AP Calculus AB Quiz: Connecting Infinite Limits And Vertical Asymptotes
20 questions · exam conditions
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Connecting Infinite Limits And Vertical AsymptotesQuestion 1 of 20

The graph shows a vertical asymptote at x=0x=0; as x→0−x\to0^-, f(x)→+∞f(x)\to +\infty. Which limit notation?

lim⁡x→0−f(x)=+∞\displaystyle \lim_{x\to 0^-} f(x)=+\infty
lim⁡x→0+f(x)=+∞\displaystyle \lim_{x\to 0^+} f(x)=+\infty
f(0)=+∞\displaystyle f(0)=+\infty
lim⁡x→0−f(x)=−∞\displaystyle \lim_{x\to 0^-} f(x)=-\infty
lim⁡x→0f(x)=0\displaystyle \lim_{x\to 0} f(x)=0
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AP Calculus AB Quiz

AP Calculus AB Quiz: Connecting Infinite Limits And Vertical Asymptotes

Practice Connecting Infinite Limits And Vertical Asymptotes in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Connecting Infinite Limits And Vertical Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The graph shows a vertical asymptote at x=0x=0; as x→0−x\to0^-, f(x)→+∞f(x)\to +\infty. Which limit notation?

  1. lim⁡x→0−f(x)=+∞\displaystyle \lim_{x\to 0^-} f(x)=+\infty (correct answer)
  2. lim⁡x→0+f(x)=+∞\displaystyle \lim_{x\to 0^+} f(x)=+\infty
  3. f(0)=+∞\displaystyle f(0)=+\infty
  4. lim⁡x→0−f(x)=−∞\displaystyle \lim_{x\to 0^-} f(x)=-\infty
  5. lim⁡x→0f(x)=0\displaystyle \lim_{x\to 0} f(x)=0
Explanation: 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.

Question 2

For f(x)=2(x+5)2f(x)=\dfrac{2}{(x+5)^2}, which limit notation describes the behavior as x→−5x\to -5?

  1. lim⁡x→−5f(x)=+∞\displaystyle \lim_{x\to -5} f(x)=+\infty (correct answer)
  2. lim⁡x→−5−f(x)=−∞\displaystyle \lim_{x\to -5^-} f(x)=-\infty
  3. f(−5)=+∞\displaystyle f(-5)=+\infty
  4. lim⁡x→−5+f(x)=−∞\displaystyle \lim_{x\to -5^+} f(x)=-\infty
  5. lim⁡x→−5f(x)=2\displaystyle \lim_{x\to -5} f(x)=2
Explanation: 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.

Question 3

The graph of ff has a vertical asymptote at x=1x=1 and rises on both sides. Which limit notation matches?

  1. lim⁡x→1f(x)=+∞\displaystyle \lim_{x\to 1} f(x)=+\infty (correct answer)
  2. lim⁡x→1−f(x)=−∞\displaystyle \lim_{x\to 1^-} f(x)=-\infty
  3. f(1)=+∞\displaystyle f(1)=+\infty
  4. lim⁡x→1+f(x)=−∞\displaystyle \lim_{x\to 1^+} f(x)=-\infty
  5. lim⁡x→1f(x)=1\displaystyle \lim_{x\to 1} f(x)=1
Explanation: 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.

Question 4

A graph shows a vertical asymptote at x=−4x=-4; as x→−4−x\to-4^-, f(x)→−∞f(x)\to -\infty. Which limit notation?

  1. lim⁡x→−4−f(x)=−∞\displaystyle \lim_{x\to -4^-} f(x)=-\infty (correct answer)
  2. lim⁡x→−4+f(x)=−∞\displaystyle \lim_{x\to -4^+} f(x)=-\infty
  3. f(−4)=−∞\displaystyle f(-4)=-\infty
  4. lim⁡x→−4−f(x)=+∞\displaystyle \lim_{x\to -4^-} f(x)=+\infty
  5. lim⁡x→−4f(x)=−4\displaystyle \lim_{x\to -4} f(x)=-4
Explanation: Graph descriptions of asymptotes require precise limit notation to match the behavior. The graph shows f(x) → -∞ as x → -4^-, so lim⁡x→−4−f(x)=−∞\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)=−∞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.

Question 5

A graph shows a vertical asymptote at x=3x=3; as x→3+x\to3^+, f(x)→+∞f(x)\to +\infty. Which limit notation?

  1. lim⁡x→3+f(x)=+∞\displaystyle \lim_{x\to 3^+} f(x)=+\infty (correct answer)
  2. lim⁡x→3−f(x)=+∞\displaystyle \lim_{x\to 3^-} f(x)=+\infty
  3. f(3)=+∞\displaystyle f(3)=+\infty
  4. lim⁡x→3+f(x)=−∞\displaystyle \lim_{x\to 3^+} f(x)=-\infty
  5. lim⁡x→3f(x)=3\displaystyle \lim_{x\to 3} f(x)=3
Explanation: Graphs with vertical asymptotes use limit notation to describe unbounded growth or decay. The graph indicates f(x) → +∞ as x → 3^+, so lim⁡x→3+f(x)=+∞\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.

Question 6

The graph of gg has a vertical asymptote at x=−1x=-1; as x→−1+x\to-1^+, g(x)g(x) decreases without bound. Which limit?

  1. lim⁡x→−1+g(x)=−∞\displaystyle \lim_{x\to -1^+} g(x)=-\infty (correct answer)
  2. lim⁡x→−1+g(x)=+∞\displaystyle \lim_{x\to -1^+} g(x)=+\infty
  3. lim⁡x→−1−g(x)=−∞\displaystyle \lim_{x\to -1^-} g(x)=-\infty
  4. g(−1)=−∞\displaystyle g(-1)=-\infty
  5. lim⁡x→−1g(x)=0\displaystyle \lim_{x\to -1} g(x)=0
Explanation: 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.

Question 7

For r(x)=4(x−3)3r(x)=\dfrac{4}{(x-3)^3}, which limit expression matches the behavior of r(x)r(x) as x→3−x\to3^-?

  1. lim⁡x→3−r(x)=−∞\displaystyle \lim_{x\to3^-} r(x)=-\infty (correct answer)
  2. lim⁡x→3−r(x)=+∞\displaystyle \lim_{x\to3^-} r(x)=+\infty
  3. lim⁡x→3r(x)=+∞\displaystyle \lim_{x\to3} r(x)=+\infty
  4. r(3)=−∞\displaystyle r(3)=-\infty
  5. lim⁡x→−3−r(x)=−∞\displaystyle \lim_{x\to-3^-} r(x)=-\infty
Explanation: For r(x) = 4/(x-3)³, we analyze the behavior as x approaches 3 from the left (x→3⁻). When x is slightly less than 3, the denominator (x-3) is a small negative number. Since we're cubing this negative value, (x-3)³ remains negative. The fraction becomes 4/(small negative) = large negative value. Therefore, lim[x→3⁻] r(x) = -∞ correctly represents this behavior. A common error is forgetting that odd powers preserve the sign of negative numbers, unlike even powers. Another mistake is writing r(3) = -∞, which incorrectly suggests the function has a value at x = 3. Notation checklist: Pay attention to the power of the denominator, verify signs carefully for odd vs even powers, and use proper limit notation.

Question 8

Let u(x)=6(x+4)2u(x)=\dfrac{6}{(x+4)^2}. Which limit notation represents the behavior of u(x)u(x) as x→−4x\to-4?

  1. lim⁡x→−4u(x)=+∞\displaystyle \lim_{x\to-4} u(x)=+\infty (correct answer)
  2. lim⁡x→−4u(x)=−∞\displaystyle \lim_{x\to-4} u(x)=-\infty
  3. lim⁡x→−4−u(x)=−∞\displaystyle \lim_{x\to-4^-} u(x)=-\infty
  4. u(−4)=+∞\displaystyle u(-4)=+\infty
  5. lim⁡x→4u(x)=+∞\displaystyle \lim_{x\to4} u(x)=+\infty
Explanation: 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.

Question 9

For g(x)=−5x+1g(x)=\dfrac{-5}{x+1}, which limit expression matches the behavior of g(x)g(x) as xx approaches −1-1 from the left?

  1. lim⁡x→−1−g(x)=+∞\displaystyle \lim_{x\to-1^-} g(x)=+\infty (correct answer)
  2. lim⁡x→−1+g(x)=+∞\displaystyle \lim_{x\to-1^+} g(x)=+\infty
  3. lim⁡x→−1g(x)=−∞\displaystyle \lim_{x\to-1} g(x)=-\infty
  4. g(−1)=+∞\displaystyle g(-1)=+\infty
  5. lim⁡x→1−g(x)=+∞\displaystyle \lim_{x\to1^-} g(x)=+\infty
Explanation: For g(x) = -5/(x+1), we analyze the behavior as x approaches -1 from the left. The notation x→-1⁻ means x approaches -1 from values less than -1. When x is slightly less than -1, the denominator (x+1) is a small negative number. The fraction becomes -5/(small negative) = -5/(-small) = large positive value. Therefore, lim[x→-1⁻] g(x) = +∞ correctly represents this behavior. A common error is confusing the direction of approach or the sign of infinity. Another mistake is writing g(-1) = +∞, which incorrectly suggests the function has a value at x = -1. Notation checklist: Verify the sign of the numerator and denominator separately, use proper limit notation with arrows, and distinguish between left (-) and right (+) approach.

Question 10

For h(x)=1x2−4xh(x)=\dfrac{1}{x^2-4x}, which limit notation describes the behavior as x→0−x\to 0^-?

  1. lim⁡x→0−h(x)=+∞\displaystyle \lim_{x\to 0^-} h(x)=+\infty (correct answer)
  2. lim⁡x→0+h(x)=+∞\displaystyle \lim_{x\to 0^+} h(x)=+\infty
  3. h(0)=+∞\displaystyle h(0)=+\infty
  4. lim⁡x→0−h(x)=−∞\displaystyle \lim_{x\to 0^-} h(x)=-\infty
  5. lim⁡x→0h(x)=10\displaystyle \lim_{x\to 0} h(x)=\tfrac{1}{0}
Explanation: For h(x) = 1/(x(x-4)), as x→0^-, x negative small, (x-4) negative, denominator positive small (negative*negative), 1/positive = +∞. The notation lim_{x→0^-} h(x) = +∞ matches. This is valid via sign analysis. A symbolic mistake is predicting -∞ or using h(0) = +∞. Confusing sides 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.

Question 11

Let q(x)=x+4x+4q(x)=\dfrac{x+4}{x+4}. Which limit notation represents the behavior as x→−4x\to -4?

  1. lim⁡x→−4q(x)=1\displaystyle \lim_{x\to -4} q(x)=1 (correct answer)
  2. q(−4)=1\displaystyle q(-4)=1
  3. lim⁡x→−4q(x)=+∞\displaystyle \lim_{x\to -4} q(x)=+\infty
  4. lim⁡x→−4−q(x)=−∞\displaystyle \lim_{x\to -4^-} q(x)=-\infty
  5. lim⁡x→−4+q(x)=−∞\displaystyle \lim_{x\to -4^+} q(x)=-\infty
Explanation: 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.

Question 12

For f(x)=1x2−1f(x)=\dfrac{1}{x^2-1}, which limit notation represents the behavior as x→−1+x\to -1^+?

  1. lim⁡x→−1+f(x)=−∞\displaystyle \lim_{x\to -1^+} f(x)=-\infty (correct answer)
  2. lim⁡x→−1−f(x)=−∞\displaystyle \lim_{x\to -1^-} f(x)=-\infty
  3. f(−1)=−∞\displaystyle f(-1)=-\infty
  4. lim⁡x→−1+f(x)=+∞\displaystyle \lim_{x\to -1^+} f(x)=+\infty
  5. lim⁡x→−1f(x)=−1\displaystyle \lim_{x\to -1} f(x)=-1
Explanation: 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→−1+f(x)=−∞\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.

Question 13

Let u(x)=−1(x−1)2u(x)=\dfrac{-1}{(x-1)^2}. Which limit notation represents the behavior as x→1x\to 1?

  1. lim⁡x→1u(x)=+∞\displaystyle \lim_{x\to 1} u(x)=+\infty
  2. lim⁡x→1u(x)=−∞\displaystyle \lim_{x\to 1} u(x)=-\infty (correct answer)
  3. lim⁡x→1−u(x)=+∞\displaystyle \lim_{x\to 1^-} u(x)=+\infty
  4. u(1)=−∞\displaystyle u(1)=-\infty
  5. lim⁡x→1+u(x)=+∞\displaystyle \lim_{x\to 1^+} u(x)=+\infty
Explanation: The function u(x) = -1/(x-1)^2 has an even-powered denominator, always positive near x=1, with negative numerator leading to -∞ from both sides. As x approaches 1, (x-1)^2 positive small, u(x) = -1/(positive) = large negative. The two-sided notation lim_{x→1} u(x) = -∞ is correct since sides match. This is valid for symmetric negative infinite behavior. A common symbolic error is predicting +∞ by overlooking the negative sign, or writing u(1) = -∞, but limits aren't point evaluations. Ignoring the even power's positivity 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.

Question 14

For k(x)=2xx2−16k(x)=\dfrac{2x}{x^2-16}, which limit notation represents the behavior as x→4+x\to 4^+?

  1. lim⁡x→4+k(x)=+∞\displaystyle \lim_{x\to 4^+} k(x)=+\infty (correct answer)
  2. lim⁡x→4−k(x)=+∞\displaystyle \lim_{x\to 4^-} k(x)=+\infty
  3. k(4)=+∞\displaystyle k(4)=+\infty
  4. lim⁡x→4+k(x)=−∞\displaystyle \lim_{x\to 4^+} k(x)=-\infty
  5. lim⁡x→−4+k(x)=+∞\displaystyle \lim_{x\to -4^+} k(x)=+\infty
Explanation: Sign analysis for k(x) = 2x/(x²-16) near x=4 involves checking numerator and denominator. As x → 4^+, denominator positive small, numerator positive, so k(x) → +∞, and lim⁡x→4+k(x)=+∞\lim_{x\to 4^+} k(x) = +\infty is correct. This notation is valid because (x-4)>0, (x+4)>0, overall positive. A common symbolic error is incorrect sign from misfactoring. Using wrong direction like -4^+ is a distraction. 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.

Question 15

For k(x)=2xx2−16k(x)=\dfrac{2x}{x^2-16}, which limit notation represents the behavior as x→4−x\to 4^-?

  1. lim⁡x→4−k(x)=−∞\displaystyle \lim_{x\to 4^-} k(x)=-\infty (correct answer)
  2. lim⁡x→4+k(x)=−∞\displaystyle \lim_{x\to 4^+} k(x)=-\infty
  3. k(4)=−∞\displaystyle k(4)=-\infty
  4. lim⁡x→4−k(x)=+∞\displaystyle \lim_{x\to 4^-} k(x)=+\infty
  5. lim⁡x→4k(x)=0\displaystyle \lim_{x\to 4} k(x)=0
Explanation: 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→4−k(x)=−∞\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.

Question 16

For f(x)=1x−8f(x)=\dfrac{1}{x-8}, which limit notation represents the behavior as x→8+x\to 8^+?

  1. lim⁡x→8+f(x)=+∞\displaystyle \lim_{x\to 8^+} f(x)=+\infty (correct answer)
  2. lim⁡x→8−f(x)=+∞\displaystyle \lim_{x\to 8^-} f(x)=+\infty
  3. f(8)=+∞\displaystyle f(8)=+\infty
  4. lim⁡x→8+f(x)=−∞\displaystyle \lim_{x\to 8^+} f(x)=-\infty
  5. lim⁡x→8f(x)=0\displaystyle \lim_{x\to 8} f(x)=0
Explanation: Analyzing signs helps identify if the limit is to +∞ or -∞. For f(x) = 1/(x-8), as x → 8^+, denominator positive small, f(x) → +∞, so lim⁡x→8+f(x)=+∞\lim_{x\to 8^+} f(x) = +\infty is the valid notation. This is correct because from the right, x-8 >0, reciprocal positive large. A common symbolic error is swapping ^+ and ^-, altering the sign. Writing the limit as 0 is a misunderstanding of infinite behavior. 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.

Question 17

For g(x)=x2x2−1g(x)=\dfrac{x^2}{x^2-1}, which limit notation represents the behavior as x→1−x\to 1^-?

  1. lim⁡x→1−g(x)=−∞\displaystyle \lim_{x\to 1^-} g(x)=-\infty (correct answer)
  2. lim⁡x→1+g(x)=−∞\displaystyle \lim_{x\to 1^+} g(x)=-\infty
  3. g(1)=−∞\displaystyle g(1)=-\infty
  4. lim⁡x→1−g(x)=+∞\displaystyle \lim_{x\to 1^-} g(x)=+\infty
  5. lim⁡x→−1−g(x)=−∞\displaystyle \lim_{x\to -1^-} g(x)=-\infty
Explanation: Infinite limits are denoted when a function approaches positive or negative infinity near a point, often indicating a vertical asymptote. For g(x) = x²/(x²-1), as x approaches 1 from the left, the denominator approaches 0 from the negative side while the numerator is positive, resulting in g(x) approaching −∞-\infty, so lim⁡x→1−g(x)=−∞\lim_{x\to 1^-} g(x) = -\infty is correct. This representation is valid because near x=1^-, (x-1) is negative and (x+1) positive, making the denominator negative, and positive over negative yields negative values growing in magnitude. A common symbolic error is writing g(1) = -∞, but limits do not assign a value to the function at the point. Confusing left and right directions is another frequent mistake, which changes the sign analysis. 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.

Question 18

Let g(x)=−4x+6g(x)=\dfrac{-4}{x+6}. Which limit notation represents the behavior as x→−6−x\to -6^-?

  1. lim⁡x→−6−g(x)=−∞\displaystyle \lim_{x\to -6^-} g(x)=-\infty
  2. lim⁡x→−6−g(x)=+∞\displaystyle \lim_{x\to -6^-} g(x)=+\infty (correct answer)
  3. g(−6)=−∞\displaystyle g(-6)=-\infty
  4. lim⁡x→−6+g(x)=−∞\displaystyle \lim_{x\to -6^+} g(x)=-\infty
  5. lim⁡x→−6g(x)=0\displaystyle \lim_{x\to -6} g(x)=0
Explanation: The coefficient's sign affects the infinity direction in rational functions. For g(x)=−4x+6g(x) = \dfrac{-4}{x+6}, as x→−6−x \to -6^-, x+6x+6 negative small, −4$/negative=positivelarge,so-4$/negative = positive large, so \lim_{x\to -6^-} g(x) = +\inftycorrectlyrepresentsthis.Thisnotationisvalidduetothenegativenumeratorovernegativedenominatoryieldingpositive.Acommonerrorisignoringthenumerator′ssign,leadingtoincorrectcorrectly represents this. This notation is valid due to the negative numerator over negative denominator yielding positive. A common error is ignoring the numerator's sign, leading to incorrect-\infty.Usingtwo−sidedlimitherewouldbewrongsincesidesdiffer.Toensurecorrectnotation,usethistransferablenotationchecklist:1.Use′lim′toindicatealimit,notthefunctionequalsnotation.2.Specifythedirectionwith+or−forone−sidedlimits.3.Determinethesign(. Using two-sided limit here would be wrong since sides differ. 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 (+\inftyoror-\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.

Question 19

Let g(x)=6x2−9g(x)=\dfrac{6}{x^2-9}. Which limit notation represents the behavior as x→3−x\to 3^-?

  1. lim⁡x→3−g(x)=−∞\displaystyle \lim_{x\to 3^-} g(x)=-\infty (correct answer)
  2. lim⁡x→3+g(x)=−∞\displaystyle \lim_{x\to 3^+} g(x)=-\infty
  3. g(3)=−∞\displaystyle g(3)=-\infty
  4. lim⁡x→3g(x)=0\displaystyle \lim_{x\to 3} g(x)=0
  5. lim⁡x→−3−g(x)=−∞\displaystyle \lim_{x\to -3^-} g(x)=-\infty
Explanation: For g(x) = 6/(x2−9x^2-9), as x→3^-, denominator approaches negative small (since x2x^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.

Question 20

Let f(x)=2x(x−4)f(x)=\dfrac{2}{x(x-4)}. Which limit notation matches the behavior as x→4−x\to 4^-?

  1. lim⁡x→4−f(x)=−∞\displaystyle \lim_{x\to 4^-} f(x)=-\infty (correct answer)
  2. lim⁡x→4+f(x)=−∞\displaystyle \lim_{x\to 4^+} f(x)=-\infty
  3. f(4)=−∞\displaystyle f(4)=-\infty
  4. lim⁡x→4−f(x)=+∞\displaystyle \lim_{x\to 4^-} f(x)=+\infty
  5. lim⁡x→4f(x)=0\displaystyle \lim_{x\to 4} f(x)=0
Explanation: For f(x) = 2/(x(x-4)), as x→4^-, x positive ≈4, (x-4) negative small, denominator negative small, 2/negative = -∞. The notation lim_{x→4^-} f(x) = -∞ fits. This is valid based on left-hand signs. A frequent mistake is using +∞ or equating to f(4) = -∞. Failing to check the (x-4) negativity 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.