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  2. AP Calculus AB
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AP Calculus AB Flashcards: Confirming Continuity Over An Interval

Study Confirming Continuity Over An Interval 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 Confirming Continuity Over An Interval, 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: Confirming Continuity Over An Interval

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QUESTION

Find the limit: limx→3(2x+1)\text{lim}_{x \to 3} (2x + 1)limx→3​(2x+1).

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ANSWER

777. Direct substitution works for polynomial functions.

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Flashcard 1: Find the limit: limx→3(2x+1)\text{lim}_{x \to 3} (2x + 1)limx→3​(2x+1).

Answer: 777. Direct substitution works for polynomial functions.

Flashcard 2: Find limx→2(x2−4)/(x−2)\text{lim}_{x \to 2} (x^2 - 4)/(x - 2)limx→2​(x2−4)/(x−2). Is it continuous at x=2x = 2x=2?

Answer: Limit is 444, not continuous without f(2)f(2)f(2) defined. Use L'Hôpital's rule or factor: limit is 444, but function undefined.

Flashcard 3: Is f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ continuous at x=0x = 0x=0? Why or why not?

Answer: No, f(x)f(x)f(x) is undefined at x=0x = 0x=0. Division by zero makes the function undefined at that point.

Flashcard 4: Find the value of ccc that makes f(x)=cx+1f(x) = cx + 1f(x)=cx+1 continuous at x=2x = 2x=2 if f(2)=5f(2) = 5f(2)=5.

Answer: c=2c = 2c=2. Set 2c+1=52c + 1 = 52c+1=5 and solve: c=2c = 2c=2.

Flashcard 5: What is the definition of continuity at a point x=ax = ax=a?

Answer: A function f(x)f(x)f(x) is continuous at x=ax = ax=a if limx→af(x)=f(a)\text{lim}_{x \to a} f(x) = f(a)limx→a​f(x)=f(a). The limit must exist and equal the function value at that point.

Flashcard 6: Is h(x)=1x−1h(x) = \frac{1}{x-1}h(x)=x−11​ continuous at x=1x = 1x=1?

Answer: No, h(x)h(x)h(x) is undefined at x=1x = 1x=1. Vertical asymptote creates an infinite discontinuity.

Flashcard 7: State the Intermediate Value Theorem.

Answer: If f(x)f(x)f(x) is continuous on [a,b][a, b][a,b], f(a)<N<f(b)f(a) < N < f(b)f(a)<N<f(b) then exists c such that f(c)=N\text{exists } c \text{ such that } f(c) = Nexists c such that f(c)=N. Guarantees all intermediate values exist for continuous functions on closed intervals.

Flashcard 8: What is the first step in confirming continuity at a point?

Answer: Verify f(a)f(a)f(a) is defined. Cannot check continuity if the function value doesn't exist.

Flashcard 9: What is the definition of a jump discontinuity?

Answer: When lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)limx→a−​f(x)=limx→a+​f(x). Left and right approaches yield different limit values.

Flashcard 10: Determine if f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​ is continuous at x=3x = 3x=3.

Answer: No, f(x)f(x)f(x) has a removable discontinuity. 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, but f(3)f(3)f(3) undefined.

Flashcard 11: What must be true for f(x)f(x)f(x) to be continuous from the right at x=ax = ax=a?

Answer: limx→a+f(x)=f(a)\text{lim}_{x \to a^+} f(x) = f(a)limx→a+​f(x)=f(a). Right-hand limit must equal the function value.

Flashcard 12: What is a point of discontinuity?

Answer: A point where f(x)f(x)f(x) is not continuous. Where any of the three continuity conditions fail.

Flashcard 13: If f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​, what value must f(2)f(2)f(2) be for continuity at x=2x = 2x=2?

Answer: 444. Factor and simplify: (x+2)(x−2)x−2=x+2\frac{(x+2)(x-2)}{x-2} = x+2x−2(x+2)(x−2)​=x+2, so limit is 444.

Flashcard 14: What must be true about limx→a−f(x)\text{lim}_{x \to a^-} f(x)limx→a−​f(x) and limx→a+f(x)\text{lim}_{x \to a^+} f(x)limx→a+​f(x) for continuity at x=ax = ax=a?

Answer: They must both exist and be equal to f(a)f(a)f(a). Left and right limits must converge to the same value as the function.

Flashcard 15: Is g(x)=ln(x)g(x) = \text{ln}(x)g(x)=ln(x) continuous for x>0x > 0x>0?

Answer: Yes, g(x)g(x)g(x) is continuous for x>0x > 0x>0. Natural logarithm is continuous on its entire domain.

Flashcard 16: What is the limit limx→23x+1\text{lim}_{x \to 2} 3x + 1limx→2​3x+1?

Answer: 777. Direct substitution: 3(2)+1=73(2) + 1 = 73(2)+1=7.

Flashcard 17: Identify the type of discontinuity if limx→af(x)\text{lim}_{x \to a} f(x)limx→a​f(x) does not exist.

Answer: This is an infinite or jump discontinuity. When limits don't exist, the function has a break or approaches infinity.

Flashcard 18: If f(x)f(x)f(x) is continuous on [0,1][0, 1][0,1] and f(0)=3f(0) = 3f(0)=3, can f(1)=0f(1) = 0f(1)=0?

Answer: Yes, if f(x)f(x)f(x) decreases continuously. IVT guarantees all values between f(0)f(0)f(0) and f(1)f(1)f(1) are achieved.

Flashcard 19: If limx→af(x)=L\text{lim}_{x \to a} f(x) = Llimx→a​f(x)=L and f(a)=Lf(a) = Lf(a)=L, is f(x)f(x)f(x) continuous at x=ax = ax=a?

Answer: Yes, f(x)f(x)f(x) is continuous at x=ax = ax=a. All three conditions for continuity are satisfied.

Flashcard 20: Find limx→0sin(x)x\text{lim}_{x \to 0} \frac{\text{sin}(x)}{x}limx→0​xsin(x)​.

Answer: 111. Standard trigonometric limit used in derivative of sine.

Flashcard 21: For f(x)=x3−8x−2f(x) = \frac{x^3 - 8}{x - 2}f(x)=x−2x3−8​, does f(x)f(x)f(x) have a removable discontinuity at x=2x = 2x=2?

Answer: Yes, f(x)f(x)f(x) has a removable discontinuity. 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, limit exists.

Flashcard 22: What type of discontinuity is present if f(x)f(x)f(x) is undefined at x=ax = ax=a?

Answer: It could be a removable or infinite discontinuity. Depends on whether the function approaches infinity or has a finite limit.

Flashcard 23: State the condition for f(x)f(x)f(x) to be continuous at an endpoint aaa of [a,b][a, b][a,b].

Answer: limx→a+f(x)=f(a)\text{lim}_{x \to a^+} f(x) = f(a)limx→a+​f(x)=f(a). Only right-hand continuity needed at left endpoint.

Flashcard 24: Which theorem guarantees a root exists in [a,b][a, b][a,b] if f(a)×f(b)<0f(a) \times f(b) < 0f(a)×f(b)<0?

Answer: The Intermediate Value Theorem. Sign change guarantees a zero by IVT for continuous functions.

Flashcard 25: What is the condition for continuity from the left at x=ax = ax=a?

Answer: limx→a−f(x)=f(a)\text{lim}_{x \to a^-} f(x) = f(a)limx→a−​f(x)=f(a). Left-hand limit must equal the function value.

Flashcard 26: For g(x)=x2−1x−1g(x) = \frac{x^2 - 1}{x - 1}g(x)=x−1x2−1​, does g(x)g(x)g(x) have a removable discontinuity at x=1x = 1x=1?

Answer: Yes, g(x)g(x)g(x) has a removable discontinuity. 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, limit exists at x=1x = 1x=1.

Flashcard 27: Identify the discontinuity: limx→af(x)\text{lim}_{x \to a} f(x)limx→a​f(x) exists but f(a)f(a)f(a) is not defined.

Answer: Removable discontinuity. The 'hole' can be filled by defining the function at that point.

Flashcard 28: If f(x)=x3f(x) = x^3f(x)=x3, is f(x)f(x)f(x) continuous on (−\textinfty,\textinfty)(-\text{\textinfty}, \text{\textinfty})(−\textinfty,\textinfty)?

Answer: Yes, f(x)f(x)f(x) is continuous everywhere. Polynomial functions are continuous everywhere in their domain.

Flashcard 29: What is a removable discontinuity?

Answer: A point where a function is not defined, but the limit exists. Can be 'fixed' by defining or redefining the function at that point.

Flashcard 30: What ensures a function is continuous on a closed interval [a,b][a, b][a,b]?

Answer: Continuous on (a,b)(a, b)(a,b), right at aaa, left at bbb. Requires continuity at all interior points plus endpoint conditions.