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  3. AP Calculus AB

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AP Calculus AB Flashcards: Determining Limits Using Algebraic Manipulation

Study Determining Limits Using Algebraic Manipulation in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Determining Limits Using Algebraic Manipulation, 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.

AP Calculus AB Flashcards: Determining Limits Using Algebraic Manipulation

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QUESTION

Determine the limit of f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ as xxx approaches 1.

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ANSWER
  1. Factor: (x−1)(x+1)x−1=x+1\frac{(x-1)(x+1)}{x-1} = x+1x−1(x−1)(x+1)​=x+1, then substitute x=1x = 1x=1.

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Flashcard 1: Determine the limit of f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ as xxx approaches 1.

Answer:

  1. Factor: (x−1)(x+1)x−1=x+1\frac{(x-1)(x+1)}{x-1} = x+1x−1(x−1)(x+1)​=x+1, then substitute x=1x = 1x=1.

Flashcard 2: Evaluate x2−1x−1\frac{x^2 - 1}{x - 1}x−1x2−1​ as xxx approaches 1.

Answer:

  1. Factor: (x−1)(x+1)x−1=x+1\frac{(x-1)(x+1)}{x-1} = x+1x−1(x−1)(x+1)​=x+1, then substitute x=1x = 1x=1.

Flashcard 3: State the limit of f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​ as xxx approaches 3.

Answer:

  1. Factor: (x−3)(x+3)x−3=x+3\frac{(x-3)(x+3)}{x-3} = x+3x−3(x−3)(x+3)​=x+3, then substitute x=3x = 3x=3.

Flashcard 4: Evaluate x2−25x−5\frac{x^2 - 25}{x - 5}x−5x2−25​ as xxx approaches 5.

Answer:

  1. Factor: (x−5)(x+5)x−5=x+5\frac{(x-5)(x+5)}{x-5} = x+5x−5(x−5)(x+5)​=x+5, then substitute x=5x = 5x=5.

Flashcard 5: Evaluate the limit of x2−25x−5\frac{x^2 - 25}{x - 5}x−5x2−25​ as xxx approaches 5.

Answer:

  1. Factor: (x−5)(x+5)x−5=x+5\frac{(x-5)(x+5)}{x-5} = x+5x−5(x−5)(x+5)​=x+5, then substitute x=5x = 5x=5.

Flashcard 6: Find the limit of x2−4x+4x−2\frac{x^2 - 4x + 4}{x - 2}x−2x2−4x+4​ as xxx approaches 2.

Answer:

  1. Factor: (x−2)2x−2=x−2\frac{(x-2)^2}{x-2} = x-2x−2(x−2)2​=x−2, then substitute x=2x = 2x=2.

Flashcard 7: Evaluate x2−4x−2\frac{x^2 - 4}{x - 2}x−2x2−4​ as xxx approaches 2.

Answer:

  1. Factor: (x−2)(x+2)x−2=x+2\frac{(x-2)(x+2)}{x-2} = x+2x−2(x−2)(x+2)​=x+2, then substitute x=2x = 2x=2.

Flashcard 8: Evaluate the limit of f(x)=x2−1x−1f(x) = \frac{x^2 - 1}{x - 1}f(x)=x−1x2−1​ as xxx approaches 1.

Answer:

  1. Factor: (x−1)(x+1)x−1=x+1\frac{(x-1)(x+1)}{x-1} = x+1x−1(x−1)(x+1)​=x+1, then substitute x=1x = 1x=1.

Flashcard 9: What is the limit of f(x)=x3f(x) = x^3f(x)=x3 as xxx approaches 3?

Answer:

  1. Direct substitution: 33=273^3 = 2733=27.

Flashcard 10: What is the limit of x3−8x−2\frac{x^3 - 8}{x - 2}x−2x3−8​ as xxx approaches 2?

Answer:

  1. Factor: (x−2)(x2+2x+4)x−2=x2+2x+4\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4x−2(x−2)(x2+2x+4)​=x2+2x+4, then substitute x=2x = 2x=2.

Flashcard 11: Evaluate the limit of x2−9x−3\frac{x^2 - 9}{x - 3}x−3x2−9​ as xxx approaches 3.

Answer:

  1. Factor: (x−3)(x+3)x−3=x+3\frac{(x-3)(x+3)}{x-3} = x+3x−3(x−3)(x+3)​=x+3, then substitute x=3x = 3x=3.

Flashcard 12: Find the limit of x2−25x−5\frac{x^2 - 25}{x - 5}x−5x2−25​ as xxx approaches 5.

Answer:

  1. Factor: (x−5)(x+5)x−5=x+5\frac{(x-5)(x+5)}{x-5} = x+5x−5(x−5)(x+5)​=x+5, then substitute x=5x = 5x=5.

Flashcard 13: Find the limit of f(x)=x2−4x+4f(x) = x^2 - 4x + 4f(x)=x2−4x+4 as xxx approaches 2.

Answer:

  1. Direct substitution: 22−4(2)+4=02^2 - 4(2) + 4 = 022−4(2)+4=0.

Flashcard 14: Determine the limit of x2−4x+4x^2 - 4x + 4x2−4x+4 as xxx approaches 4.

Answer:

  1. Direct substitution: 42−4(4)+4=44^2 - 4(4) + 4 = 442−4(4)+4=4.

Flashcard 15: What is the limit of f(x)=x3f(x) = x^3f(x)=x3 as xxx approaches -1?

Answer: -1. Direct substitution: (−1)3=−1(-1)^3 = -1(−1)3=−1.

Flashcard 16: State the limit of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ as xxx approaches negative infinity.

Answer:

  1. As x→−∞x \to -\inftyx→−∞, 1x→0\frac{1}{x} \to 0x1​→0.

Flashcard 17: Determine the limit of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ as xxx approaches infinity.

Answer:

  1. As x→∞x \to \inftyx→∞, 1x→0\frac{1}{x} \to 0x1​→0.

Flashcard 18: What is the limit of h(x)=x2+x−6x−2h(x) = \frac{x^2 + x - 6}{x - 2}h(x)=x−2x2+x−6​ as xxx approaches 2?

Answer:

  1. Factor: (x+3)(x−2)x−2=x+3\frac{(x+3)(x-2)}{x-2} = x+3x−2(x+3)(x−2)​=x+3, then substitute x=2x = 2x=2.

Flashcard 19: Determine the limit of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1}f(x)=x2+1x2−1​ as xxx approaches ∞\infty∞.

Answer:

  1. Divide by highest power: 1−1x21+1x2→1\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 11+x21​1−x21​​→1.

Flashcard 20: Find the limit of f(x)=x3f(x) = x^3f(x)=x3 as xxx approaches -2.

Answer: -8. Direct substitution: (−2)3=−8(-2)^3 = -8(−2)3=−8.

Flashcard 21: Find the limit of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1}f(x)=x2+1x2−1​ as xxx approaches infinity.

Answer:

  1. Divide by highest power: 1−1x21+1x2→1\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 11+x21​1−x21​​→1.

Flashcard 22: Evaluate x2−4x+4x−2\frac{x^2 - 4x + 4}{x - 2}x−2x2−4x+4​ as xxx approaches 2.

Answer:

  1. Factor: (x−2)2x−2=x−2\frac{(x-2)^2}{x-2} = x-2x−2(x−2)2​=x−2, then substitute x=2x = 2x=2.

Flashcard 23: What is the limit of f(x)=x2−16x−4f(x) = \frac{x^2 - 16}{x - 4}f(x)=x−4x2−16​ as xxx approaches 4?

Answer:

  1. Factor: (x−4)(x+4)x−4=x+4\frac{(x-4)(x+4)}{x-4} = x+4x−4(x−4)(x+4)​=x+4, then substitute x=4x = 4x=4.

Flashcard 24: State the limit of f(x)=x2+2x+1x+1f(x) = \frac{x^2 + 2x + 1}{x + 1}f(x)=x+1x2+2x+1​ as xxx approaches -1.

Answer:

  1. Factor: (x+1)2x+1=x+1\frac{(x+1)^2}{x+1} = x+1x+1(x+1)2​=x+1, then substitute x=−1x = -1x=−1.

Flashcard 25: What is the limit of x2−9x−3\frac{x^2 - 9}{x - 3}x−3x2−9​ as xxx approaches 3?

Answer:

  1. Factor: (x−3)(x+3)x−3=x+3\frac{(x-3)(x+3)}{x-3} = x+3x−3(x−3)(x+3)​=x+3, then substitute x=3x = 3x=3.

Flashcard 26: Find the limit of f(x)=x3f(x) = x^3f(x)=x3 as xxx approaches 2.

Answer:

  1. Direct substitution: 23=82^3 = 823=8.

Flashcard 27: What is the limit of f(x)=x2−1x2+1f(x) = \frac{x^2 - 1}{x^2 + 1}f(x)=x2+1x2−1​ as xxx approaches infinity?

Answer:

  1. Divide by highest power: 1−1x21+1x2→1\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 11+x21​1−x21​​→1.

Flashcard 28: What is the limit of f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​ as xxx approaches 3?

Answer:

  1. Factor: (x−3)(x+3)x−3=x+3\frac{(x-3)(x+3)}{x-3} = x+3x−3(x−3)(x+3)​=x+3, then substitute x=3x = 3x=3.

Flashcard 29: What is the limit of g(x)=1xg(x) = \frac{1}{x}g(x)=x1​ as xxx approaches 0?

Answer: Does not exist. Function undefined at x=0x = 0x=0; left and right limits differ.

Flashcard 30: What is the limit of f(x)=2x−6x−3f(x) = \frac{2x - 6}{x - 3}f(x)=x−32x−6​ as xxx approaches 3?

Answer:

  1. Factor: 2(x−3)x−3=2\frac{2(x-3)}{x-3} = 2x−32(x−3)​=2, then substitute x=3x = 3x=3.