \frac{x^2 - 4x}{x - 4}$ .

Flashcard 26: What must be true for $f(x)$ to be continuous at $x = a$ ?

Flashcard 27: Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$ .

Flashcard 31: Determine the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$ .

Flashcard 35: What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ ?

Flashcard 37: Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$ ?

Flashcard 38: Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$ .

Flashcard 39: Identify the discontinuity at $x = 0$ for $f(x) = \frac{1}{x^2}$ .

Flashcard 40: What happens to $f(x)$ at an essential discontinuity?

Flashcard 41: What is the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ ?

Flashcard 42: Identify the discontinuity type: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$ .

Flashcard 43: What happens to $f(x)$ at an essential discontinuity?

Flashcard 44: What is the limit as $x \to 2$ of $\frac{x^2 - 4}{x - 2}$ ?

Flashcard 46: Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Flashcard 49: What must be true for $f(x)$ to be continuous at $x = a$ ?

Flashcard 50: Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$ .

Flashcard 51: Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$ .

Flashcard 52: What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$ ?

Flashcard 53: What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$ ?

Flashcard 55: What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$ ?

Flashcard 56: What is the limit definition of continuity at $x = a$ ?

Flashcard 59: Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$ .

Flashcard 60: Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$ .

Flashcard 61: Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$ .

Flashcard 62: What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$ ?

Flashcard 63: Find and remove the discontinuity: $f(x) = \frac{x^2 - 4x}{x - 4}$ .

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

Flashcard 65: What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$ ?

Flashcard 67: Determine the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$ .

Flashcard 69: What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$ ?

Flashcard 72: Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$ .