AP Calculus AB Flashcards: Determining Limits Using The Squeeze Theorem

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

AP Calculus AB

Determining Limits Using The Squeeze Theorem

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QUESTION
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Determine limx0x5sin(1x6)\lim_{x \to 0} x^5 \sin(\frac{1}{x^6}) using the Squeeze Theorem.

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ANSWER

00. Since x5x5sin(1x6)x5-x^5 \leq x^5 \sin(\frac{1}{x^6}) \leq x^5 and both bounds approach 00.

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This deck focuses on Determining Limits Using The Squeeze Theorem, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

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Flashcard 1: Determine limx0x5sin(1x6)\lim_{x \to 0} x^5 \sin(\frac{1}{x^6}) using the Squeeze Theorem.

Answer: 00. Since x5x5sin(1x6)x5-x^5 \leq x^5 \sin(\frac{1}{x^6}) \leq x^5 and both bounds approach 00.

Flashcard 2: What is the limit limx0x2sin(x2)\lim_{x \to 0} x^2 \sin(x^2)?

Answer: 00. Since x2sin(x2)x2|x^2 \sin(x^2)| \leq x^2 and x20x^2 \to 0.

Flashcard 3: Find limx0x2cos(1x)\lim_{x \to 0} x^2 \cos(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2cos(1x)x2-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2 and both bounds approach 00.

Flashcard 4: Find limx0x2sin(1x)\lim_{x \to 0} x^2 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2sin(1x)x2-|x^2| \leq x^2 \sin(\frac{1}{x}) \leq |x^2| and both bounds approach 00.

Flashcard 5: Determine limx0x3cos(1x2)\lim_{x \to 0} x^3 \cos(\frac{1}{x^2}) using the Squeeze Theorem.

Answer: 00. Since x3x3cos(1x2)x3-x^3 \leq x^3 \cos(\frac{1}{x^2}) \leq x^3 and both bounds approach 00.

Flashcard 6: Determine limx0x4sin(1x)\lim_{x \to 0} x^4 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x4x4sin(1x)x4-x^4 \leq x^4 \sin(\frac{1}{x}) \leq x^4 and both bounds approach 00.

Flashcard 7: What is the limit of xsin(1x)x \sin(\frac{1}{x}) as x0x \to 0?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 8: What should g(x)g(x) and h(x)h(x) approach for limxcf(x)\lim_{x \to c}f(x) to exist?

Answer: The same limit LL. This ensures the squeezed function approaches the unique limit.

Flashcard 9: What is limx0xsin(2x)\lim_{x \to 0} x \sin(\frac{2}{x})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 10: Determine limx0x5sin(1x6)\lim_{x \to 0} x^5 \sin(\frac{1}{x^6}) using the Squeeze Theorem.

Answer: 00. Since x5x5sin(1x6)x5-x^5 \leq x^5 \sin(\frac{1}{x^6}) \leq x^5 and both bounds approach 00.

Flashcard 11: What is limx0xcos(1x3)\lim_{x \to 0} x \cos(\frac{1}{x^3})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 12: What is limx0xsin(2x)\lim_{x \to 0} x \sin(\frac{2}{x})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 13: What must the limits of g(x)g(x) and h(x)h(x) equal for the Squeeze Theorem to apply?

Answer: Both must equal LL. Required for the theorem to conclude f(x)f(x) has the same limit.

Flashcard 14: What is limx0x3sin(x2)\lim_{x \to 0} x^3 \sin(x^2) using the Squeeze Theorem?

Answer: 00. Since x3sin(x2)x3|x^3 \sin(x^2)| \leq x^3 and x30x^3 \to 0.

Flashcard 15: What is limx0x2sinx\lim_{x \to 0} x^2 \sin x?

Answer: 00. Since x2sinxx2|x^2 \sin x| \leq x^2 and x20x^2 \to 0.

Flashcard 16: What must the limits of g(x)g(x) and h(x)h(x) equal for the Squeeze Theorem to apply?

Answer: Both must equal LL. Required for the theorem to conclude f(x)f(x) has the same limit.

Flashcard 17: Find limx0x3sin(1x5)\lim_{x \to 0} x^3 \sin(\frac{1}{x^5}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(1x5)x3-x^3 \leq x^3 \sin(\frac{1}{x^5}) \leq x^3 and both bounds approach 00.

Flashcard 18: What inequality must f(x)f(x) satisfy in the Squeeze Theorem?

Answer: g(x)f(x)h(x)g(x) \leq f(x) \leq h(x). This defines the sandwich relationship between the three functions.

Flashcard 19: Find limx0x2cosx\lim_{x \to 0} x^2 \cos x using the Squeeze Theorem.

Answer: 00. Since x2cosxx2|x^2 \cos x| \leq x^2 and x20x^2 \to 0.

Flashcard 20: What is the Squeeze Theorem used for in calculus?

Answer: To determine the limit of a function. Forces a function between two bounds that approach the same limit.

Flashcard 21: What is limx0x4cos(1x5)\lim_{x \to 0} x^4 \cos(\frac{1}{x^5})?

Answer: 00. Since x4x4cos(1x5)x4-x^4 \leq x^4 \cos(\frac{1}{x^5}) \leq x^4 and both bounds approach 00.

Flashcard 22: Is limx0x2sin(1x)\lim_{x \to 0} x^2 \sin(\frac{1}{x}) finite?

Answer: Yes, it is 00. The limit equals 00, which is a finite value.

Flashcard 23: What is the limit of x2sin(1x2)x^2 \sin(\frac{1}{x^2}) as x0x \to 0?

Answer: 00. Bounded between x2-x^2 and x2x^2, both approaching 00.

Flashcard 24: What should g(x)g(x) and h(x)h(x) approach for limxcf(x)\lim_{x \to c}f(x) to exist?

Answer: The same limit LL. This ensures the squeezed function approaches the unique limit.

Flashcard 25: What is limx0x3sin(x2)\lim_{x \to 0} x^3 \sin(x^2) using the Squeeze Theorem?

Answer: 00. Since x3sin(x2)x3|x^3 \sin(x^2)| \leq x^3 and x30x^3 \to 0.

Flashcard 26: What is limx0xcos(1x3)\lim_{x \to 0} x \cos(\frac{1}{x^3})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 27: Is limx0x2sin(x3)\lim_{x \to 0} x^2 \sin(x^3) zero?

Answer: Yes. Since x2sin(x3)x2|x^2 \sin(x^3)| \leq x^2 and x20x^2 \to 0.

Flashcard 28: Find limx0x3sin(1x5)\lim_{x \to 0} x^3 \sin(\frac{1}{x^5}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(1x5)x3-x^3 \leq x^3 \sin(\frac{1}{x^5}) \leq x^3 and both bounds approach 00.

Flashcard 29: What is the limit limx0x2sin(1x4)\lim_{x \to 0} x^2 \sin(\frac{1}{x^4})?

Answer: 00. Since x2x2sin(1x4)x2-x^2 \leq x^2 \sin(\frac{1}{x^4}) \leq x^2 and both bounds approach 00.

Flashcard 30: Determine limx0x4cos(1x)\lim_{x \to 0} x^4 \cos(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x4x4cos(1x)x4-x^4 \leq x^4 \cos(\frac{1}{x}) \leq x^4 and both bounds approach 00.

Flashcard 31: Find limx0x2sin(1x3)\lim_{x \to 0} x^2 \sin(\frac{1}{x^3}) using the Squeeze Theorem.

Answer: 00. Since x2x2sin(1x3)x2-x^2 \leq x^2 \sin(\frac{1}{x^3}) \leq x^2 and both bounds approach 00.

Flashcard 32: Is limx0xsinx\lim_{x \to 0} x \sin x equal to zero?

Answer: Yes. Since xsinxx|x \sin x| \leq |x| and x0|x| \to 0.

Flashcard 33: Determine limx0x3cos(1x3)\lim_{x \to 0} x^3 \cos(\frac{1}{x^3}) using the Squeeze Theorem.

Answer: 00. Since x3x3cos(1x3)x3-x^3 \leq x^3 \cos(\frac{1}{x^3}) \leq x^3 and both bounds approach 00.

Flashcard 34: Determine limx0x2cos(x2)\lim_{x \to 0} x^2 \cos(x^2) using the Squeeze Theorem.

Answer: 00. Since x2cos(x2)x2|x^2 \cos(x^2)| \leq x^2 and x20x^2 \to 0.

Flashcard 35: What is the limit of xsin(1x)x \sin(\frac{1}{x}) as x0x \to 0?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 36: What is limx0x2sinx\lim_{x \to 0} x^2 \sin x?

Answer: 00. Since x2sinxx2|x^2 \sin x| \leq x^2 and x20x^2 \to 0.

Flashcard 37: Is limx0x2sin(1x)\lim_{x \to 0} x^2 \sin(\frac{1}{x}) finite?

Answer: Yes, it is 00. The limit equals 00, which is a finite value.

Flashcard 38: Determine limx0x2cos(x2)\lim_{x \to 0} x^2 \cos(x^2) using the Squeeze Theorem.

Answer: 00. Since x2cos(x2)x2|x^2 \cos(x^2)| \leq x^2 and x20x^2 \to 0.

Flashcard 39: What is the limit limx0x2sin(1x4)\lim_{x \to 0} x^2 \sin(\frac{1}{x^4})?

Answer: 00. Since x2x2sin(1x4)x2-x^2 \leq x^2 \sin(\frac{1}{x^4}) \leq x^2 and both bounds approach 00.

Flashcard 40: Find limxsinxx\lim_{x \to \infty} \frac{\sin x}{x} using the Squeeze Theorem.

Answer: 00. Since 1xsinxx1x-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} and both bounds approach 00.

Flashcard 41: Find limx0x3sin(1x)\lim_{x \to 0} x^3 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(1x)x3-|x^3| \leq x^3 \sin(\frac{1}{x}) \leq |x^3| and both bounds approach 00.

Flashcard 42: Determine limx0x3cos(1x2)\lim_{x \to 0} x^3 \cos(\frac{1}{x^2}) using the Squeeze Theorem.

Answer: 00. Since x3x3cos(1x2)x3-x^3 \leq x^3 \cos(\frac{1}{x^2}) \leq x^3 and both bounds approach 00.

Flashcard 43: Is limx0xsinx\lim_{x \to 0} x \sin x equal to zero?

Answer: Yes. Since xsinxx|x \sin x| \leq |x| and x0|x| \to 0.

Flashcard 44: Find limx0x5sin(1x4)\lim_{x \to 0} x^5 \sin(\frac{1}{x^4}) using the Squeeze Theorem.

Answer: 00. Since x5x5sin(1x4)x5-x^5 \leq x^5 \sin(\frac{1}{x^4}) \leq x^5 and both bounds approach 00.

Flashcard 45: Find limx0x3sin(1x)\lim_{x \to 0} x^3 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(1x)x3-|x^3| \leq x^3 \sin(\frac{1}{x}) \leq |x^3| and both bounds approach 00.

Flashcard 46: Find limx0x2cosx\lim_{x \to 0} x^2 \cos x using the Squeeze Theorem.

Answer: 00. Since x2cosxx2|x^2 \cos x| \leq x^2 and x20x^2 \to 0.

Flashcard 47: State the Squeeze Theorem in terms of f(x)f(x), g(x)g(x), h(x)h(x).

Answer: If g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) and limxcg(x)=limxch(x)=L\lim_{x \to c}g(x) = \lim_{x \to c}h(x) = L, then limxcf(x)=L\lim_{x \to c}f(x) = L. The fundamental statement of the Squeeze Theorem with three functions.

Flashcard 48: Find limx0x2sin(1x)\lim_{x \to 0} x^2 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2sin(1x)x2-|x^2| \leq x^2 \sin(\frac{1}{x}) \leq |x^2| and both bounds approach 00.

Flashcard 49: Determine limx0x3cos(x4)\lim_{x \to 0} x^3 \cos(x^4) using the Squeeze Theorem.

Answer: 00. Since x3cos(x4)x3|x^3 \cos(x^4)| \leq x^3 and x30x^3 \to 0.

Flashcard 50: Find limx0x2cos(2x)\lim_{x \to 0} x^2 \cos(\frac{2}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2cos(2x)x2-x^2 \leq x^2 \cos(\frac{2}{x}) \leq x^2 and both bounds approach 00.

Flashcard 51: Find limx0xsin(1x2)\lim_{x \to 0} x \sin(\frac{1}{x^2}) using the Squeeze Theorem.

Answer: 00. Since xxsin(1x2)x-|x| \leq x \sin(\frac{1}{x^2}) \leq |x| and both bounds approach 00.

Flashcard 52: Identify the conditions necessary for the Squeeze Theorem to apply.

Answer: Functions g(x)g(x), f(x)f(x), h(x)h(x) must satisfy g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) near cc. The inequality must hold in a neighborhood around cc.

Flashcard 53: Find limx0x3sin(2x)\lim_{x \to 0} x^3 \sin(\frac{2}{x}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(2x)x3-x^3 \leq x^3 \sin(\frac{2}{x}) \leq x^3 and both bounds approach 00.

Flashcard 54: Find limxsinxx\lim_{x \to \infty} \frac{\sin x}{x} using the Squeeze Theorem.

Answer: 00. Since 1xsinxx1x-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} and both bounds approach 00.

Flashcard 55: What inequality must f(x)f(x) satisfy in the Squeeze Theorem?

Answer: g(x)f(x)h(x)g(x) \leq f(x) \leq h(x). This defines the sandwich relationship between the three functions.

Flashcard 56: Determine limx0x4sin(1x)\lim_{x \to 0} x^4 \sin(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x4x4sin(1x)x4-x^4 \leq x^4 \sin(\frac{1}{x}) \leq x^4 and both bounds approach 00.

Flashcard 57: Can the Squeeze Theorem be used if limxcg(x)limxch(x)\lim_{x \to c}g(x) \neq \lim_{x \to c}h(x)?

Answer: No. The bounding functions must approach the same value for the theorem to work.

Flashcard 58: Determine limx0x3cos(1x3)\lim_{x \to 0} x^3 \cos(\frac{1}{x^3}) using the Squeeze Theorem.

Answer: 00. Since x3x3cos(1x3)x3-x^3 \leq x^3 \cos(\frac{1}{x^3}) \leq x^3 and both bounds approach 00.

Flashcard 59: What is limx0x4sin(1x3)\lim_{x \to 0} x^4 \sin(\frac{1}{x^3})?

Answer: 00. Since x4x4sin(1x3)x4-x^4 \leq x^4 \sin(\frac{1}{x^3}) \leq x^4 and both bounds approach 00.

Flashcard 60: Determine limx0x3cos(x4)\lim_{x \to 0} x^3 \cos(x^4) using the Squeeze Theorem.

Answer: 00. Since x3cos(x4)x3|x^3 \cos(x^4)| \leq x^3 and x30x^3 \to 0.

Flashcard 61: What is the result of limx0xcos(1x)\lim_{x \to 0} x \cos(\frac{1}{x})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 62: What is the Squeeze Theorem used for in calculus?

Answer: To determine the limit of a function. Forces a function between two bounds that approach the same limit.

Flashcard 63: Find limx0x2sin(1x3)\lim_{x \to 0} x^2 \sin(\frac{1}{x^3}) using the Squeeze Theorem.

Answer: 00. Since x2x2sin(1x3)x2-x^2 \leq x^2 \sin(\frac{1}{x^3}) \leq x^2 and both bounds approach 00.

Flashcard 64: Find limx0x2cos(1x)\lim_{x \to 0} x^2 \cos(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2cos(1x)x2-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2 and both bounds approach 00.

Flashcard 65: What is the result of limx0xcos(1x)\lim_{x \to 0} x \cos(\frac{1}{x})?

Answer: 00. Bounded between x-|x| and x|x|, both approaching 00.

Flashcard 66: Find limx0xsin(1x2)\lim_{x \to 0} x \sin(\frac{1}{x^2}) using the Squeeze Theorem.

Answer: 00. Since xxsin(1x2)x-|x| \leq x \sin(\frac{1}{x^2}) \leq |x| and both bounds approach 00.

Flashcard 67: Find limx0x3sin(2x)\lim_{x \to 0} x^3 \sin(\frac{2}{x}) using the Squeeze Theorem.

Answer: 00. Since x3x3sin(2x)x3-x^3 \leq x^3 \sin(\frac{2}{x}) \leq x^3 and both bounds approach 00.

Flashcard 68: What is the limit limx0x2sin(x2)\lim_{x \to 0} x^2 \sin(x^2)?

Answer: 00. Since x2sin(x2)x2|x^2 \sin(x^2)| \leq x^2 and x20x^2 \to 0.

Flashcard 69: Determine limx0x4cos(1x)\lim_{x \to 0} x^4 \cos(\frac{1}{x}) using the Squeeze Theorem.

Answer: 00. Since x4x4cos(1x)x4-x^4 \leq x^4 \cos(\frac{1}{x}) \leq x^4 and both bounds approach 00.

Flashcard 70: State the Squeeze Theorem in terms of f(x)f(x), g(x)g(x), h(x)h(x).

Answer: If g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) and limxcg(x)=limxch(x)=L\lim_{x \to c}g(x) = \lim_{x \to c}h(x) = L, then limxcf(x)=L\lim_{x \to c}f(x) = L. The fundamental statement of the Squeeze Theorem with three functions.

Flashcard 71: Find limx0x2cos(2x)\lim_{x \to 0} x^2 \cos(\frac{2}{x}) using the Squeeze Theorem.

Answer: 00. Since x2x2cos(2x)x2-x^2 \leq x^2 \cos(\frac{2}{x}) \leq x^2 and both bounds approach 00.

Flashcard 72: Find limx0x5sin(1x4)\lim_{x \to 0} x^5 \sin(\frac{1}{x^4}) using the Squeeze Theorem.

Answer: 00. Since x5x5sin(1x4)x5-x^5 \leq x^5 \sin(\frac{1}{x^4}) \leq x^5 and both bounds approach 00.

Flashcard 73: What is limx0x4sin(1x3)\lim_{x \to 0} x^4 \sin(\frac{1}{x^3})?

Answer: 00. Since x4x4sin(1x3)x4-x^4 \leq x^4 \sin(\frac{1}{x^3}) \leq x^4 and both bounds approach 00.

Flashcard 74: Can the Squeeze Theorem be used if limxcg(x)limxch(x)\lim_{x \to c}g(x) \neq \lim_{x \to c}h(x)?

Answer: No. The bounding functions must approach the same value for the theorem to work.

Flashcard 75: What is limx0x4cos(1x5)\lim_{x \to 0} x^4 \cos(\frac{1}{x^5})?

Answer: 00. Since x4x4cos(1x5)x4-x^4 \leq x^4 \cos(\frac{1}{x^5}) \leq x^4 and both bounds approach 00.

Flashcard 76: What is the limit of x2sin(1x2)x^2 \sin(\frac{1}{x^2}) as x0x \to 0?

Answer: 00. Bounded between x2-x^2 and x2x^2, both approaching 00.

Flashcard 77: Identify the conditions necessary for the Squeeze Theorem to apply.

Answer: Functions g(x)g(x), f(x)f(x), h(x)h(x) must satisfy g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) near cc. The inequality must hold in a neighborhood around cc.