Removing Discontinuities - AP Calculus BC
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Determine the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
Determine the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
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Removable discontinuity. The factor $ (x-2) $ cancels, leaving a hole at $x=2$.
Removable discontinuity. The factor $ (x-2) $ cancels, leaving a hole at $x=2$.
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How do you remove a removable discontinuity?
How do you remove a removable discontinuity?
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Redefine the function at the discontinuity point. Set the function value equal to the limit at that point.
Redefine the function at the discontinuity point. Set the function value equal to the limit at that point.
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What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$?
What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$?
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Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
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What is the behavior of $f(x)$ at a jump discontinuity?
What is the behavior of $f(x)$ at a jump discontinuity?
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Differing left and right-hand limits. Left and right limits exist but are unequal.
Differing left and right-hand limits. Left and right limits exist but are unequal.
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What must be true for $f(x)$ to be continuous at $x = a$?
What must be true for $f(x)$ to be continuous at $x = a$?
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$\lim_{x \to a} f(x) = f(a)$. All three continuity conditions must hold at the point.
$\lim_{x \to a} f(x) = f(a)$. All three continuity conditions must hold at the point.
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What condition creates a removable discontinuity?
What condition creates a removable discontinuity?
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A factor cancels in the numerator and denominator. Common factors in numerator and denominator create holes.
A factor cancels in the numerator and denominator. Common factors in numerator and denominator create holes.
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Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$.
Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$.
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Redefine $f(3) = 6$. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.
Redefine $f(3) = 6$. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.
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Identify the discontinuity type: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
Identify the discontinuity type: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
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Removable discontinuity. The factor $(x-2)$ cancels, creating a hole at $x=2$.
Removable discontinuity. The factor $(x-2)$ cancels, creating a hole at $x=2$.
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Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$.
Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$.
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$x = 3$. The denominator equals zero when $x=3$, creating a hole.
$x = 3$. The denominator equals zero when $x=3$, creating a hole.
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What is the necessary condition for a function to be continuous?
What is the necessary condition for a function to be continuous?
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The function is defined and limits match the function value. Function exists, limit exists, and both values are equal.
The function is defined and limits match the function value. Function exists, limit exists, and both values are equal.
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Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$.
Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$.
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$x = 0$. The factor $x$ in denominator creates a hole when $x=0$.
$x = 0$. The factor $x$ in denominator creates a hole when $x=0$.
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What is the condition for a removable discontinuity?
What is the condition for a removable discontinuity?
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A cancelable factor in the expression. A common factor exists in both numerator and denominator.
A cancelable factor in the expression. A common factor exists in both numerator and denominator.
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What happens to $f(x)$ at an essential discontinuity?
What happens to $f(x)$ at an essential discontinuity?
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The limits don't exist or are undefined. The function oscillates wildly or has no pattern near the point.
The limits don't exist or are undefined. The function oscillates wildly or has no pattern near the point.
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Determine the discontinuity type: $f(x) = \frac{1}{x}$ at $x = 0$.
Determine the discontinuity type: $f(x) = \frac{1}{x}$ at $x = 0$.
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Infinite discontinuity. Division by zero with no cancellation creates a vertical asymptote.
Infinite discontinuity. Division by zero with no cancellation creates a vertical asymptote.
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State the condition for a function to have no discontinuities.
State the condition for a function to have no discontinuities.
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The function must be continuous everywhere in its domain. No holes, jumps, or vertical asymptotes anywhere.
The function must be continuous everywhere in its domain. No holes, jumps, or vertical asymptotes anywhere.
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What is the limit definition of continuity at $x = a$?
What is the limit definition of continuity at $x = a$?
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$\lim_{x \to a} f(x) = f(a)$. The limit as $x$ approaches $a$ equals the function value.
$\lim_{x \to a} f(x) = f(a)$. The limit as $x$ approaches $a$ equals the function value.
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What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$?
What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$?
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- Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, so limit is $1+1=2$.
- Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, so limit is $1+1=2$.
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What is an infinite discontinuity?
What is an infinite discontinuity?
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A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote at that point.
A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote at that point.
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What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$?
What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$?
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Infinite discontinuity at $x = a$. The function approaches infinity, creating a vertical asymptote.
Infinite discontinuity at $x = a$. The function approaches infinity, creating a vertical asymptote.
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Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.
Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.
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Redefine $f(2) = 4$. The limit at $x=2$ is 4, so define $f(2)=4$ for continuity.
Redefine $f(2) = 4$. The limit at $x=2$ is 4, so define $f(2)=4$ for continuity.
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Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.
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- Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $2+2=4$.
- Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $2+2=4$.
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How is a jump discontinuity characterized?
How is a jump discontinuity characterized?
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Differing left and right-hand limits. The function has different values approaching from each side.
Differing left and right-hand limits. The function has different values approaching from each side.
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Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$.
Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$.
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$x = 5$. The denominator becomes zero when $x=5$, creating a hole.
$x = 5$. The denominator becomes zero when $x=5$, creating a hole.
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Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$?
Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$?
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Removable. The factor $(x-1)$ cancels, leaving a hole at $x=1$.
Removable. The factor $(x-1)$ cancels, leaving a hole at $x=1$.
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What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$?
What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$?
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Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
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Identify the discontinuity at $x = 0$ for $f(x) = \frac{1}{x^2}$.
Identify the discontinuity at $x = 0$ for $f(x) = \frac{1}{x^2}$.
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Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.
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What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$?
What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$?
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- Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.
- Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.
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Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
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Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.
Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.
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Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
Tap to reveal answer
Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.
Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.
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What is the condition for a removable discontinuity?
What is the condition for a removable discontinuity?
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A cancelable factor in the expression. A common factor exists in both numerator and denominator.
A cancelable factor in the expression. A common factor exists in both numerator and denominator.
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