AP Calculus BC Quiz: Connecting Infinite Limits And Vertical Asymptotes

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 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.

Question 1

1. $\lim_{x\to4} q(x)=1$, so $y=1$ is a vertical asymptote
2. $\lim_{x\to\infty} q(x)=1$, so $x=1$ is a vertical asymptote
3. $\lim_{x\to4^-} q(x)=-\infty$ and $\lim_{x\to4^+} q(x)=+\infty$, so $x=4$ is a vertical asymptote (correct answer)
4. $\lim_{x\to4^-} q(x)=+\infty$ and $\lim_{x\to4^+} q(x)=-\infty$, so $y=4$ is a vertical asymptote
5. $\lim_{x\to4} q(x)=0$, so there is a removable discontinuity at $x=4$

Explanation: This problem tests linking infinite limits to vertical asymptotes in functions with added constants. For q(x) = 2/(x-4) + 1, the denominator zero at x=4 with nonzero numerator term drives the behavior. Approaching from the left yields negative infinity, and from the right positive infinity, despite the +1 not affecting the infinite trend. These opposing infinite limits confirm a vertical asymptote at x=4. A common distractor confuses this with a horizontal asymptote at y=1 due to the added constant, but horizontal asymptotes relate to limits at infinity, not vertical ones. A general strategy is to isolate terms causing division by zero and evaluate one-sided limits for infinity.

Question 2

1. Vertical asymptote at $y=0$ and $\lim_{x\to0} u(x)=-2$.
2. Horizontal asymptote at $x=0$ and $\lim_{x\to0} u(x)=\infty$.
3. Vertical asymptote at $x=0$ and $\lim_{x\to0^+} u(x)=\infty$, $\lim_{x\to0^-} u(x)=-\infty$. (correct answer)
4. No vertical asymptote because subtracting 2 shifts the graph down.
5. Vertical asymptote at $x=-2$ and $\lim_{x\to-2} u(x)=\infty$.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes in transformed reciprocal functions. For u(x) = 5/x - 2, the 5/x term causes infinite behavior at x=0: positive infinity from the right and negative infinity from the left. Subtracting 2 shifts the graph vertically but doesn't prevent the limits from being infinite. Thus, there's a vertical asymptote at x=0 with opposite-signed infinite limits. Choice D is tempting but fails because the shift affects horizontal asymptotes, not the vertical one from the 1/x core. Always isolate the term causing division by zero and assess its infinite limits, regardless of added constants.

Question 3

1. There is a vertical asymptote at $x=1$.
2. There is a vertical asymptote at $x=4$. (correct answer)
3. $\lim_{x\to 4} q(x)=1$.
4. There is a horizontal asymptote at $x=4$.
5. $\lim_{x\to \infty} q(x)=\infty$.

Explanation: This question tests your ability to identify vertical asymptotes in transformed rational functions. For $q(x)=\frac{2}{x-4}+1$, the rational part $\frac{2}{x-4}$ has a denominator that equals zero when $x=4$. As $x\to 4^-$, we have $(x-4)\to 0^-$, so $\frac{2}{x-4}\to -\infty$, making $q(x)\to -\infty$. As $x\to 4^+$, we have $(x-4)\to 0^+$, so $\frac{2}{x-4}\to +\infty$, making $q(x)\to +\infty$. This confirms a vertical asymptote at $x=4$. Choice D incorrectly calls this a horizontal asymptote, confusing the direction of the asymptote line. Remember that vertical asymptotes are vertical lines $x=a$ where the function has infinite limits, while horizontal asymptotes are horizontal lines $y=b$ describing end behavior.

Question 4

1. There is a vertical asymptote at $x=0$.
2. $\lim_{x\to 2} f(x)$ exists and is finite.
3. There are vertical asymptotes at $x=-2$ and $x=2$. (correct answer)
4. The graph has a horizontal asymptote $y=\dfrac{3}{2}$.
5. $\lim_{x\to -2} f(x)=0$.

Explanation: This question tests your ability to identify vertical asymptotes by analyzing when denominators equal zero and limits become infinite. For $f(x)=\frac{3x+1}{x^2-4}$, we need to find where the denominator $x^2-4=0$, which factors as $(x-2)(x+2)=0$, giving us $x=2$ and $x=-2$. At both these values, the numerator $3x+1$ is non-zero (it equals 7 and -5 respectively), so the function has infinite limits at these points. Since the denominator changes sign as we cross each zero, we get vertical asymptotes at both $x=-2$ and $x=2$. Choice A incorrectly suggests an asymptote at $x=0$ where the function is actually well-defined. When finding vertical asymptotes, always check that the numerator doesn't also equal zero at the same point, as that could indicate a removable discontinuity instead.

Question 5

1. $\lim_{x\to1^+} u(x)=\infty$, so $x=1$ is a vertical asymptote
2. $\lim_{x\to1^+} u(x)=0$, so $x=1$ is not a vertical asymptote (correct answer)
3. $\lim_{x\to1^+} u(x)=1$, so $y=1$ is a vertical asymptote
4. $\lim_{x\to1^+} u(x)=-\infty$, so $x=1$ is a vertical asymptote
5. $\lim_{x\to1^+} u(x)$ does not exist, so $y=0$ is a vertical asymptote

Explanation: This problem tests simplifying algebraic expressions before taking limits. For $u(x)=\frac{x-1}{\sqrt{x-1}}$ with $x>1$, we can rewrite this as $u(x)=\frac{x-1}{(x-1)^{1/2}}=(x-1)^{1-1/2}=(x-1)^{1/2}=\sqrt{x-1}$. As $x\to1^+$, we have $\sqrt{x-1}\to\sqrt{0^+}=0$. Since the limit is finite (zero), there is no vertical asymptote at $x=1$. Choice A incorrectly assumes an infinite limit without simplifying. Always simplify expressions algebraically before evaluating limits to avoid missing cancellations.

Question 6

1. Vertical asymptote at $x=7$ with $\lim_{x\to 7^+}t(x)=+\infty$ (correct answer)
2. Horizontal asymptote at $x=7$ with $\lim_{x\to 7^+}t(x)=+\infty$
3. Vertical asymptote at $y=7$ with $\lim_{x\to 7^+}t(x)=+\infty$
4. Finite limit $\lim_{x\to 7}t(x)=5$ and no vertical asymptote
5. Vertical asymptote at $x=5$ with $\lim_{x\to 5^+}t(x)=+\infty$

Explanation: This problem requires recognizing vertical asymptotes in transformed rational functions through infinite limit behavior. The function t(x) = 2/(x-7) + 5 is a transformation of the basic rational function 1/x. The term 2/(x-7) has a vertical asymptote at x = 7, where the denominator equals zero. As x approaches 7 from the right (x → 7⁺), (x-7) approaches 0 through positive values, making 2/(x-7) approach +∞, and thus t(x) approaches +∞. The constant term +5 shifts the graph vertically but doesn't affect the location of the vertical asymptote. Choice C incorrectly places the asymptote at y = 7, confusing vertical and horizontal asymptotes. For functions of the form a/(x-h) + k, vertical asymptotes always occur at x = h.

Question 7

1. $\lim_{x\to2^-}f(x)=\infty$ and $\lim_{x\to2^+}f(x)=\infty$
2. $\lim_{x\to2^-}f(x)=-\infty$ and $\lim_{x\to2^+}f(x)=\infty$ (correct answer)
3. $\lim_{x\to2^-}f(x)=0$ and $\lim_{x\to2^+}f(x)=0$
4. $\lim_{x\to2^-}f(x)=\infty$ and $\lim_{x\to2^+}f(x)=-\infty$
5. $\lim_{x\to2}f(x)=\dfrac{7}{6}$

Explanation: This question tests your ability to connect infinite limits with vertical asymptote behavior. For $f(x)=\frac{3x+1}{(x-2)(x+4)}$, as $x$ approaches 2, the denominator approaches 0 while the numerator approaches $3(2)+1=7$. When $x$ approaches 2 from the left ($x<2$), the factor $(x-2)$ is negative and $(x+4)$ is positive, making the denominator negative overall, so $f(x)\to-\infty$. When $x$ approaches 2 from the right ($x>2$), the factor $(x-2)$ is positive and $(x+4)$ is positive, making the denominator positive, so $f(x)\to+\infty$. Choice A incorrectly claims both one-sided limits approach $+\infty$, failing to account for the sign change. The key strategy is to analyze the sign of each factor near the asymptote to determine whether the function approaches $+\infty$ or $-\infty$ from each side.

Question 8

1. $\lim_{x\to2^-}r(x)=\infty$ and $\lim_{x\to2^+}r(x)=-\infty$
2. $\lim_{x\to2^-}r(x)=0$ and $\lim_{x\to2^+}r(x)=0$
3. $\lim_{x\to2^-}r(x)=-\infty$ and $\lim_{x\to2^+}r(x)=\infty$ (correct answer)
4. $\lim_{x\to2}r(x)=\dfrac{1}{2}$
5. $\lim_{x\to2^-}r(x)=\infty$ and $\lim_{x\to2^+}r(x)=\infty$

Explanation: This problem requires analyzing infinite limits for a rational function with a factorable denominator. For $r(x)=\frac{2}{x^2-4}=\frac{2}{(x-2)(x+2)}$, near $x=2$, the factor $(x+2)$ is approximately 4 (positive). When approaching from the left ($x<2$), the factor $(x-2)$ is negative, making the denominator negative overall, so $r(x)\to-\infty$. When approaching from the right ($x>2$), the factor $(x-2)$ is positive, making the denominator positive, so $r(x)\to+\infty$. This matches choice C, confirming that $x=2$ is a vertical asymptote with different one-sided infinite limits. Choice E incorrectly claims both limits are $+\infty$, missing the sign change. The strategy is to determine the sign of each factor in the denominator on each side of the potential asymptote.

Question 9

1. $x=-3$ is a vertical asymptote because at least one one-sided limit is infinite. (correct answer)
2. $x=-3$ is a horizontal asymptote because one limit equals 5.
3. $x=-3$ is removable because one one-sided limit is finite.
4. $x=-3$ is neither an asymptote nor a discontinuity because limits disagree.
5. $x=-3$ is a vertical asymptote only if both one-sided limits are infinite.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes when one-sided limits differ. If the left-hand limit at x=-3 is negative infinity and the right-hand is finite (5), the infinite side indicates unbounded behavior approaching x=-3. This supports a vertical asymptote, even if only one side is infinite. The finite side means the graph approaches 5 from the right, but the infinite side dominates for asymptote classification. Choice E is tempting but fails because vertical asymptotes require at least one infinite one-sided limit, not necessarily both. Always evaluate both one-sided limits separately; an infinite result from either confirms a vertical asymptote.

Question 10

1. $y=-2$ is a vertical asymptote because the function is unbounded
2. $x=-2$ is a vertical asymptote because the one-sided limits are infinite (correct answer)
3. $\lim_{x\to-2} f(x)$ exists and equals $0$, so there is no asymptote
4. $y=0$ is a vertical asymptote because the limits differ
5. $x=-2$ is a horizontal asymptote because $f(x)$ is unbounded

Explanation: This question evaluates recognizing vertical asymptotes from described infinite limit behavior on graphs. The graph approaching positive infinity from the left of x=-2 and negative infinity from the right indicates unbounded behavior. These opposing infinite one-sided limits define a vertical asymptote at x=-2. The function is undefined or discontinuous there, consistent with asymptotic behavior. A tempting distractor confuses vertical with horizontal asymptotes due to the unbounded mention, but horizontal ones involve limits as x approaches infinity. Generally, confirm vertical asymptotes by verifying if one-sided limits at a point diverge to plus or minus infinity.

Question 11

1. Vertical asymptotes at $x=-2$ and $x=5$ (correct answer)
2. Horizontal asymptote at $x=-2$ and $x=5$
3. Finite limits at $x=-2$ and $x=5$
4. Vertical asymptote at $y=-2$ and $y=5$
5. No vertical asymptotes because the degree of numerator is less

Explanation: This question tests your ability to identify vertical asymptotes by analyzing infinite limits of rational functions. The function f(x) = (3x-1)/[(x+2)(x-5)] has its denominator equal to zero when x = -2 and x = 5, while the numerator is non-zero at these values. As x approaches -2 or 5, the denominator approaches 0 while the numerator approaches finite non-zero values, causing the function to approach ±∞. This creates vertical asymptotes at x = -2 and x = 5. Choice B incorrectly identifies these as horizontal asymptotes, which would be lines of the form y = k, not x = k. When a rational function's denominator has zeros that don't cancel with the numerator, vertical asymptotes occur at those x-values.

Question 12

1. Vertical asymptote at $x=3$ with $\lim_{x\to 3^+}h(x)=-\infty$ (correct answer)
2. Vertical asymptote at $y=3$ with $\lim_{x\to 3^+}h(x)=-\infty$
3. Horizontal asymptote at $y=3$ with $\lim_{x\to 3^+}h(x)=-\infty$
4. Finite limit $\lim_{x\to 3^+}h(x)=0$ and no vertical asymptote
5. Vertical asymptote at $x=0$ with $\lim_{x\to 0^+}h(x)=-\infty$

Explanation: This question examines vertical asymptotes in logarithmic functions through infinite limit behavior. The function h(x) = ln(x-3) is defined only for x > 3, where the argument (x-3) must be positive. As x approaches 3 from the right (x → 3⁺), the argument (x-3) approaches 0 through positive values, and ln(x-3) approaches -∞. This creates a vertical asymptote at x = 3 with the specific one-sided limit behavior described in choice A. Choice B incorrectly places the asymptote at y = 3 instead of x = 3, confusing vertical and horizontal asymptotes. For logarithmic functions ln(x-a), vertical asymptotes always occur at x = a where the argument becomes zero.

Question 13

1. Vertical asymptotes at $x=4$ and $x=2$
2. Vertical asymptote at $x=2$ only (correct answer)
3. Horizontal asymptote at $x=2$ only
4. Finite limits at $x=2$ and $x=4$
5. Vertical asymptote at $x=4$ only

Explanation: This problem requires recognizing how factor cancellation affects vertical asymptotes through infinite limit analysis. The function g(x) = [(x-4)(x+1)]/[(x-4)(x-2)] has a common factor of (x-4) in both numerator and denominator that cancels, leaving g(x) = (x+1)/(x-2) for x ≠ 4. After cancellation, the only zero of the denominator is x = 2, where the simplified function approaches ±∞. Therefore, there is a vertical asymptote only at x = 2, not at x = 4. Choice A incorrectly assumes vertical asymptotes at both x = 4 and x = 2, failing to recognize that the cancellation at x = 4 creates a removable discontinuity instead. Remember that vertical asymptotes occur only where the denominator approaches zero without a corresponding zero in the numerator after all possible cancellations.

Question 14

1. $\lim_{x\to2} f(x)=\dfrac{7}{0}$, so there is no asymptote
2. $\lim_{x\to2^-} f(x)=-\infty$ and $\lim_{x\to2^+} f(x)=+\infty$, so $x=2$ is a vertical asymptote (correct answer)
3. $\lim_{x\to2^-} f(x)=+\infty$ and $\lim_{x\to2^+} f(x)=-\infty$, so $y=2$ is a vertical asymptote
4. $\lim_{x\to2} f(x)=\dfrac{7}{0}$, so $y=\dfrac{7}{0}$ is a horizontal asymptote
5. $\lim_{x\to2} f(x)=\dfrac{7}{4}$, so there is a removable discontinuity at $x=2$

Explanation: This question tests the skill of using infinite limits to identify vertical asymptotes in rational functions. The function f(x) = (3x+1)/(x²-4) has a denominator that factors to (x-2)(x+2), which is zero at x=2 while the numerator is 7, indicating a potential vertical asymptote. As x approaches 2 from the left, the function goes to negative infinity because the denominator is negative and the numerator positive, creating a large negative value. From the right, it approaches positive infinity as both are positive, confirming the asymptote at x=2. A tempting distractor might suggest a removable discontinuity because the limit appears as 7/0, but this undefined form actually signals infinite behavior rather than a finite limit after simplification. To generally identify vertical asymptotes, evaluate one-sided limits where the denominator is zero and the numerator is not, checking if they approach infinity.

Question 15

1. $\lim_{x\to -1} p(x)=0$.
2. There is a horizontal asymptote at $x=-1$.
3. $\lim_{x\to -1^-} p(x)=-\infty$ and $\lim_{x\to -1^+} p(x)=\infty$.
4. $\lim_{x\to -1^-} p(x)=\infty$ and $\lim_{x\to -1^+} p(x)=\infty$. (correct answer)
5. There is a vertical asymptote at $y=0$.

Explanation: This problem involves analyzing infinite limits for a rational function with a squared denominator. For $p(x)=\frac{1}{(x+1)^2}$, the denominator equals zero when $x=-1$, while the numerator remains 1 (non-zero). Since $(x+1)^2$ is always positive for $x\neq -1$, as $x$ approaches $-1$ from either direction, the denominator approaches $0^+$. This means $\lim_{x\to -1^-} p(x)=\frac{1}{0^+}=\infty$ and $\lim_{x\to -1^+} p(x)=\frac{1}{0^+}=\infty$, confirming a vertical asymptote at $x=-1$ with the function approaching positive infinity from both sides. Choice C incorrectly suggests the limits have opposite signs, which would require the denominator to change sign. For functions with squared factors in the denominator, the vertical asymptote behavior is always the same from both sides.

Question 16

1. $x=1$ is a vertical asymptote because the denominator is zero there.
2. $x=1$ is a horizontal asymptote and $\lim_{x\to1} g(x)=\infty$.
3. $x=1$ is a removable discontinuity and $\lim_{x\to1} g(x)=\dfrac{5}{-2}$. (correct answer)
4. $x=1$ is a vertical asymptote and $\lim_{x\to1} g(x)=0$.
5. $x=1$ is neither a discontinuity nor an asymptote because factors cancel.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes through simplification of rational expressions. For g(x) = (x-1)(x+4)/((x-1)(x-3)), canceling the (x-1) factor for x ≠ 1 simplifies to (x+4)/(x-3), showing a hole at x=1 rather than unbounded behavior. The limit as x approaches 1 is (1+4)/(1-3) = 5/(-2) = -2.5, which is finite from both sides. Thus, there is no infinite limit, indicating a removable discontinuity instead of an asymptote. Choice A is tempting but fails because after cancellation, the denominator is not zero at x=1, so no asymptote forms. Always simplify rational functions before assessing limits to distinguish asymptotes from removable discontinuities.

Question 17

1. There is a horizontal asymptote at $x=-1$ and $\lim_{x\to-1} p(x)=\infty$.
2. There is a vertical asymptote at $x=-1$ and $\lim_{x\to-1} p(x)=\infty$. (correct answer)
3. There is a vertical asymptote at $y=-1$ and $\lim_{x\to-1} p(x)=-\infty$.
4. There is no vertical asymptote because the exponent makes the limit finite.
5. There is a vertical asymptote at $x=-1$ and $\lim_{x\to-1} p(x)=0$.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes in functions with even-powered denominators. For p(x) = 2/(x+1)^2, the denominator approaches zero at x=-1, and since it's squared, it approaches from the positive side regardless of direction. Thus, the function approaches positive infinity from both sides as x nears -1. The numerator is constant and positive, reinforcing the infinite limit. Choice E is tempting but fails because the limit is infinite, not zero; zero would require the numerator to approach zero as well. Always consider the sign and power of terms near potential asymptotes to determine limit directions.

Question 18

1. Vertical asymptotes at $x=\pm2$; $\lim_{x\to2^-} s(x)=-\infty$ and $\lim_{x\to2^+} s(x)=\infty$. (correct answer)
2. Horizontal asymptotes at $x=\pm2$; $\lim_{x\to2} s(x)=0$.
3. Vertical asymptotes at $y=\pm2$; $\lim_{x\to2} s(x)=\infty$.
4. No vertical asymptotes because the denominator is quadratic.
5. Vertical asymptote only at $x=2$ because $x=-2$ makes the limit finite.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes in rational functions with quadratic denominators. For s(x) = 1/(x^2-4), the denominator zeros at x=±2, and since the numerator is 1, limits are infinite there. At x=2, from the left the denominator approaches negative (leading to -∞), and from the right positive (leading to +∞); similar opposite behaviors occur at x=-2. These infinite one-sided limits confirm vertical asymptotes at both points. Choice D is tempting but fails because a quadratic denominator still causes infinite limits at its roots, just like linear ones. Always find denominator roots and evaluate one-sided limits to pinpoint asymptotes and their directions.

Question 19

1. $\lim_{x\to1^+} g(x)=-\infty$, so $x=1$ is a vertical asymptote (correct answer)
2. $\lim_{x\to1^+} g(x)=\infty$, so $y=1$ is a vertical asymptote
3. $\lim_{x\to1^+} g(x)=0$, so $x=1$ is a vertical asymptote
4. $\lim_{x\to1^+} g(x)=-\infty$, so $y=-\infty$ is a horizontal asymptote
5. $\lim_{x\to1^+} g(x)=1$, so $x=1$ is not an asymptote

Explanation: This question requires understanding infinite limits for logarithmic functions and their connection to vertical asymptotes. For $g(x)=\ln(x-1)$, the domain requires $x>1$, so we can only approach $x=1$ from the right. As $x\to1^+$, the argument $(x-1)$ approaches $0^+$, and since $\ln(0^+)=-\infty$, we have $\lim_{x\to1^+} g(x)=-\infty$. This infinite limit confirms that $x=1$ is a vertical asymptote. Choice B incorrectly identifies $y=1$ as the asymptote location instead of $x=1$. For vertical asymptotes, always identify the x-value where the function approaches infinity, not a y-value.

Question 20

1. There is a vertical asymptote at $x=\pi$ and $\lim_{x\to\pi} t(x)=\infty$.
2. There is a removable discontinuity at $x=\pi$ and $\lim_{x\to\pi} t(x)=-1$. (correct answer)
3. There is a horizontal asymptote at $y=\pi$ and $\lim_{x\to\pi} t(x)=0$.
4. There is a vertical asymptote at $y=\pi$ and $\lim_{x\to\pi} t(x)=-1$.
5. There is no discontinuity because $\sin(x)$ is continuous.

Explanation: This question tests your ability to connect infinite limits with vertical asymptotes in trigonometric rational functions. For t(x) = sin(x)/(x-π), as x approaches π, it's 0/0 indeterminate, but simplifying or using L'Hôpital's rule gives limit cos(π) = -1, which is finite. Thus, there's no infinite behavior, indicating a removable discontinuity. The sine function's continuity doesn't prevent the hole from the zero denominator. Choice A is tempting but fails because the numerator also approaches zero, making the limit finite after analysis. Always check if both numerator and denominator approach zero; if so, simplify to see if the limit is finite, ruling out asymptotes.