AP Calculus AB Flashcards: Confirming Continuity Over An Interval

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.

Flashcard 1: Find $\text{lim}_{x \to 2} (x^2 - 4)/(x - 2)$. Is it continuous at $x = 2$?

Answer: Limit is $4$, not continuous without $f(2)$ defined. Use L'Hôpital's rule or factor: limit is $4$, but function undefined.

Flashcard 2: If $f(x) = x^3$, is $f(x)$ continuous on $(-\text{\textinfty}, \text{\textinfty})$?

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

Flashcard 3: 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 4: Find the limit: $\lim_{x \to 3} (2x + 1)$.

Answer: $7$. Direct substitution works for polynomial functions.

Flashcard 5: What is the condition for a function to be continuous over an open interval $(a, b)$?

Answer: $f(x)$ is continuous at every point in $(a, b)$. No endpoint conditions needed for open intervals.

Flashcard 6: Find $\lim_{x \to 0} \frac{\sin(x)}{x}$

Answer: $1$. Standard trigonometric limit used in derivative of sine.

Flashcard 7: Identify the type of discontinuity if $\text{lim}_{x \to a} f(x)$ 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 8: If $f(x) = \frac{x^2 - 4}{x - 2}$, what value must $f(2)$ be for continuity at $x = 2$?

Answer: $4$. Factor and simplify: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $4$.

Flashcard 9: What is the definition of continuity at a point $x = a$?

Answer: A function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. The limit must exist and equal the function value at that point.

Flashcard 10: Is $h(x) = \frac{1}{x-1}$ continuous at $x = 1$?

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

Flashcard 11: What is the limit $\text{lim}_{x \to 2} 3x + 1$?

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

Flashcard 12: What is a point of discontinuity?

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

Flashcard 13: What kind of discontinuity occurs if $\text{lim}_{x \to a^-} f(x) \neq \text{lim}_{x \to a^+} f(x)$?

Answer: A jump discontinuity. One-sided limits exist but differ, creating a vertical 'jump'.

Flashcard 14: Find $\text{lim}_{x \to 2} (x^2 - 4)/(x - 2)$. Is it continuous at $x = 2$?

Answer: Limit is $4$, not continuous without $f(2)$ defined. Use L'Hôpital's rule or factor: limit is $4$, but function undefined.

Flashcard 15: What type of discontinuity is present if $f(x)$ is undefined at $x = a$?

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

Flashcard 16: What must be true for $f(x)$ to be continuous from the right at $x = a$?

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

Flashcard 17: If $f(x) = \frac{x^2 - 4}{x - 2}$, what value must $f(2)$ be for continuity at $x = 2$?

Answer: $4$. Factor and simplify: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $4$.

Flashcard 18: Identify the discontinuity: $\text{lim}_{x \to a} f(x)$ exists but $f(a)$ is not defined.

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

Flashcard 19: State the condition for $f(x)$ to be continuous at an endpoint $a$ of $[a, b]$.

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

Flashcard 20: Is $g(x) = \text{ln}(x)$ continuous for $x > 0$?

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

Flashcard 21: What must be true about $\text{lim}_{x \to a^-} f(x)$ and $\text{lim}_{x \to a^+} f(x)$ for continuity at $x = a$?

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

Flashcard 22: Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.

Answer: No, $f(x)$ has a removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, but $f(3)$ undefined.

Flashcard 23: Which theorem guarantees a root exists in $[a, b]$ if $f(a) \times f(b) < 0$?

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

Flashcard 24: What is the definition of a jump discontinuity?

Answer: When $\text{lim}_{x \to a^-} f(x) \neq \text{lim}_{x \to a^+} f(x)$. Left and right approaches yield different limit values.

Flashcard 25: If $\text{lim}_{x \to a} f(x) = L$ and $f(a) = L$, is $f(x)$ continuous at $x = a$?

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

Flashcard 26: What ensures a function is continuous on a closed interval $[a, b]$?

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

Flashcard 27: Determine if $f(x) = |x|$ is continuous at $x = 0$.

Answer: Yes, $f(x)$ is continuous at $x = 0$. Left and right limits both equal $0$, matching $f(0) = |0| = 0$.

Flashcard 28: What is the definition of continuity at a point $x = a$?

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

Flashcard 29: Which theorem guarantees a root exists in $[a, b]$ if $f(a) \times f(b) < 0$?

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

Flashcard 30: What must be true for $f(x)$ to be continuous from the right at $x = a$?

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

Flashcard 31: For $f(x) = \frac{x^3 - 8}{x - 2}$, does $f(x)$ have a removable discontinuity at $x = 2$?

Answer: Yes, $f(x)$ has a removable discontinuity. Factor: $\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$, limit exists.

Flashcard 32: Find $\text{lim}_{x \to 0} \frac{\text{sin}(x)}{x}$.

Answer: $1$. Standard trigonometric limit used in derivative of sine.

Flashcard 33: 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 34: What is the first step in confirming continuity at a point?

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

Flashcard 35: What ensures a function is continuous on a closed interval $[a, b]$?

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

Flashcard 36: What is the condition for continuity from the left at $x = a$?

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

Flashcard 37: If $f(x)$ is continuous on $[0, 1]$ and $f(0) = 3$, can $f(1) = 0$?

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

Flashcard 38: Which condition must be checked first to confirm continuity on an interval $[a, b]$?

Answer: Check if $f(x)$ is defined for all $x \text{ in } [a, b]$. Domain restrictions are the most common source of discontinuities.

Flashcard 39: For $g(x) = \frac{x^2 - 1}{x - 1}$, does $g(x)$ have a removable discontinuity at $x = 1$?

Answer: Yes, $g(x)$ has a removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, limit exists at $x = 1$.

Flashcard 40: What is the limit $\text{lim}_{x \to 2} 3x + 1$?

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

Flashcard 41: Is $h(x) = \frac{1}{x-1}$ continuous at $x = 1$?

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

Flashcard 42: For $g(x) = \frac{x^2 - 1}{x - 1}$, does $g(x)$ have a removable discontinuity at $x = 1$?

Answer: Yes, $g(x)$ has a removable discontinuity. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, limit exists at $x = 1$.

Flashcard 43: What type of discontinuity is present if $f(x)$ is undefined at $x = a$?

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

Flashcard 44: If $f(x) = x^3$, is $f(x)$ continuous on $(-\text{\textinfty}, \text{\textinfty})$?

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

Flashcard 45: What is the first step in confirming continuity at a point?

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

Flashcard 46: Is $f(x) = \frac{1}{x}$ continuous at $x = 0$? Why or why not?

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

Flashcard 47: Find the limit: $\text{lim}_{x \to 3} (2x + 1)$.

Answer: $7$. Direct substitution works for polynomial functions.

Flashcard 48: What is a point of discontinuity?

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

Flashcard 49: If $f(x)$ is continuous on $[0, 1]$ and $f(0) = 3$, can $f(1) = 0$?

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

Flashcard 50: Find the value of $c$ that makes $f(x) = cx + 1$ continuous at $x = 2$ if $f(2) = 5$.

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

Flashcard 51: What kind of discontinuity occurs if $\text{lim}_{x \to a^-} f(x) \neq \text{lim}_{x \to a^+} f(x)$?

Answer: A jump discontinuity. One-sided limits exist but differ, creating a vertical 'jump'.

Flashcard 52: What is the definition of a jump discontinuity?

Answer: When $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$. Left and right approaches yield different limit values.

Flashcard 53: State the condition for $f(x)$ to be continuous at an endpoint $a$ of $[a, b]$.

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

Flashcard 54: Is $f(x) = \frac{1}{x}$ continuous at $x = 0$? Why or why not?

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

Flashcard 55: State the Intermediate Value Theorem.

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

Flashcard 56: Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.

Answer: No, $f(x)$ has a removable discontinuity. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, but $f(3)$ undefined.

Flashcard 57: Is $g(x) = \text{ln}(x)$ continuous for $x > 0$?

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

Flashcard 58: Determine if $f(x) = |x|$ is continuous at $x = 0$.

Answer: Yes, $f(x)$ is continuous at $x = 0$. Left and right limits both equal $0$, matching $f(0) = |0| = 0$.

Flashcard 59: Identify the discontinuity: $\text{lim}_{x \to a} f(x)$ exists but $f(a)$ is not defined.

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

Flashcard 60: What is the condition for a function to be continuous over an open interval $(a, b)$?

Answer: $f(x)$ is continuous at every point in $(a, b)$. No endpoint conditions needed for open intervals.

Flashcard 61: Find the value of $c$ that makes $f(x) = cx + 1$ continuous at $x = 2$ if $f(2) = 5$.

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

Flashcard 62: If $\text{lim}_{x \to a} f(x) = L$ and $f(a) = L$, is $f(x)$ continuous at $x = a$?

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

Flashcard 63: State the Intermediate Value Theorem.

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

Flashcard 64: Which condition must be checked first to confirm continuity on an interval $[a, b]$?

Answer: Check if $f(x)$ is defined for all $x \text{ in } [a, b]$. Domain restrictions are the most common source of discontinuities.

Flashcard 65: What is the condition for continuity from the left at $x = a$?

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

Flashcard 66: What must be true about $\text{lim}_{x \to a^-} f(x)$ and $\text{lim}_{x \to a^+} f(x)$ for continuity at $x = a$?

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

Flashcard 67: For $f(x) = \frac{x^3 - 8}{x - 2}$, does $f(x)$ have a removable discontinuity at $x = 2$?

Answer: Yes, $f(x)$ has a removable discontinuity. Factor: $\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$, limit exists.

Flashcard 68: Identify the type of discontinuity if $\text{lim}_{x \to a} f(x)$ does not exist.

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