AP Calculus AB Flashcards: Connecting Infinite Limits And Vertical Asymptotes

What this deck covers

This deck focuses on Connecting Infinite Limits And Vertical Asymptotes, 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: What does $\text{lim}_{x \to a} f(x) = -\text{infinity}$ imply about $f(x)$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Negative infinite limit still indicates vertical asymptote.

Flashcard 2: Identify the asymptote of $f(x) = \frac{2}{x+3}$.

Answer: Vertical asymptote at $x = -3$. Denominator equals zero when $x + 3 = 0$.

Flashcard 3: What happens if $\text{lim}_{x \to a^-} f(x) = -\text{infinity}$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Left-sided negative limit confirms vertical asymptote.

Flashcard 4: Determine the vertical asymptote in $g(x) = \frac{x^3}{x^2 - x - 6}$.

Answer: Vertical asymptotes at $x = 3$ and $x = -2$. Factor $x^2 - x - 6 = (x-3)(x+2) = 0$.

Flashcard 5: What does it mean if $\text{lim}_{x \to a^+} f(x) = \text{infinity}$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Right-sided infinite limit confirms vertical asymptote exists.

Flashcard 6: What does $\text{lim}_{x \to a^+} f(x) = -\text{infinity}$ tell us?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Right-sided negative limit confirms vertical asymptote.

Flashcard 7: Find the vertical asymptotes in $g(x) = \frac{x^2}{x^2 + 4x + 4}$.

Answer: Vertical asymptote at $x = -2$. Denominator $(x+2)^2 = 0$ only when $x = -2$.

Flashcard 8: What does it mean if $\text{lim}_{x \to a^-} f(x) = \text{infinity}$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Left-sided infinite limit confirms vertical asymptote exists.

Flashcard 9: What does $\text{lim}_{x \to a} f(x) = \text{infinity}$ imply for $f(x)$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Two-sided infinite limit confirms vertical asymptote.

Flashcard 10: Determine the vertical asymptote of $f(x) = \frac{1}{(x-1)(x+2)}$.

Answer: Vertical asymptotes at $x = 1$ and $x = -2$. Each factor in denominator creates a separate asymptote.

Flashcard 11: What does $\text{lim}_{x \to a^+} f(x) = -\text{infinity}$ indicate?

Answer: $f(x)$ has a vertical asymptote at $x = a$. Right-sided negative infinite limit confirms asymptote.

Flashcard 12: Identify the vertical asymptotes of $f(x) = \frac{2}{x^2 - 9x + 18}$.

Answer: Vertical asymptotes at $x = 3$ and $x = 6$. Factor $x^2 - 9x + 18 = (x-3)(x-6) = 0$.

Flashcard 13: Find the vertical asymptote of $g(x) = \frac{1}{x^2 - 1}$.

Answer: Vertical asymptotes at $x = 1$ and $x = -1$. Factor denominator: $x^2 - 1 = (x-1)(x+1) = 0$.

Flashcard 14: Find the vertical asymptotes of $f(x) = \frac{x}{x^2 - 9}$.

Answer: Vertical asymptotes at $x = 3$ and $x = -3$. Factor denominator: $x^2 - 9 = (x-3)(x+3) = 0$.

Flashcard 15: Find the vertical asymptote of $f(x) = \frac{x}{x^2 - x - 2}$.

Answer: Vertical asymptotes at $x = 2$ and $x = -1$. Factor $x^2 - x - 2 = (x-2)(x+1) = 0$.

Flashcard 16: Determine the asymptote of $f(x) = \frac{3x}{x^2 - 2x}$.

Answer: Vertical asymptote at $x = 0$ and $x = 2$. Factor out $x$: denominator $x(x-2) = 0$.

Flashcard 17: Find the vertical asymptotes of $f(x) = \frac{x}{x^2 - 16}$.

Answer: Vertical asymptotes at $x = 4$ and $x = -4$. Factor denominator: $x^2 - 16 = (x-4)(x+4) = 0$.

Flashcard 18: What does $\text{lim}_{x \to a^+} f(x) = \text{infinity}$ tell about $f(x)$?

Answer: $f(x)$ approaches infinity at $x = a$ from the right. Describes function behavior approaching from the right.

Flashcard 19: Identify the vertical asymptotes of $f(x) = \frac{2x}{x^2 - 4}$.

Answer: Vertical asymptotes at $x = 2$ and $x = -2$. Factor denominator: $x^2 - 4 = (x-2)(x+2) = 0$.

Flashcard 20: Identify the asymptotes of $f(x) = \frac{2x}{x^2 - 4x + 4}$.

Answer: Vertical asymptote at $x = 2$. Denominator $x^2 - 4x + 4 = (x-2)^2 = 0$ at $x = 2$.

Flashcard 21: Find the vertical asymptotes of $f(x) = \frac{3x}{x^2 + 2x - 3}$.

Answer: Vertical asymptotes at $x = 1$ and $x = -3$. Factor $x^2 + 2x - 3 = (x-1)(x+3) = 0$.

Flashcard 22: Identify the vertical asymptotes of $f(x) = \frac{x^2}{x^2 - 4}$.

Answer: Vertical asymptotes at $x = 2$ and $x = -2$. Factor denominator: $(x-2)(x+2) = 0$ gives both roots.

Flashcard 23: What is the behavior of $f(x)$ if $\text{lim}_{x \to a^-} f(x) = \text{infinity}$?

Answer: $f(x)$ approaches infinity at $x = a$ from the left. Left-sided limit describes approach from negative direction.

Flashcard 24: What does $\text{lim}_{x \to a} f(x) = \text{infinity}$ imply about $f(x)$?

Answer: $f(x)$ has a vertical asymptote at $x = a$. An infinite limit indicates unbounded growth near that point.

Flashcard 25: Determine the vertical asymptotes of $f(x) = \frac{x^2 + 2}{x^2 - 4x + 4}$.

Answer: Vertical asymptote at $x = 2$. Denominator $(x-2)^2 = 0$ only when $x = 2$.

Flashcard 26: Calculate the vertical asymptotes of $f(x) = \frac{x^2 + 1}{x - 5}$.

Answer: Vertical asymptote at $x = 5$. Denominator equals zero only when $x - 5 = 0$.

Flashcard 27: What is the behavior of $f(x)$ near a vertical asymptote?

Answer: $f(x)$ approaches infinity or negative infinity. Function values grow without bound near the asymptote.

Flashcard 28: What is the condition for a vertical asymptote at $x = a$ in $f(x) = \frac{p(x)}{q(x)}$?

Answer: $q(a) = 0$ and $p(a) \neq 0$. Denominator zero with non-zero numerator creates asymptote.

Flashcard 29: Identify the asymptotes of $f(x) = \frac{1}{x^2 - 4x + 3}$.

Answer: Vertical asymptotes at $x = 1$ and $x = 3$. Factor $x^2 - 4x + 3 = (x-1)(x-3) = 0$.

Flashcard 30: Identify the vertical asymptote in $f(x) = \frac{1}{x-2}$.

Answer: Vertical asymptote at $x = 2$. Denominator equals zero when $x - 2 = 0$.