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  2. AP Calculus AB
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AP Calculus AB Flashcards: Algebraic Properties Of Limits

Study Algebraic Properties Of Limits 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 Algebraic Properties Of Limits, 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: Algebraic Properties Of Limits

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QUESTION

What is the limit of f(x)=xf(x) = xf(x)=x as xxx approaches aaa?

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ANSWER

The limit is aaa. The identity function equals its input value.

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Flashcard 1: What is the limit of f(x)=xf(x) = xf(x)=x as xxx approaches aaa?

Answer: The limit is aaa. The identity function equals its input value.

Flashcard 2: Find limx→3x2−9x−3\text{lim}_{x \to 3} \frac{x^2 - 9}{x - 3}limx→3​x−3x2−9​. Use algebraic properties.

Answer: The limit is 666. Factor difference of squares: (x+3)(x−3)x−3=x+3\frac{(x+3)(x-3)}{x-3} = x+3x−3(x+3)(x−3)​=x+3.

Flashcard 3: Find limx→2sqrt(x+2)−2x−2\text{lim}_{x \to 2} \frac{\text{sqrt}(x+2) - 2}{x - 2}limx→2​x−2sqrt(x+2)−2​. Use algebraic properties.

Answer: The limit is 14\frac{1}{4}41​. Rationalize by multiplying by x+2+2x+2+2\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}x+2​+2x+2​+2​.

Flashcard 4: Find limx→−2x2+4x+4x+2\text{lim}_{x \to -2} \frac{x^2 + 4x + 4}{x + 2}limx→−2​x+2x2+4x+4​. Use algebraic properties.

Answer: The limit is −4-4−4. Factor perfect square: (x+2)2x+2=x+2\frac{(x+2)^2}{x+2} = x+2x+2(x+2)2​=x+2, then substitute.

Flashcard 5: Find limx→1x2−1x2+x−2\text{lim}_{x \to 1} \frac{x^2 - 1}{x^2 + x - 2}limx→1​x2+x−2x2−1​. Use algebraic properties.

Answer: The limit is 000. Factor both: (x−1)(x+1)(x−1)(x+2)=x+1x+2\frac{(x-1)(x+1)}{(x-1)(x+2)} = \frac{x+1}{x+2}(x−1)(x+2)(x−1)(x+1)​=x+2x+1​, then substitute.

Flashcard 6: Find limx→1(3x3−2x2+x−5)\text{lim}_{x \to 1} (3x^3 - 2x^2 + x - 5)limx→1​(3x3−2x2+x−5). Use algebraic properties.

Answer: The limit is −3-3−3. Direct substitution: 3(1)3−2(1)2+1−5=−33(1)^3 - 2(1)^2 + 1 - 5 = -33(1)3−2(1)2+1−5=−3.

Flashcard 7: Find limx→0x2cos(x)\text{lim}_{x \to 0} x^2 \text{cos}(x)limx→0​x2cos(x). Use algebraic properties.

Answer: The limit is 000. Product of limits: lim x2×lim cos(x)=0×1\text{lim } x^2 \times \text{lim } \text{cos}(x) = 0 \times 1lim x2×lim cos(x)=0×1.

Flashcard 8: Find limx→−1(x3+4x2+x−6)\text{lim}_{x \to -1} (x^3 + 4x^2 + x - 6)limx→−1​(x3+4x2+x−6). Use algebraic properties.

Answer: The limit is −2-2−2. Direct substitution: (−1)3+4(−1)2+(−1)−6=−2(-1)^3 + 4(-1)^2 + (-1) - 6 = -2(−1)3+4(−1)2+(−1)−6=−2.

Flashcard 9: Find limx→0sin(2x)x\text{lim}_{x \to 0} \frac{\text{sin}(2x)}{x}limx→0​xsin(2x)​. Use known results.

Answer: The limit is 222. Use sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x) and known limit sin⁡(x)/x=1\sin(x)/x = 1sin(x)/x=1.

Flashcard 10: Find limx→2x2−5x+6x−2\text{lim}_{x \to 2} \frac{x^2 - 5x + 6}{x - 2}limx→2​x−2x2−5x+6​. Use algebraic properties.

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

Flashcard 11: What is the limit property for a constant multiple: If limx→af(x)=L\text{If }\text{lim}_{x \to a} f(x) = LIf limx→a​f(x)=L, what is limx→a[cf(x)]\text{lim}_{x \to a} [cf(x)]limx→a​[cf(x)]?

Answer: The limit is cLcLcL. Constant multiple property: factor out constants.

Flashcard 12: Find limx→2x3−8x−2\text{lim}_{x \to 2} \frac{x^3 - 8}{x - 2}limx→2​x−2x3−8​. Use algebraic properties.

Answer: The limit is 121212. Factor difference of cubes: (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.

Flashcard 13: Find limx→1x2−1x−1\text{lim}_{x \to 1} \frac{x^2 - 1}{x - 1}limx→1​x−1x2−1​. Use algebraic properties.

Answer: The limit is 222. Factor difference of squares: (x+1)(x−1)x−1=x+1\frac{(x+1)(x-1)}{x-1} = x+1x−1(x+1)(x−1)​=x+1.

Flashcard 14: What is the limit of a constant function f(x)=cf(x) = cf(x)=c as xxx approaches any value aaa?

Answer: The limit is ccc. Constants remain unchanged regardless of the input value.

Flashcard 15: What is the algebraic property of limits for the absolute value function?

Answer: lim⁡x→a∣f(x)∣=∣lim⁡x→af(x)∣\lim_{x \to a} |f(x)| = |\lim_{x \to a} f(x)|limx→a​∣f(x)∣=∣limx→a​f(x)∣. Absolute value of the limit equals limit of absolute value.

Flashcard 16: Find limx→2x2−4x−2\text{lim}_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​. Use algebraic properties.

Answer: The limit is 444. Factor and cancel: (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.

Flashcard 17: Find limx→4x2−16x−4\text{lim}_{x \to 4} \frac{x^2 - 16}{x - 4}limx→4​x−4x2−16​. Use algebraic properties.

Answer: The limit is 888. Factor difference of squares: (x+4)(x−4)x−4=x+4\frac{(x+4)(x-4)}{x-4} = x+4x−4(x+4)(x−4)​=x+4.

Flashcard 18: Find limx→4x−4sqrt(x)−2\text{lim}_{x \to 4} \frac{x - 4}{\text{sqrt}(x) - 2}limx→4​sqrt(x)−2x−4​. Use algebraic properties.

Answer: The limit is 444. Rationalize by multiplying by x+2x+2\frac{\sqrt{x}+2}{\sqrt{x}+2}x​+2x​+2​.

Flashcard 19: If limx→af(x)=L\text{lim}_{x \to a} f(x) = Llimx→a​f(x)=L, what is limx→a(f(x)+c)\text{lim}_{x \to a} (f(x) + c)limx→a​(f(x)+c)?

Answer: The limit is L+cL + cL+c. Sum property: limit of sum equals sum of limits.

Flashcard 20: Find limx→1x3−1x−1\text{lim}_{x \to 1} \frac{x^3 - 1}{x - 1}limx→1​x−1x3−1​. Use algebraic properties.

Answer: The limit is 333. Factor difference of cubes: (x−1)(x2+x+1)x−1=x2+x+1\frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1x−1(x−1)(x2+x+1)​=x2+x+1.

Flashcard 21: Find limx→0xsin(x)\text{lim}_{x \to 0} \frac{x}{\text{sin}(x)}limx→0​sin(x)x​. Use known results.

Answer: The limit is 111. Reciprocal of the standard limit sin⁡(x)/x=1\sin(x)/x = 1sin(x)/x=1.

Flashcard 22: Find limx→0sin(3x)sin(x)\text{lim}_{x \to 0} \frac{\text{sin}(3x)}{\text{sin}(x)}limx→0​sin(x)sin(3x)​. Use algebraic properties.

Answer: The limit is 333. Rewrite as sin⁡(3x)3x⋅3xx⋅xsin⁡(x)=1⋅3⋅1\frac{\sin(3x)}{3x} \cdot \frac{3x}{x} \cdot \frac{x}{\sin(x)} = 1 \cdot 3 \cdot 13xsin(3x)​⋅x3x​⋅sin(x)x​=1⋅3⋅1.

Flashcard 23: What is the limit of a power function: If lim⁡x→af(x)=L\text{If }\lim_{x \to a} f(x) = LIf limx→a​f(x)=L, what is lim⁡x→a[f(x)]n\lim_{x \to a} [f(x)]^nlimx→a​[f(x)]n?

Answer: The limit is LnL^nLn. Power property: raise the limit to the same power.

Flashcard 24: Find limx→−2x2+4x+4x+2\text{lim}_{x \to -2} \frac{x^2 + 4x + 4}{x + 2}limx→−2​x+2x2+4x+4​. Use algebraic properties.

Answer: The limit is −4-4−4. Factor perfect square: (x+2)2x+2=x+2\frac{(x+2)^2}{x+2} = x+2x+2(x+2)2​=x+2, then substitute.

Flashcard 25: What is the limit property for a constant multiple: If limx→af(x)=L\text{If }\text{lim}_{x \to a} f(x) = LIf limx→a​f(x)=L, what is limx→a[cf(x)]\text{lim}_{x \to a} [cf(x)]limx→a​[cf(x)]?

Answer: The limit is cLcLcL. Constant multiple property: factor out constants.

Flashcard 26: Find limx→2x2−4x−2\text{lim}_{x \to 2} \frac{x^2 - 4}{x - 2}limx→2​x−2x2−4​. Use algebraic properties.

Answer: The limit is 444. Factor and cancel: (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.

Flashcard 27: What is the limit of a constant function f(x)=cf(x) = cf(x)=c as xxx approaches any value aaa?

Answer: The limit is ccc. Constants remain unchanged regardless of the input value.

Flashcard 28: What is the algebraic property of limits for the absolute value function?

Answer: lim⁡x→a∣f(x)∣=∣lim⁡x→af(x)∣\lim_{x \to a} |f(x)| = |\lim_{x \to a} f(x)|limx→a​∣f(x)∣=∣limx→a​f(x)∣. Absolute value of the limit equals limit of absolute value.

Flashcard 29: Find limx→4x−4sqrt(x)−2\text{lim}_{x \to 4} \frac{x - 4}{\text{sqrt}(x) - 2}limx→4​sqrt(x)−2x−4​. Use algebraic properties.

Answer: The limit is 444. Rationalize by multiplying by x+2x+2\frac{\sqrt{x}+2}{\sqrt{x}+2}x​+2x​+2​.

Flashcard 30: Find limx→1x2−1x2+x−2\text{lim}_{x \to 1} \frac{x^2 - 1}{x^2 + x - 2}limx→1​x2+x−2x2−1​. Use algebraic properties.

Answer: The limit is 000. Factor both: (x−1)(x+1)(x−1)(x+2)=x+1x+2\frac{(x-1)(x+1)}{(x-1)(x+2)} = \frac{x+1}{x+2}(x−1)(x+2)(x−1)(x+1)​=x+2x+1​, then substitute.