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  3. AP Calculus BC

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AP Calculus BC Flashcards: Algebraic Properties Of Limits

Study Algebraic Properties Of Limits in AP Calculus BC 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 Algebraic Properties Of Limits, 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.

AP Calculus BC Flashcards: Algebraic Properties Of Limits

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QUESTION

What is the limit of f(x)=x×g(x)f(x) = x \times g(x)f(x)=x×g(x) as xxx approaches 0 and g(x)g(x)g(x) is bounded?

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ANSWER
  1. If g(x)g(x)g(x) is bounded, then x⋅g(x)→0x \cdot g(x) \to 0x⋅g(x)→0 as x→0x \to 0x→0.

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Flashcard 1: What is the limit of f(x)=x×g(x)f(x) = x \times g(x)f(x)=x×g(x) as xxx approaches 0 and g(x)g(x)g(x) is bounded?

Answer:

  1. If g(x)g(x)g(x) is bounded, then x⋅g(x)→0x \cdot g(x) \to 0x⋅g(x)→0 as x→0x \to 0x→0.

Flashcard 2: Find the limit of 3x2−x2x2+x\frac{3x^2 - x}{2x^2 + x}2x2+x3x2−x​ as xxx approaches infinity.

Answer: 32\frac{3}{2}23​. For rational functions, the limit equals the ratio of leading coefficients.

Flashcard 3: Evaluate: limx→−2x2−4x+2\text{lim}_{x \to -2} \frac{x^2 - 4}{x + 2}limx→−2​x+2x2−4​.

Answer: -4. Factor as (x−2)(x+2)/(x+2)=x−2(x-2)(x+2)/(x+2) = x-2(x−2)(x+2)/(x+2)=x−2, then substitute.

Flashcard 4: Find the limit: limx→1x2−1x−1\text{lim}_{x \to 1} \frac{x^2 - 1}{x - 1}limx→1​x−1x2−1​.

Answer:

  1. Factor as (x−1)(x+1)/(x−1)=x+1(x-1)(x+1)/(x-1) = x+1(x−1)(x+1)/(x−1)=x+1, then substitute.

Flashcard 5: What is the limit of 1x2\frac{1}{x^2}x21​ as xxx approaches infinity?

Answer:

  1. Higher powers in denominators approach zero faster.

Flashcard 6: Evaluate: limx→0tan⁡(x)x\text{lim}_{x \to 0} \frac{\tan(x)}{x}limx→0​xtan(x)​.

Answer:

  1. Since tan⁡(x)=sin⁡(x)/cos⁡(x)\tan(x) = \sin(x)/\cos(x)tan(x)=sin(x)/cos(x) and cos⁡(0)=1\cos(0) = 1cos(0)=1.

Flashcard 7: Find the limit: limx→2x2−4x−2\text{lim}_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​.

Answer:

  1. Factor as (x−2)(x+2)/(x−2)=x+2(x-2)(x+2)/(x-2) = x+2(x−2)(x+2)/(x−2)=x+2, then substitute.

Flashcard 8: What is the limit of 1x\frac{1}{x}x1​ as xxx approaches infinity?

Answer:

  1. As xxx grows large, 1x\frac{1}{x}x1​ approaches zero.

Flashcard 9: Evaluate: limx→0x3x2\text{lim}_{x \to 0} \frac{x^3}{x^2}limx→0​x2x3​.

Answer:

  1. Simplify to x3x2=x\frac{x^3}{x^2} = xx2x3​=x, then substitute x=0x=0x=0.

Flashcard 10: Find the limit: limx→0sin(5x)x\text{lim}_{x \to 0} \frac{\text{sin}(5x)}{x}limx→0​xsin(5x)​.

Answer:

  1. Use sin⁡(5x)x=5⋅sin⁡(5x)5x\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}xsin(5x)​=5⋅5xsin(5x)​ and the fundamental limit.

Flashcard 11: What is the limit of xnx^nxn as xxx approaches aaa?

Answer: ana^nan. Power functions are continuous, so direct substitution works.

Flashcard 12: Identify the limit: limx→ac×f(x)\text{lim}_{x \to a} c \times f(x)limx→a​c×f(x).

Answer: c×limx→af(x)c \times \text{lim}_{x \to a} f(x)c×limx→a​f(x). Constants factor out of limit expressions.

Flashcard 13: Evaluate the limit: limx→−1(x2+2x+1)\text{lim}_{x \to -1} (x^2 + 2x + 1)limx→−1​(x2+2x+1).

Answer:

  1. Direct substitution: (−1)2+2(−1)+1=1−2+1=0(-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0(−1)2+2(−1)+1=1−2+1=0.

Flashcard 14: Find the limit: limx→3(5x2−2x+1)\text{lim}_{x \to 3}(5x^2 - 2x + 1)limx→3​(5x2−2x+1).

Answer:

  1. Direct substitution: 5(9)−2(3)+1=45−6+1=405(9) - 2(3) + 1 = 45 - 6 + 1 = 405(9)−2(3)+1=45−6+1=40.

Flashcard 15: Find the limit: limx→2(5x−3)\text{lim}_{x \to 2} (5x - 3)limx→2​(5x−3).

Answer:

  1. Direct substitution: 5(2)−3=10−3=75(2) - 3 = 10 - 3 = 75(2)−3=10−3=7.

Flashcard 16: Evaluate: limx→0sin(3x)x\text{lim}_{x \to 0} \frac{\text{sin}(3x)}{x}limx→0​xsin(3x)​.

Answer:

  1. Use sin⁡(3x)x=3⋅sin⁡(3x)3x\frac{\sin(3x)}{x} = 3 \cdot \frac{\sin(3x)}{3x}xsin(3x)​=3⋅3xsin(3x)​ and the fundamental limit.

Flashcard 17: State the property for the limit of a sum: limx→a(f(x)+g(x))\text{lim}_{x \to a} (f(x) + g(x))limx→a​(f(x)+g(x)).

Answer: limx→af(x)+limx→ag(x)\text{lim}_{x \to a} f(x) + \text{lim}_{x \to a} g(x)limx→a​f(x)+limx→a​g(x). The limit of a sum equals the sum of the individual limits.

Flashcard 18: State the property for the limit of a product: limx→a(f(x)×g(x))\text{lim}_{x \to a} (f(x) \times g(x))limx→a​(f(x)×g(x)).

Answer: limx→af(x)×limx→ag(x)\text{lim}_{x \to a} f(x) \times \text{lim}_{x \to a} g(x)limx→a​f(x)×limx→a​g(x). The limit of a product equals the product of individual limits.

Flashcard 19: Find the limit of x2−1x−1\frac{x^2 - 1}{x - 1}x−1x2−1​ as xxx approaches 1.

Answer:

  1. Factor as (x−1)(x+1)/(x−1)=x+1(x-1)(x+1)/(x-1) = x+1(x−1)(x+1)/(x−1)=x+1, then substitute.

Flashcard 20: Find the limit of a polynomial f(x)f(x)f(x) as xxx approaches aaa.

Answer: f(a)f(a)f(a). Polynomials are continuous, so use direct substitution.

Flashcard 21: Evaluate: limx→3x2−9x−3\text{lim}_{x \to 3} \frac{x^2 - 9}{x - 3}limx→3​x−3x2−9​.

Answer:

  1. Factor as (x−3)(x+3)/(x−3)=x+3(x-3)(x+3)/(x-3) = x+3(x−3)(x+3)/(x−3)=x+3, then substitute.

Flashcard 22: Evaluate: limx→0(1+x)n−1x\text{lim}_{x \to 0} \frac{(1+x)^n - 1}{x}limx→0​x(1+x)n−1​ for n>0n > 0n>0.

Answer: nnn. This follows from the binomial theorem and L'Hôpital's rule.

Flashcard 23: Evaluate: limx→1(x3−3x2+3x−1)\text{lim}_{x \to 1} (x^3 - 3x^2 + 3x - 1)limx→1​(x3−3x2+3x−1).

Answer:

  1. Direct substitution: 1−3+3−1=01 - 3 + 3 - 1 = 01−3+3−1=0.

Flashcard 24: What is the limit of x2−3x+2x−2\frac{x^2 - 3x + 2}{x - 2}x−2x2−3x+2​ as xxx approaches 2?

Answer:

  1. Factor as (x−1)(x−2)/(x−2)=x−1(x-1)(x-2)/(x-2) = x-1(x−1)(x−2)/(x−2)=x−1, then substitute.

Flashcard 25: Find the limit: limx→0(1+x)1x\text{lim}_{x \to 0} (1 + x)^\frac{1}{x}limx→0​(1+x)x1​.

Answer: eee. This is the definition of eee as a limit.

Flashcard 26: State the property for the limit of a difference: limx→a(f(x)−g(x))\text{lim}_{x \to a} (f(x) - g(x))limx→a​(f(x)−g(x)).

Answer: limx→af(x)−limx→ag(x)\text{lim}_{x \to a} f(x) - \text{lim}_{x \to a} g(x)limx→a​f(x)−limx→a​g(x). The limit of a difference equals the difference of individual limits.

Flashcard 27: What is the limit of sin⁡(x)x\frac{\sin(x)}{x}xsin(x)​ as xxx approaches 0?

Answer:

  1. This is a fundamental trigonometric limit identity.

Flashcard 28: Find the limit: limx→pisin(x)\text{lim}_{x \to \text{pi}} \text{sin}(x)limx→pi​sin(x).

Answer:

  1. Direct substitution: sin⁡(π)=0\sin(\pi) = 0sin(π)=0.

Flashcard 29: What is the limit of exx\frac{e^x}{x}xex​ as xxx approaches infinity?

Answer: Infinity. Exponential growth dominates polynomial growth.

Flashcard 30: Evaluate: limx→1x3−1x−1\text{lim}_{x \to 1} \frac{x^3 - 1}{x - 1}limx→1​x−1x3−1​.

Answer:

  1. Factor as (x2−1)/(x−1)=x+1(x^2-1)/(x-1) = x+1(x2−1)/(x−1)=x+1, then substitute x=1x=1x=1.