Connecting Infinite Limits and Vertical Asymptotes - AP Calculus AB

Let $f(x)=\dfrac{2}{x(x-4)}$ . Which limit notation matches the behavior as $x\to 4^+$ ?
Answer: $\displaystyle \lim_{x\to 4^+} f(x)=+\infty$ The function f(x) = 2/(x(x-4)), as x→ $4^+$ , x positive, (x-4) positive small, denominator positive small, 2/positive = +∞. The notation $\lim_{x→4^+}$ f(x) = +∞ is correct. This is valid for the right-hand sign. A common error is assuming -∞ or writing f(4) = +∞. Overlooking the positive product is typical. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
A graph shows a vertical asymptote at $x=-4$ ; as $x\to-4^-$ , $f(x)\to -\infty$ . Which limit notation?
Answer: $\displaystyle \lim_{x\to -4^-} f(x)=-\infty$ Graph descriptions of asymptotes require precise limit notation to match the behavior. The graph shows f(x) → -∞ as x → - $4^-$ , so $\lim_{x\to -4^-} f(x) = -\infty$ correctly captures this left-side decrease to negative infinity. This is valid as it specifies the direction and sign matching the description. A common error is using ^+ instead of ^-, describing the wrong side. Writing $f(-4) = -\infty$ mistakes the limit for a point value. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign ( $+\infty$ or $-\infty$ ) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.
A graph shows $f(x)$ approaches $-\infty$ as $x\to 2^+$ near a vertical asymptote at $x=2$ . Which limit?
Answer: $\displaystyle \lim_{x\to 2^+} f(x)=-\infty$ The limit notation identifies the behavior of the function as x approaches a specific value from a particular side, here as x approaches 2 from the right, denoted by the superscript +. The correct representation is $\lim_{x\to 2^+} f(x) = -\infty$ , which is valid because the graph shows the function values decreasing without bound as x gets closer to 2 from values greater than 2, indicating a vertical asymptote at x=2. This one-sided limit is necessary since the behavior is specified only from the right side. A common symbolic error is writing f(2) = -\infty instead of the limit, but this is incorrect because infinity is not a value the function can take at a point; limits describe approaching behavior, not the function's value. Another error might be confusing the direction, such as using $2^-$ instead of $2^+$ , which would describe the left-side approach. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign (+∞ or -∞) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.
For $f(x)=\dfrac{2}{(x+5)^2}$ , which limit notation describes the behavior as $x\to -5$ ?
Answer: $\displaystyle \lim_{x\to -5} f(x)=+\infty$ For f(x) = 2/ $(x+5)^2$ , the even power ensures positive denominator near x=-5 from both sides, leading to +∞. The two-sided limit $\lim_{x→-5}$ f(x) = +∞ is correct. This is valid for symmetric positive behavior. A common error is predicting differing sides or using f(-5) = +∞. Overlooking the square's effect is typical. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
Let $u(x)=\dfrac{6}{(x+4)^2}$ . Which limit notation represents the behavior of $u(x)$ as $x\to-4$ ?
Answer: $\displaystyle \lim_{x\to-4} u(x)=+\infty$ For u(x) = 6/(x+4)², we examine the behavior as x approaches -4. Since the denominator is squared, (x+4)² is always positive for x ≠ -4. As x approaches -4 from either direction, the denominator approaches 0 while remaining positive. The fraction becomes 6/(small positive) = large positive value. Since this happens from both sides, lim[x→-4] u(x) = +∞ correctly represents this behavior. A common error is thinking that squared denominators might lead to different one-sided limits, but squares are always non-negative. Another mistake is writing u(-4) = +∞ instead of using limit notation. Notation checklist: Remember that even powers create the same behavior from both sides, two-sided limits exist when both one-sided limits equal +∞, and always use limit notation for infinite values.
A graph shows a vertical asymptote at $x=3$ ; as $x\to3^+$ , $f(x)\to +\infty$ . Which limit notation?
Answer: $\displaystyle \lim_{x\to 3^+} f(x)=+\infty$ Graphs with vertical asymptotes use limit notation to describe unbounded growth or decay. The graph indicates f(x) → +∞ as x → $3^+$ , so $\lim_{x\to 3^+} f(x) = +\infty$ is the correct representation. This is valid as it matches the right-side increase without bound. A common error is using left-side notation for right-side behavior. Writing f(3) = +∞ confuses limit with value. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign (+∞ or -∞) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.
For $f(x)=\dfrac{3}{x-2}$ , which limit notation represents the behavior of $f(x)$ as $x$ approaches $2$ from the right?
Answer: $\displaystyle \lim_{x\to2^+} f(x)=+\infty$ For f(x) = 3/(x-2), we need to identify the limit notation for the behavior as x approaches 2 from the right. The notation x→2⁺ means x approaches 2 from values greater than 2. As x gets closer to 2 from the right, the denominator (x-2) becomes a small positive number, making the fraction 3/(small positive) = large positive value. Therefore, lim[x→2⁺] f(x) = +∞ correctly represents this behavior. A common error is using an equals sign instead of an arrow (x=2⁺), which is incorrect notation. Another mistake is evaluating f(2), which is undefined since division by zero is not allowed. Notation checklist: Use → not =, include superscript + or - for one-sided limits, and remember that limits describe approach behavior, not function values.
The graph of $f$ has a vertical asymptote at $x=1$ and rises on both sides. Which limit notation matches?
Answer: $\displaystyle \lim_{x\to 1} f(x)=+\infty$ The graph with a vertical asymptote at x=1 rising on both sides means f(x)→+∞ from left and right. The two-sided notation $\lim_{x→1}$ f(x) = +∞ is appropriate since behaviors match. This is valid for symmetric positive infinite limits. A common error is using one-sided notation unnecessarily or writing f(1) = +∞. Misinterpreting 'rising' as -∞ occurs sometimes. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
The graph shows a vertical asymptote at $x=0$ ; as $x\to0^-$ , $f(x)\to +\infty$ . Which limit notation?
Answer: $\displaystyle \lim_{x\to 0^-} f(x)=+\infty$ The graph indicates a vertical asymptote at x=0 with f(x)→+∞ as x→0 from the left. The notation $\lim_{x→0^-}$ f(x) = +∞ matches this left-hand behavior. This is valid for graphical infinite limits. A common symbolic error is using right-side or writing f(0) = +∞. Misreading the direction of approach is frequent. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
For $h(x)=\dfrac{1}{x^2-4x}$ , which limit notation describes the behavior as $x\to 0^+$ ?
Answer: $\displaystyle \lim_{x\to 0^+} h(x)=-\infty$ The function h(x) = 1/( $x^2-4x$ ) = 1/(x(x-4)), as x→ $0^+$ , x positive small, (x-4) negative, denominator negative small, 1/negative = -∞. The notation $\lim_{x→0^+}$ h(x) = -∞ is accurate. This is valid from factoring and sign check. A common error is ignoring factoring or writing h(0) = -∞. Mixing with left-side +∞ is frequent. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
The graph of $g$ has a vertical asymptote at $x=-1$ ; as $x\to-1^+$ , $g(x)$ decreases without bound. Which limit?
Answer: $\displaystyle \lim_{x\to -1^+} g(x)=-\infty$ The graph shows a vertical asymptote at x=-1 with g(x) decreasing without bound as x→- $1^+$ , meaning large negative. The notation $\lim_{x→-1^+}$ g(x) = -∞ correctly specifies this. This is valid for one-sided graphical behavior. A frequent mistake is using -∞ for +∞ or equating to g(-1) = -∞. Confusing decrease with positive direction is common. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
For $t(x)=\dfrac{2}{1-x}$ , which limit expression represents the behavior of $t(x)$ as $x\to1^-$ ?
Answer: $\displaystyle \lim_{x\to1^-} t(x)=+\infty$ For t(x) = 2/(1-x), we analyze the behavior as x approaches 1 from the left (x→1⁻). When x is slightly less than 1, the denominator 1-x is a small positive number. The fraction becomes 2/(small positive) = large positive value. Therefore, lim[x→1⁻] t(x) = +∞ correctly represents this behavior. A common error is misidentifying the sign of (1-x) when x < 1; remember that if x < 1, then 1-x > 0. Another mistake is confusing the direction of approach or using incorrect notation like t(1) = +∞. Notation checklist: Carefully determine the sign of expressions like (a-x), use superscript - for left approach, and verify your sign analysis.
Let $q(x)=\dfrac{x+4}{x+4}$ . Which limit notation represents the behavior as $x\to -4$ ?
Answer: $\displaystyle \lim_{x\to -4} q(x)=1$ The function q(x) = (x+4)/(x+4) simplifies to 1 for x ≠ -4, so there's a removable discontinuity at x=-4, but the limit exists. As x approaches -4 from both sides, q(x) approaches 1. The correct notation is $\lim_{x→-4}$ q(x) = 1, indicating a finite two-sided limit. This is valid because after simplification, the function is constant near x=-4. A common symbolic error is failing to simplify and assuming infinite limit like +∞, or writing q(-4) = 1 when it's undefined. Mixing it with asymptotic behavior without checking is frequent. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
Let $g(x)=\dfrac{-4}{x+6}$ . Which limit notation represents the behavior as $x\to -6^+$ ?
Answer: $\displaystyle \lim_{x\to -6^+} g(x)=-\infty$ Direction matters in determining the sign for limits near asymptotes. For g(x) = -4/(x+6), as $x \to -6^+$ , $x+6$ positive small, $-4/\text{positive} = \text{negative large}$ , so $\lim_{x\to -6^+} g(x) = -\infty$ is the correct notation. This is valid because the negative numerator over positive denominator results in large negative values. A common symbolic error is misplacing the direction superscript, changing the behavior described. Writing $g(-6) = -\infty$ confuses limit with function value. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign (+∞ or -∞) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.
Let $g(x)=\dfrac{6}{x^2-9}$ . Which limit notation represents the behavior as $x\to 3^-$ ?
Answer: $\displaystyle \lim_{x\to 3^-} g(x)=-\infty$ For g(x) = 6/( $x^2-9$ ), as x→ $3^-$ , denominator approaches negative small (since $x^2$ -9 negative near 3 from left), 6/negative small = -∞. The notation $\lim_{x→3^-}$ g(x) = -∞ fits. This is valid based on the quadratic's sign. A frequent mistake is predicting +∞ or using g(3) = -∞. Overlooking the denominator's negativity inside roots is common. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
For $f(x)=\dfrac{1}{x-2}$ , which limit notation represents the behavior as $x$ approaches $2$ from the right?
Answer: $\displaystyle \lim_{x\to 2^+} f(x)=+\infty$ The function f(x) = 1/(x-2) has a vertical asymptote at x=2, where the denominator becomes zero. As x approaches 2 from the right (x > 2), x-2 is positive and small, so f(x) becomes a large positive number, indicating the limit is +∞. The correct notation is $\lim_{x→2^+}$ f(x) = +∞, which specifies the right-hand side and the direction to positive infinity. This is valid because the one-sided limit captures the behavior without the function being defined at x=2. A common symbolic error is writing f(2) = +∞ instead of using limit notation, as the function value at x=2 is undefined, not infinite. Another mistake is confusing the sides, such as using x→ $2^-$ which would yield -∞ for this function. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
Let $s(x)=\dfrac{x^2}{x^2-9}$ . Which limit notation represents the behavior of $s(x)$ as $x\to3^+$ ?
Answer: $\displaystyle \lim_{x\to3^+} s(x)=+\infty$ For s(x) = x²/(x²-9), we examine the behavior as x approaches 3 from the right (x→3⁺). First, factor the denominator: x²-9 = (x-3)(x+3). When x is slightly greater than 3, we have x² ≈ 9 (positive), x-3 is a small positive number, and x+3 ≈ 6 (positive). The fraction becomes 9/[(small positive)(6)] = large positive value. Therefore, lim[x→3⁺] s(x) = +∞ correctly represents this behavior. A common error is not factoring the denominator to identify all vertical asymptotes (at x = 3 and x = -3). Another mistake is writing s(3) = +∞ instead of using limit notation. Notation checklist: Factor when possible to identify asymptotes, check signs of all factors, and use proper one-sided limit notation.
For $f(x)=\dfrac{1}{x^2-1}$ , which limit notation represents the behavior as $x\to -1^+$ ?
Answer: $\displaystyle \lim_{x\to -1^+} f(x)=-\infty$ Limit notation specifies direction and infinity type to describe asymptotic behavior accurately. For f(x) = 1/(x²-1), as x approaches -1 from the right, the denominator approaches 0 from the negative side, so f(x) approaches -∞, making $\lim_{x\to -1^+} f(x) = -\infty$ the correct choice. This is valid since (x+1)>0 small and (x-1)<0, product negative, reciprocal large negative. A common symbolic error is omitting the direction superscript, implying a two-sided limit that doesn't exist here. Misjudging the sign by not evaluating factor signs is another frequent issue. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign (+∞ or -∞) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.
Let $h(x)=\dfrac{-2}{x-5}$ . Which limit notation represents the behavior as $x\to 5^+$ ?
Answer: $\displaystyle \lim_{x\to 5^+} h(x)=-\infty$ The function h(x) = -2/(x-5) has a vertical asymptote at x=5. As x approaches 5 from the right (x > 5), x-5 is positive small, so h(x) = -2/(positive small) = large negative, approaching -∞. The notation $\lim_{x→5^+}$ h(x) = -∞ correctly indicates the right-hand limit to negative infinity. This is valid because it captures the behavior as x nears 5 without being defined there. A frequent symbolic mistake is omitting the negative sign from the numerator, leading to incorrect +∞, or using function value like h(5) = -∞, which is improper since infinity isn't a value. Confusing directions, such as using left-side for right-side behavior, is also common. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.
For $k(x)=\dfrac{2x}{x^2-16}$ , which limit notation represents the behavior as $x\to 4^-$ ?
Answer: $\displaystyle \lim_{x\to 4^-} k(x)=-\infty$ For k(x) = 2x/(x²-16), as x → $4^-$ , (x-4) negative, (x+4) positive, denominator negative, numerator positive, so k(x) → -∞, making $\lim_{x\to 4^-} k(x) = -\infty$ the valid choice. This represents the left-side behavior accurately. The sign is determined by the negative denominator dominating. A common error is assuming +∞ without sign check. Writing the two-sided limit as 0 ignores the asymptote. To ensure correct notation, use this transferable notation checklist: 1. Use 'lim' to indicate a limit, not the function equals notation. 2. Specify the direction with ^+ or ^- for one-sided limits. 3. Determine the sign (+∞ or -∞) based on whether the function increases or decreases without bound. 4. Remember that vertical asymptotes correspond to limits to infinity, and the function is undefined at that point.

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