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AP Calculus BC Quiz

AP Calculus BC Quiz: Limits At Infinity And Horizontal Asymptotes

Practice Limits At Infinity And Horizontal Asymptotes in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Let t(x)=exex+5t(x)=\dfrac{e^x}{e^x+5}t(x)=ex+5ex​. What is the horizontal asymptote of ttt as x→∞x\to\inftyx→∞?

Select an answer to continue

What this quiz covers

This quiz focuses on Limits At Infinity And Horizontal Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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

Let t(x)=exex+5t(x)=\dfrac{e^x}{e^x+5}t(x)=ex+5ex​. What is the horizontal asymptote of ttt as x→∞x\to\inftyx→∞?

  1. y=0y=0y=0
  2. x=0x=0x=0
  3. y=1y=1y=1 (correct answer)
  4. y=5y=5y=5
  5. x=1x=1x=1

Explanation: This question involves finding the horizontal asymptote of t(x) = e^x/(e^x + 5) as x approaches infinity. As x approaches infinity, e^x grows exponentially and dominates the constant 5 in the denominator. We can divide both numerator and denominator by e^x to get 1/(1 + 5/e^x). Since e^x approaches infinity, 5/e^x approaches 0, giving us a limit of 1/(1 + 0) = 1. Therefore, the horizontal asymptote is y = 1. Choice A (y = 0) would be correct if x approached negative infinity instead. For exponential functions, identify which terms dominate as x approaches infinity.

Question 2

For F(x)=3x2+76x2−5F(x)=\dfrac{3x^2+7}{6x^2-5}F(x)=6x2−53x2+7​, what is the horizontal asymptote of FFF?

  1. y=12y=\dfrac{1}{2}y=21​ (correct answer)
  2. y=2y=2y=2
  3. x=56x=\sqrt{\dfrac{5}{6}}x=65​​
  4. x=0x=0x=0
  5. y=0y=0y=0

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For F(x) = (3x² + 7)/(6x² - 5), degrees are both 2, so the horizontal asymptote is y = 3/6 = 1/2. Even degree maintains the ratio. Negligible lower terms. A tempting distractor is y = 2, inverting the fraction. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

Question 3

Let J(x)=2x+1x−3J(x)=\dfrac{2x+1}{x-3}J(x)=x−32x+1​. What is the horizontal asymptote of J(x)J(x)J(x)?

  1. y=2y=2y=2 (correct answer)
  2. x=3x=3x=3
  3. y=12y=\dfrac{1}{2}y=21​
  4. y=0y=0y=0
  5. x=0x=0x=0

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For J(x) = (2x + 1)/(x - 3), degrees both 1, horizontal asymptote is y = 2/1 = 2. Odd degree but same limit from both sides. Leading terms dominate. A tempting distractor is y = 1/2, inverting the ratio. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

Question 4

For v(x)=ln⁡(x)xv(x)=\dfrac{\ln(x)}{x}v(x)=xln(x)​ with x>0x>0x>0, what is the value of lim⁡x→∞v(x)\lim_{x\to\infty} v(x)limx→∞​v(x)?

  1. ∞\infty∞
  2. 111
  3. 000 (correct answer)
  4. x=0x=0x=0
  5. −∞-\infty−∞

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes for logarithmic functions. As x approaches infinity, ln(x) grows much slower than x, so v(x) = ln(x)/x approaches 0. This can be confirmed using L'Hôpital's rule since it's ∞/∞ form: derivative is (1/x)/1 = 1/x → 0. Thus, the limit is 0. A tempting distractor like ∞ might arise if someone thinks ln(x) grows faster than x, which is incorrect. A transferable strategy for horizontal asymptotes with logs and polynomials is to recognize that logarithms grow slower than any positive power of x, leading to a limit of zero when divided by x.

Question 5

Let r(x)=ln⁡(x)xr(x)=\dfrac{\ln(x)}{x}r(x)=xln(x)​. Which value equals lim⁡x→∞r(x)\lim_{x\to\infty} r(x)limx→∞​r(x)?

  1. 111
  2. 000 (correct answer)
  3. ∞\infty∞
  4. x=0x=0x=0
  5. y=1y=1y=1

Explanation: This question assesses understanding of limits at infinity, particularly for logarithmic over polynomial growth. For r(x) = ln(x)/x, as x approaches infinity, ln(x) grows slower than any positive power of x, including x itself. Therefore, the ratio approaches 0 by known limit properties or L'Hôpital's rule, since it's ∞/∞ form: derivative is (1/x)/1 = 1/x → 0. This means the function approaches the x-axis. A tempting distractor could be infinity, possibly confusing the growth rates of log and linear functions. In general, compare growth rates of functions to determine limits at infinity, using tools like L'Hôpital for indeterminate forms.

Question 6

Let t(x)=x2−4x+12x2+7t(x)=\dfrac{x^2-4x+1}{2x^2+7}t(x)=2x2+7x2−4x+1​. What is lim⁡x→−∞t(x)\lim_{x\to -\infty} t(x)limx→−∞​t(x)?

  1. 12\dfrac{1}{2}21​ (correct answer)
  2. 222
  3. −12-\dfrac{1}{2}−21​
  4. x=72x=\dfrac{7}{2}x=27​
  5. y=0y=0y=0

Explanation: This question assesses understanding of limits at infinity, for rational functions as x approaches negative infinity. For t(x) = (x² - 4x + 1)/(2x² + 7), divide by x²: (1 - 4/x + 1/x²)/(2 + 7/x²). As x approaches negative infinity, the fractional terms approach zero, yielding 1/2. The sign consistency arises because both leading terms are positive and degrees match. A tempting distractor could be -1/2, maybe from incorrectly introducing a negative sign for negative infinity. In general, for equal-degree rationals, the asymptote is the leading coefficient ratio, independent of infinity's direction.

Question 7

For s(x)=x2+9xs(x)=\dfrac{\sqrt{x^2+9}}{x}s(x)=xx2+9​​, what is the horizontal asymptote as x→∞x\to\inftyx→∞?

  1. y=−1y=-1y=−1
  2. y=0y=0y=0
  3. y=1y=1y=1 (correct answer)
  4. x=0x=0x=0
  5. x=±3ix=\pm 3ix=±3i

Explanation: This question assesses understanding of limits at infinity, for functions involving square roots. For s(x) = √(x² + 9)/x, factor x² out of the square root: √(x²(1 + 9/x²))/x = |x| √(1 + 9/x²)/x. As x approaches positive infinity, |x|/x = 1, and √(1 + 9/x²) → 1, so the limit is 1, giving y=1 as the asymptote. Note that for x → -∞, it would be -1, but the question specifies x → ∞. A tempting distractor might be y=-1, perhaps overlooking the direction of infinity. In general, simplify expressions by factoring the dominant term inside roots or powers to find horizontal asymptotes.

Question 8

For f(x)=3x2−5x+12x2+7f(x)=\dfrac{3x^2-5x+1}{2x^2+7}f(x)=2x2+73x2−5x+1​, what is the horizontal asymptote of fff as x→∞x\to\inftyx→∞?

  1. y=32y=\dfrac{3}{2}y=23​ (correct answer)
  2. x=72x=\dfrac{7}{2}x=27​
  3. y=23y=\dfrac{2}{3}y=32​
  4. y=0y=0y=0
  5. x=0x=0x=0

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes for rational functions. To find the horizontal asymptote of f(x) as x approaches infinity, divide both the numerator and denominator by x², the highest power in the denominator, yielding (3 - 5/x + 1/x²)/(2 + 7/x²). As x approaches infinity, the terms with 1/x and 1/x² approach zero, so the limit simplifies to 3/2. Therefore, the horizontal asymptote is y = 3/2. A tempting distractor like y = 2/3 might arise if someone mistakenly swaps the leading coefficients of the numerator and denominator. A transferable strategy for horizontal asymptotes in rational functions is to compare the degrees of the numerator and denominator; if equal, the asymptote is the ratio of the leading coefficients.

Question 9

For s(x)=x2+4xxs(x)=\dfrac{\sqrt{x^2+4x}}{x}s(x)=xx2+4x​​, what is lim⁡x→∞s(x)\lim_{x\to\infty} s(x)limx→∞​s(x)?​

  1. 000
  2. 111 (correct answer)
  3. −1-1−1
  4. ∞\infty∞
  5. x=−4x=-4x=−4

Explanation: This problem tests limits at infinity involving square roots, requiring careful algebraic manipulation. For s(x) = √(x² + 4x)/x, we need to factor out x² from under the radical: √(x²(1 + 4/x))/x = |x|√(1 + 4/x)/x. As x → ∞, x is positive so |x| = x, giving us x√(1 + 4/x)/x = √(1 + 4/x). As x approaches infinity, 4/x approaches 0, so the limit is √(1 + 0) = 1. Students often incorrectly think the answer is 0 or infinity without properly simplifying the radical expression. When dealing with radicals in limits at infinity, always factor out the highest power of x from under the radical first.

Question 10

Let g(x)=4x−9x+3g(x)=\dfrac{4x-9}{x+3}g(x)=x+34x−9​. What is lim⁡x→∞g(x)\lim_{x\to\infty} g(x)limx→∞​g(x)?​

  1. 444 (correct answer)
  2. 000
  3. −3-3−3
  4. 14\dfrac{1}{4}41​
  5. 333

Explanation: This question asks for the limit at infinity of a rational function, which determines its horizontal asymptote. For g(x) = (4x - 9)/(x + 3), the numerator has degree 1 and the denominator has degree 1. When degrees are equal, we find the limit by dividing the leading coefficients: 4/1 = 4. As x approaches infinity, the constant terms -9 and +3 become negligible compared to the x terms. Students might incorrectly choose 0 (choice B) thinking the degree in the denominator is higher, or -3 (choice C) by looking at the denominator's constant. Remember: for rational functions with equal degrees in numerator and denominator, the horizontal asymptote is the ratio of leading coefficients.

Question 11

A graph of v(x)v(x)v(x) approaches the line y=−2y=-2y=−2 as x→±∞x\to\pm\inftyx→±∞; what is lim⁡x→∞v(x)\lim_{x\to\infty} v(x)limx→∞​v(x)?

  1. 222
  2. −2-2−2 (correct answer)
  3. 000
  4. x=−2x=-2x=−2
  5. ∞\infty∞

Explanation: This question assesses understanding of limits at infinity, relating graph behavior to limit values. The graph of v(x) approaching y=−2y=-2y=−2 as x goes to both positive and negative infinity means the horizontal asymptote is y=−2y=-2y=−2. Therefore, specifically for x approaching infinity, the limit is −2-2−2. This is directly from the definition of horizontal asymptotes. A tempting distractor might be infinity, possibly misinterpreting 'approaches' as unbounded growth. In general, identify horizontal asymptotes from limit behavior at infinity to understand long-term function trends.

Question 12

For s(x)=x2+9xs(x)=\dfrac{\sqrt{x^2+9}}{x}s(x)=xx2+9​​, what is lim⁡x→∞s(x)\lim_{x\to\infty} s(x)limx→∞​s(x)?

  1. 000
  2. 111 (correct answer)
  3. ∞\infty∞
  4. x=0x=0x=0
  5. −1-1−1

Explanation: This problem tests limits at infinity involving radicals: s(x) = √(x² + 9)/x as x approaches infinity. To evaluate this, we factor x² from inside the radical: √(x²(1 + 9/x²))/x = |x|√(1 + 9/x²)/x. For x > 0 (approaching positive infinity), |x| = x, so we get x√(1 + 9/x²)/x = √(1 + 9/x²). As x approaches infinity, 9/x² approaches 0, so the limit is √1 = 1. Choice A (0) incorrectly assumes the radical grows slower than x. When radicals contain polynomials, factor out the highest power to find limits at infinity.

Question 13

For q(x)=2x2−1x3+3xq(x)=\dfrac{2x^2-1}{x^3+3x}q(x)=x3+3x2x2−1​, what is the horizontal asymptote as x→∞x\to\inftyx→∞?

  1. x=0x=0x=0
  2. y=2y=2y=2
  3. y=0y=0y=0 (correct answer)
  4. x=−3x=-3x=−3
  5. y=23y=\dfrac{2}{3}y=32​

Explanation: This problem requires finding the horizontal asymptote of q(x) = (2x² - 1)/(x³ + 3x) as x approaches infinity. The numerator has degree 2 while the denominator has degree 3, meaning the denominator grows faster than the numerator. When the denominator's degree exceeds the numerator's degree by any amount, the horizontal asymptote is y = 0. As x approaches infinity, the function behaves like 2x²/x³ = 2/x, which approaches 0. Choice B (y = 2) incorrectly ignores the degree difference. Remember: higher denominator degree always yields horizontal asymptote y = 0.

Question 14

Given v(x)=2x3−8x3+4xv(x)=\dfrac{2x^3-8}{x^3+4x}v(x)=x3+4x2x3−8​, what is the horizontal asymptote as x→−∞x\to-\inftyx→−∞?

  1. y=2y=2y=2 (correct answer)
  2. y=12y=\dfrac{1}{2}y=21​
  3. x=−4x=-4x=−4
  4. y=0y=0y=0
  5. x=2x=2x=2

Explanation: This question asks for the horizontal asymptote of v(x) = (2x³ - 8)/(x³ + 4x) as x approaches negative infinity. Both numerator and denominator have degree 3, so we find the asymptote by dividing the leading coefficients: 2/1 = 2. As x approaches negative infinity, the highest-degree terms dominate, making the function behave like 2x³/x³ = 2. The horizontal asymptote is the same whether x approaches positive or negative infinity for rational functions with equal degrees. Choice D (y = 0) would only be correct if the denominator had higher degree. For equal-degree rational functions, the horizontal asymptote is always the ratio of leading coefficients.

Question 15

Given p(x)=4x2+9x2−16p(x)=\dfrac{4x^2+9}{x^2-16}p(x)=x2−164x2+9​, what is lim⁡x→∞p(x)\lim_{x\to\infty} p(x)limx→∞​p(x)?

  1. 444 (correct answer)
  2. 000
  3. Does not exist because x=4x=4x=4 is a vertical asymptote
  4. 14\dfrac{1}{4}41​
  5. 161616

Explanation: This question tests evaluating the limit of p(x) = (4x² + 9)/(x² - 16) as x approaches infinity. Since both numerator and denominator have degree 2, we find the limit by taking the ratio of leading coefficients: 4/1 = 4. As x becomes very large, the constant terms become negligible, and the function behaves like 4x²/x² = 4. The vertical asymptotes at x = ±4 (where the denominator equals zero) don't affect the horizontal behavior at infinity. Choice B (0) would only be correct if the denominator had higher degree. For equal-degree rational functions, the limit at infinity equals the ratio of leading coefficients.

Question 16

For u(x)=ln⁡xxu(x)=\dfrac{\ln x}{x}u(x)=xlnx​ (with x>0x>0x>0), what is lim⁡x→∞u(x)\lim_{x\to\infty} u(x)limx→∞​u(x)?

  1. 111
  2. 000 (correct answer)
  3. ∞\infty∞
  4. y=ln⁡xy=\ln xy=lnx
  5. x=0x=0x=0

Explanation: This problem requires evaluating the limit of u(x) = (ln x)/x as x approaches infinity. This is a classic limit where logarithmic growth is compared to linear growth. Since ln x grows much slower than x (logarithmic vs. linear growth), the fraction approaches 0 as x approaches infinity. This can be proven using L'Hôpital's rule: taking derivatives gives us 1/x divided by 1, which equals 1/x, confirming the limit is 0. Choice A (1) incorrectly assumes ln x and x grow at the same rate. Remember that logarithmic functions grow slower than any positive power of x.

Question 17

For h(x)=7−2x35x3+xh(x)=\dfrac{7-2x^3}{5x^3+x}h(x)=5x3+x7−2x3​, what is the horizontal asymptote as x→−∞x\to-\inftyx→−∞?

  1. y=−25y=-\dfrac{2}{5}y=−52​ (correct answer)
  2. y=25y=\dfrac{2}{5}y=52​
  3. y=0y=0y=0
  4. x=0x=0x=0
  5. x=−5x=-5x=−5

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For h(x) = (7 - 2x³)/(5x³ + x), the degrees are both 3, so the horizontal asymptote is the ratio of leading coefficients, -2/5, but we must consider the direction as x approaches negative infinity. Substituting x = -t where t approaches infinity yields (7 + 2t³)/(-5t³ - t), which simplifies to -2/5 for large t. Thus, the horizontal asymptote is y = -2/5. A tempting distractor is y = 2/5, which ignores the sign change due to approaching negative infinity. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients, accounting for signs in limits to negative infinity.

Question 18

Given u(x)=6x2−3x2+10u(x)=\dfrac{6x^2}{-3x^2+10}u(x)=−3x2+106x2​, what is the horizontal asymptote of u(x)u(x)u(x)?

  1. y=−2y=-2y=−2 (correct answer)
  2. y=2y=2y=2
  3. x=103x=\dfrac{10}{3}x=310​
  4. y=0y=0y=0
  5. x=0x=0x=0

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For u(x) = (6x²)/(-3x² + 10), the degrees are both 2, so the horizontal asymptote is y = 6/(-3) = -2. Even degree ensures consistency from both sides. Lower terms are negligible. A tempting distractor is y = 2, forgetting the negative sign. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

Question 19

For f(x)=5x2−32x2+7x+1f(x)=\dfrac{5x^2-3}{2x^2+7x+1}f(x)=2x2+7x+15x2−3​, what is the horizontal asymptote of fff as x→∞x\to\inftyx→∞?

  1. y=25y=\dfrac{2}{5}y=52​
  2. y=52y=\dfrac{5}{2}y=25​ (correct answer)
  3. x=−17x=-\dfrac{1}{7}x=−71​
  4. y=0y=0y=0
  5. x=0x=0x=0

Explanation: This problem involves finding limits at infinity to determine horizontal asymptotes. For the rational function f(x) = (5x² - 3)/(2x² + 7x + 1), the degrees of the numerator and denominator are both 2, so the horizontal asymptote is given by the ratio of the leading coefficients, which is 5/2. As x approaches infinity, the lower-degree terms become negligible, leaving the function behaving like 5x²/2x². Therefore, the limit as x approaches infinity is 5/2, corresponding to the horizontal asymptote y = 5/2. A tempting distractor is y = 2/5, which would result from incorrectly swapping the leading coefficients. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

Question 20

For f(x)=3x2−5x+12x2+7f(x)=\dfrac{3x^2-5x+1}{2x^2+7}f(x)=2x2+73x2−5x+1​, what is the horizontal asymptote as x→∞x\to\inftyx→∞?

  1. y=32y=\dfrac{3}{2}y=23​ (correct answer)
  2. y=23y=\dfrac{2}{3}y=32​
  3. x=32x=\dfrac{3}{2}x=23​
  4. y=0y=0y=0
  5. x=−72x=-\dfrac{7}{2}x=−27​

Explanation: This problem tests finding horizontal asymptotes by evaluating limits at infinity. For the rational function f(x) = (3x² - 5x + 1)/(2x² + 7), we compare the degrees of the numerator and denominator polynomials. Since both have degree 2, the horizontal asymptote is y = 3/2, found by taking the ratio of the leading coefficients (3 and 2). As x approaches infinity, the highest-degree terms dominate, making the function behave like 3x²/2x² = 3/2. Choice D (y = 0) incorrectly assumes the denominator has higher degree. When degrees are equal in rational functions, always divide the leading coefficients to find the horizontal asymptote.