Connecting Infinite Limits and Vertical Asymptotes - AP Calculus BC
Card 1 of 30
Identify the vertical asymptote for $f(x) = \frac{1}{x-3}$.
Identify the vertical asymptote for $f(x) = \frac{1}{x-3}$.
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$x = 3$. Set denominator $x-3=0$ to find where function is undefined.
$x = 3$. Set denominator $x-3=0$ to find where function is undefined.
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What is the limit of $f(x) = \frac{1}{x}$ as $x \to 0^+$?
What is the limit of $f(x) = \frac{1}{x}$ as $x \to 0^+$?
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$+\text{∞}$. As $x$ approaches 0 from right, $\frac{1}{x}$ grows positively.
$+\text{∞}$. As $x$ approaches 0 from right, $\frac{1}{x}$ grows positively.
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Determine the vertical asymptote for $f(x) = \frac{x+2}{x^2-4x+4}$.
Determine the vertical asymptote for $f(x) = \frac{x+2}{x^2-4x+4}$.
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$x = 2$. Denominator $x^2-4x+4=(x-2)^2=0$ only at $x=2$.
$x = 2$. Denominator $x^2-4x+4=(x-2)^2=0$ only at $x=2$.
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What is the behavior of $f(x)$ near a vertical asymptote at $x = a$?
What is the behavior of $f(x)$ near a vertical asymptote at $x = a$?
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$f(x) \to \text{±}\text{∞}$ as $x \to a$. Function values become infinitely large near asymptotes.
$f(x) \to \text{±}\text{∞}$ as $x \to a$. Function values become infinitely large near asymptotes.
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What is the vertical asymptote for $f(x) = \frac{\text{ln}(x)}{x-1}$?
What is the vertical asymptote for $f(x) = \frac{\text{ln}(x)}{x-1}$?
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$x = 1$. Denominator $x-1=0$ when $x=1$, creating vertical asymptote.
$x = 1$. Denominator $x-1=0$ when $x=1$, creating vertical asymptote.
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Identify the vertical asymptote for $f(x) = \frac{x}{x^2+1}$.
Identify the vertical asymptote for $f(x) = \frac{x}{x^2+1}$.
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No vertical asymptote. Denominator $x^2+1$ is always positive, never zero.
No vertical asymptote. Denominator $x^2+1$ is always positive, never zero.
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Does $f(x) = \frac{1}{x^2+1}$ have a vertical asymptote?
Does $f(x) = \frac{1}{x^2+1}$ have a vertical asymptote?
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No vertical asymptote. Denominator $x^2+1$ is never zero for real values.
No vertical asymptote. Denominator $x^2+1$ is never zero for real values.
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What indicates a vertical asymptote in a rational function's graph?
What indicates a vertical asymptote in a rational function's graph?
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The function approaches $\text{±}\text{∞}$ near a vertical line. Graph shows function values shooting to infinity near vertical lines.
The function approaches $\text{±}\text{∞}$ near a vertical line. Graph shows function values shooting to infinity near vertical lines.
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Identify the vertical asymptote for $f(x) = \frac{x^2+1}{x^2-3x+2}$.
Identify the vertical asymptote for $f(x) = \frac{x^2+1}{x^2-3x+2}$.
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$x = 1$ and $x = 2$. Factor denominator: $x^2-3x+2=(x-1)(x-2)=0$ at $x=1,2$.
$x = 1$ and $x = 2$. Factor denominator: $x^2-3x+2=(x-1)(x-2)=0$ at $x=1,2$.
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State the vertical asymptotes for $f(x) = \frac{1}{x^2-1}$.
State the vertical asymptotes for $f(x) = \frac{1}{x^2-1}$.
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$x = \text{±}1$. Factor denominator: $x^2-1=(x-1)(x+1)=0$ at $x=\pm 1$.
$x = \text{±}1$. Factor denominator: $x^2-1=(x-1)(x+1)=0$ at $x=\pm 1$.
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What does $f(x) \to \text{±}\text{∞}$ as $x \to a^\text{±}$ indicate?
What does $f(x) \to \text{±}\text{∞}$ as $x \to a^\text{±}$ indicate?
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A vertical asymptote at $x = a$. Infinite limit behavior defines a vertical asymptote location.
A vertical asymptote at $x = a$. Infinite limit behavior defines a vertical asymptote location.
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State the limit of $f(x) = \frac{1}{x}$ as $x \to 0^-$.
State the limit of $f(x) = \frac{1}{x}$ as $x \to 0^-$.
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$-\text{∞}$. As $x$ approaches 0 from left, $\frac{1}{x}$ becomes negative infinity.
$-\text{∞}$. As $x$ approaches 0 from left, $\frac{1}{x}$ becomes negative infinity.
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What behavior does $f(x) = \frac{1}{x^2}$ exhibit as $x \to 0$?
What behavior does $f(x) = \frac{1}{x^2}$ exhibit as $x \to 0$?
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Approaches $\infty$. Function grows to infinity from both sides of $x=0$.
Approaches $\infty$. Function grows to infinity from both sides of $x=0$.
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What is the limit of $f(x) = \frac{1}{x^2}$ as $x \to 0$?
What is the limit of $f(x) = \frac{1}{x^2}$ as $x \to 0$?
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$\infty$. Both one-sided limits approach positive infinity at $x=0$.
$\infty$. Both one-sided limits approach positive infinity at $x=0$.
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What happens to $f(x)$ as $x$ approaches a vertical asymptote?
What happens to $f(x)$ as $x$ approaches a vertical asymptote?
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$f(x) \to \pm \infty$. Function values become unbounded at vertical asymptotes.
$f(x) \to \pm \infty$. Function values become unbounded at vertical asymptotes.
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What is the infinite limit definition at a vertical asymptote?
What is the infinite limit definition at a vertical asymptote?
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As $x \to a$, $f(x) \to \text{±}\text{∞}$. Function grows without bound as $x$ approaches the asymptote.
As $x \to a$, $f(x) \to \text{±}\text{∞}$. Function grows without bound as $x$ approaches the asymptote.
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Does $f(x) = \frac{x^3}{x^2+4x+4}$ have a vertical asymptote?
Does $f(x) = \frac{x^3}{x^2+4x+4}$ have a vertical asymptote?
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Yes, $x = -2$ is a vertical asymptote. Denominator $x^2+4x+4=(x+2)^2=0$ only at $x=-2$.
Yes, $x = -2$ is a vertical asymptote. Denominator $x^2+4x+4=(x+2)^2=0$ only at $x=-2$.
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State the condition for a vertical asymptote at $x = a$ for $f(x) = \frac{P(x)}{Q(x)}$.
State the condition for a vertical asymptote at $x = a$ for $f(x) = \frac{P(x)}{Q(x)}$.
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$Q(a) = 0$ and $P(a) \neq 0$. Denominator zero but numerator nonzero creates division by zero.
$Q(a) = 0$ and $P(a) \neq 0$. Denominator zero but numerator nonzero creates division by zero.
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What is the definition of a vertical asymptote?
What is the definition of a vertical asymptote?
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A line $x = a$ where $f(x)$ approaches $\frac{\text{±}\text{∞}}{}$ as $x \to a$. Function value becomes infinitely large at the asymptote.
A line $x = a$ where $f(x)$ approaches $\frac{\text{±}\text{∞}}{}$ as $x \to a$. Function value becomes infinitely large at the asymptote.
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Identify the vertical asymptote for $f(x) = \frac{1}{(x+1)^2}$.
Identify the vertical asymptote for $f(x) = \frac{1}{(x+1)^2}$.
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$x = -1$. Denominator $(x+1)^2=0$ only when $x=-1$.
$x = -1$. Denominator $(x+1)^2=0$ only when $x=-1$.
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State the vertical asymptote for $f(x) = \frac{x+1}{x^2+2x}$.
State the vertical asymptote for $f(x) = \frac{x+1}{x^2+2x}$.
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$x = 0$ and $x = -2$. Factor denominator: $x^2+2x=x(x+2)=0$ at $x=0,-2$.
$x = 0$ and $x = -2$. Factor denominator: $x^2+2x=x(x+2)=0$ at $x=0,-2$.
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Identify the vertical asymptote for $f(x) = \frac{1}{x^3}$.
Identify the vertical asymptote for $f(x) = \frac{1}{x^3}$.
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$x = 0$. Denominator $x^3=0$ only when $x=0$.
$x = 0$. Denominator $x^3=0$ only when $x=0$.
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What kind of asymptote does $f(x) = \frac{x^2 + 1}{x-1}$ have at $x = 1$?
What kind of asymptote does $f(x) = \frac{x^2 + 1}{x-1}$ have at $x = 1$?
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Vertical asymptote. Denominator zero at $x=1$ but numerator nonzero.
Vertical asymptote. Denominator zero at $x=1$ but numerator nonzero.
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Identify the vertical asymptote of $f(x) = \frac{3}{x^2 - 9}$.
Identify the vertical asymptote of $f(x) = \frac{3}{x^2 - 9}$.
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$x = \text{±}3$. Factor denominator: $x^2-9=(x-3)(x+3)=0$ at $x=\pm 3$.
$x = \text{±}3$. Factor denominator: $x^2-9=(x-3)(x+3)=0$ at $x=\pm 3$.
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Does $f(x) = \frac{x^2}{x^2-1}$ have a vertical asymptote at $x = 1$?
Does $f(x) = \frac{x^2}{x^2-1}$ have a vertical asymptote at $x = 1$?
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Yes, $x = 1$ is a vertical asymptote. At $x=1$, denominator is zero but numerator is nonzero.
Yes, $x = 1$ is a vertical asymptote. At $x=1$, denominator is zero but numerator is nonzero.
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What is a sign of a vertical asymptote in the limit of a function?
What is a sign of a vertical asymptote in the limit of a function?
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Limit approaches $\text{±}\text{∞}$ as $x \to a$. Infinite limits indicate vertical asymptote presence.
Limit approaches $\text{±}\text{∞}$ as $x \to a$. Infinite limits indicate vertical asymptote presence.
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Determine the vertical asymptote for $f(x) = \frac{2x+3}{x^2-9}$.
Determine the vertical asymptote for $f(x) = \frac{2x+3}{x^2-9}$.
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$x = \text{±}3$. Factor denominator: $(x-3)(x+3)=0$ gives $x=\pm 3$.
$x = \text{±}3$. Factor denominator: $(x-3)(x+3)=0$ gives $x=\pm 3$.
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Determine the vertical asymptote for $f(x) = \frac{x^2+5}{x^2-1}$.
Determine the vertical asymptote for $f(x) = \frac{x^2+5}{x^2-1}$.
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$x = \text{±}1$. Set denominator $x^2-1=0$, giving $x=\pm 1$.
$x = \text{±}1$. Set denominator $x^2-1=0$, giving $x=\pm 1$.
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What is the vertical asymptote for $f(x) = \frac{e^x}{x-1}$?
What is the vertical asymptote for $f(x) = \frac{e^x}{x-1}$?
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$x = 1$. Denominator $x-1=0$ when $x=1$, numerator stays finite.
$x = 1$. Denominator $x-1=0$ when $x=1$, numerator stays finite.
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For $f(x) = \frac{2x}{x^2-4}$, identify a vertical asymptote.
For $f(x) = \frac{2x}{x^2-4}$, identify a vertical asymptote.
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$x = \pm 2$. Factor denominator: $x^2-4=(x-2)(x+2)=0$ at $x=\pm 2$.
$x = \pm 2$. Factor denominator: $x^2-4=(x-2)(x+2)=0$ at $x=\pm 2$.
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