Determining Limits Using Algebraic Manipulation - AP Calculus AB
Card 1 of 30
Determine the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Determine the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Tap to reveal answer
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
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Evaluate $\frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Evaluate $\frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Tap to reveal answer
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
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State the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.
State the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.
Tap to reveal answer
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
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Evaluate $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Evaluate $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Tap to reveal answer
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
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Evaluate the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Evaluate the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Tap to reveal answer
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
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Find the limit of $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.
Find the limit of $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.
Tap to reveal answer
- Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.
- Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.
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Evaluate $\frac{x^2 - 4}{x - 2}$ as $x$ approaches 2.
Evaluate $\frac{x^2 - 4}{x - 2}$ as $x$ approaches 2.
Tap to reveal answer
- Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$, then substitute $x = 2$.
- Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$, then substitute $x = 2$.
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Evaluate the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Evaluate the limit of $f(x) = \frac{x^2 - 1}{x - 1}$ as $x$ approaches 1.
Tap to reveal answer
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
- Factor: $\frac{(x-1)(x+1)}{x-1} = x+1$, then substitute $x = 1$.
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What is the limit of $f(x) = x^3$ as $x$ approaches 3?
What is the limit of $f(x) = x^3$ as $x$ approaches 3?
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- Direct substitution: $3^3 = 27$.
- Direct substitution: $3^3 = 27$.
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What is the limit of $\frac{x^3 - 8}{x - 2}$ as $x$ approaches 2?
What is the limit of $\frac{x^3 - 8}{x - 2}$ as $x$ approaches 2?
Tap to reveal answer
- Factor: $\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$, then substitute $x = 2$.
- Factor: $\frac{(x-2)(x^2+2x+4)}{x-2} = x^2+2x+4$, then substitute $x = 2$.
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Evaluate the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.
Evaluate the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3.
Tap to reveal answer
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
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Find the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Find the limit of $\frac{x^2 - 25}{x - 5}$ as $x$ approaches 5.
Tap to reveal answer
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
- Factor: $\frac{(x-5)(x+5)}{x-5} = x+5$, then substitute $x = 5$.
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Find the limit of $f(x) = x^2 - 4x + 4$ as $x$ approaches 2.
Find the limit of $f(x) = x^2 - 4x + 4$ as $x$ approaches 2.
Tap to reveal answer
- Direct substitution: $2^2 - 4(2) + 4 = 0$.
- Direct substitution: $2^2 - 4(2) + 4 = 0$.
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Determine the limit of $x^2 - 4x + 4$ as $x$ approaches 4.
Determine the limit of $x^2 - 4x + 4$ as $x$ approaches 4.
Tap to reveal answer
- Direct substitution: $4^2 - 4(4) + 4 = 4$.
- Direct substitution: $4^2 - 4(4) + 4 = 4$.
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What is the limit of $f(x) = x^3$ as $x$ approaches -1?
What is the limit of $f(x) = x^3$ as $x$ approaches -1?
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-1. Direct substitution: $(-1)^3 = -1$.
-1. Direct substitution: $(-1)^3 = -1$.
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State the limit of $f(x) = \frac{1}{x}$ as $x$ approaches negative infinity.
State the limit of $f(x) = \frac{1}{x}$ as $x$ approaches negative infinity.
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- As $x \to -\infty$, $\frac{1}{x} \to 0$.
- As $x \to -\infty$, $\frac{1}{x} \to 0$.
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Determine the limit of $f(x) = \frac{1}{x}$ as $x$ approaches infinity.
Determine the limit of $f(x) = \frac{1}{x}$ as $x$ approaches infinity.
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- As $x \to \infty$, $\frac{1}{x} \to 0$.
- As $x \to \infty$, $\frac{1}{x} \to 0$.
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What is the limit of $h(x) = \frac{x^2 + x - 6}{x - 2}$ as $x$ approaches 2?
What is the limit of $h(x) = \frac{x^2 + x - 6}{x - 2}$ as $x$ approaches 2?
Tap to reveal answer
- Factor: $\frac{(x+3)(x-2)}{x-2} = x+3$, then substitute $x = 2$.
- Factor: $\frac{(x+3)(x-2)}{x-2} = x+3$, then substitute $x = 2$.
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Determine the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches $\infty$.
Determine the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches $\infty$.
Tap to reveal answer
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
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Find the limit of $f(x) = x^3$ as $x$ approaches -2.
Find the limit of $f(x) = x^3$ as $x$ approaches -2.
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-8. Direct substitution: $(-2)^3 = -8$.
-8. Direct substitution: $(-2)^3 = -8$.
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Find the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity.
Find the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity.
Tap to reveal answer
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
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Evaluate $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.
Evaluate $\frac{x^2 - 4x + 4}{x - 2}$ as $x$ approaches 2.
Tap to reveal answer
- Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.
- Factor: $\frac{(x-2)^2}{x-2} = x-2$, then substitute $x = 2$.
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What is the limit of $f(x) = \frac{x^2 - 16}{x - 4}$ as $x$ approaches 4?
What is the limit of $f(x) = \frac{x^2 - 16}{x - 4}$ as $x$ approaches 4?
Tap to reveal answer
- Factor: $\frac{(x-4)(x+4)}{x-4} = x+4$, then substitute $x = 4$.
- Factor: $\frac{(x-4)(x+4)}{x-4} = x+4$, then substitute $x = 4$.
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State the limit of $f(x) = \frac{x^2 + 2x + 1}{x + 1}$ as $x$ approaches -1.
State the limit of $f(x) = \frac{x^2 + 2x + 1}{x + 1}$ as $x$ approaches -1.
Tap to reveal answer
- Factor: $\frac{(x+1)^2}{x+1} = x+1$, then substitute $x = -1$.
- Factor: $\frac{(x+1)^2}{x+1} = x+1$, then substitute $x = -1$.
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What is the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?
What is the limit of $\frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?
Tap to reveal answer
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
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Find the limit of $f(x) = x^3$ as $x$ approaches 2.
Find the limit of $f(x) = x^3$ as $x$ approaches 2.
Tap to reveal answer
- Direct substitution: $2^3 = 8$.
- Direct substitution: $2^3 = 8$.
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What is the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity?
What is the limit of $f(x) = \frac{x^2 - 1}{x^2 + 1}$ as $x$ approaches infinity?
Tap to reveal answer
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
- Divide by highest power: $\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x^2}} \to 1$.
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What is the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?
What is the limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3?
Tap to reveal answer
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
- Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, then substitute $x = 3$.
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What is the limit of $g(x) = \frac{1}{x}$ as $x$ approaches 0?
What is the limit of $g(x) = \frac{1}{x}$ as $x$ approaches 0?
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Does not exist. Function undefined at $x = 0$; left and right limits differ.
Does not exist. Function undefined at $x = 0$; left and right limits differ.
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What is the limit of $f(x) = \frac{2x - 6}{x - 3}$ as $x$ approaches 3?
What is the limit of $f(x) = \frac{2x - 6}{x - 3}$ as $x$ approaches 3?
Tap to reveal answer
- Factor: $\frac{2(x-3)}{x-3} = 2$, then substitute $x = 3$.
- Factor: $\frac{2(x-3)}{x-3} = 2$, then substitute $x = 3$.
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