AP Calculus AB Quiz: Limits At Infinity And Horizontal Asymptotes

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 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.

Question 1

1. $49$
2. $2000$ (correct answer)
3. $50$
4. The population grows infinitely.

Explanation: To find the long-term behavior of the population, we need to evaluate the limit of $P(t)$ as $t \to \infty$. As $t \to \infty$, the term $-0.2t$ approaches $-\infty$, so $e^{-0.2t}$ approaches 0. The limit is then $\lim_{t \to \infty} \frac{2000}{1 + 49e^{-0.2t}} = \frac{2000}{1 + 49(0)} = 2000$. This value is the carrying capacity of the population.

Question 2

1. $\lim_{x \to \infty} f(x) = 5$ or $\lim_{x \to -\infty} f(x) = 5$ (correct answer)
2. $\lim_{x \to 5} f(x) = \infty$
3. $f(5)$ is undefined.
4. The graph of $f(x)$ cannot intersect the line $y=5$.

Explanation: The definition of a horizontal asymptote at $y=c$ is that either $\lim_{x \to \infty} f(x) = c$ or $\lim_{x \to -\infty} f(x) = c$ (or both). Therefore, if $y=5$ is a horizontal asymptote, at least one of these limit statements must be true. Option B describes a vertical asymptote at $x=5$. Option C is not necessarily true. Option D is a common misconception; a function can cross its horizontal asymptote.

Question 3

1. $0$ mg/L (correct answer)
2. $150$ mg/L
3. $75$ mg/L
4. The concentration increases without bound.

Explanation: The question asks for the long-term behavior of the concentration, which is found by evaluating the limit of $C(t)$ as $t \to \infty$. We evaluate $\lim_{t \to \infty} \frac{150t}{t^2 + 4}$. Since the degree of the denominator (2) is greater than the degree of the numerator (1), the limit is 0. So, the long-term concentration of the drug approaches 0 mg/L.

Question 4

1. $\frac{1}{4}$
2. $2$
3. $\frac{1}{2}$ (correct answer)
4. $8$

Explanation: To find the limit of this rational expression at infinity, we compare the leading terms of the numerator and the denominator. The leading term of the numerator is $(2x)^3 = 8x^3$. The leading term of the denominator is $x(4x)^2 = x(16x^2) = 16x^3$. The limit is the ratio of the coefficients of these leading terms, which is $\frac{8}{16} = \frac{1}{2}$.

Question 5

1. $-\frac{3}{2}$ (correct answer)
2. $0$
3. $\frac{3}{7}$
4. The limit does not exist.

Explanation: To find the limit of a rational function as $x \to \infty$, we compare the degrees of the numerator and the denominator. Since the degrees are the same (both are 2), the limit is the ratio of the leading coefficients. The leading coefficient of the numerator is 3, and the leading coefficient of the denominator is -2. Therefore, the limit is $\frac{3}{-2} = -\frac{3}{2}$.

Question 6

1. $y = \frac{2}{3}$ only
2. $y = -\frac{2}{3}$ only
3. $y = \frac{2}{3}$ and $y = -\frac{2}{3}$ (correct answer)
4. $y = \frac{4}{3}$ and $y = -\frac{4}{3}$

Explanation: We need to evaluate the limits as $x \to \infty$ and $x \to -\infty$. For $x > 0$, $x = \sqrt{x^2}$. So, $\lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{3x - 5} = \lim_{x \to \infty} \frac{\sqrt{x^2(4 + 1/x^2)}}{x(3 - 5/x)} = \lim_{x \to \infty} \frac{x\sqrt{4 + 1/x^2}}{x(3 - 5/x)} = \frac{\sqrt{4}}{3} = \frac{2}{3}$. For $x < 0$, $x = -\sqrt{x^2}$, so $\sqrt{x^2} = -x$. So, $\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{3x - 5} = \lim_{x \to -\infty} \frac{\sqrt{x^2(4 + 1/x^2)}}{x(3 - 5/x)} = \lim_{x \to -\infty} \frac{-x\sqrt{4 + 1/x^2}}{x(3 - 5/x)} = \frac{-\sqrt{4}}{3} = -\frac{2}{3}$. Thus, there are two horizontal asymptotes: $y = \frac{2}{3}$ and $y = -\frac{2}{3}$.

Question 7

1. $0$
2. $-\frac{5}{3}$ (correct answer)
3. $1$
4. The limit does not exist.

Explanation: As $x \to -\infty$, the term $e^x$ approaches 0. Similarly, $e^{2x} = (e^x)^2$ also approaches 0. Substituting these values into the expression, we get $\lim_{x \to -\infty} \frac{e^x + 5}{e^{2x} - 3} = \frac{0 + 5}{0 - 3} = -\frac{5}{3}$.

Question 8

1. $\lim_{x \to \infty} \frac{x^{100}}{e^x}$ (correct answer)
2. $\lim_{x \to \infty} \frac{e^{2x}}{e^x}$
3. $\lim_{x \to \infty} \frac{\ln(x)}{x^{-1}}$
4. $\lim_{x \to \infty} \frac{5^x}{4^x}$

Explanation: This question concerns the relative growth rates of functions. Exponential functions like $e^x$ grow faster than any polynomial function, including $x^{100}$. Therefore, as $x \to \infty$, the denominator grows much faster than the numerator, and the limit is 0. For B, the limit is $\lim_{x \to \infty} e^x = \infty$. For C, the limit is $\lim_{x \to \infty} x \ln(x) = \infty$. For D, the limit is $\lim_{x \to \infty} (\frac{5}{4})^x = \infty$.

Question 9

1. The graph of $f$ has two distinct horizontal asymptotes. (correct answer)
2. The graph of $f$ must intersect the x-axis at least once.
3. The graph of $f$ has at least one vertical asymptote.
4. The function $f$ must be a rational function with different leading coefficients.

Explanation: The definition of a horizontal asymptote is a line $y=c$ where $\lim_{x \to \infty} f(x) = c$ or $\lim_{x \to -\infty} f(x) = c$. Since $\lim_{x \to \infty} f(x) = L$ and $\lim_{x \to -\infty} f(x) = M$ with $L \neq M$, the graph of $f$ has two distinct horizontal asymptotes, $y=L$ and $y=M$. The other options are not necessarily true.

Question 10

1. $0$
2. $\frac{3}{2}$ (correct answer)
3. $1$
4. The limit does not exist.

Explanation: To evaluate this limit, we can divide the numerator and the denominator by the highest power of $x$, which is $x$. This gives $\lim_{x \to \infty} \frac{3 + \frac{\sin(x)}{x}}{2 - \frac{\cos(x)}{x}}$. Since $\sin(x)$ and $\cos(x)$ are bounded between -1 and 1, $\lim_{x \to \infty} \frac{\sin(x)}{x} = 0$ and $\lim_{x \to \infty} \frac{\cos(x)}{x} = 0$ by the Squeeze Theorem. Therefore, the limit is $\frac{3+0}{2-0} = \frac{3}{2}$.

Question 11

1. The graph of $f$ has a horizontal asymptote at $y=4$. (correct answer)
2. The graph of $f$ has a vertical asymptote at $x=4$.
3. $\lim_{x \to -\infty} f(x) = 4$.
4. The graph of $f$ does not intersect the line $y=4$.

Explanation: The statement $\lim_{x \to \infty} f(x) = 4$ is the definition of a horizontal asymptote at $y=4$. Option B confuses horizontal and vertical asymptotes. Option C is not necessarily true; the limit as $x \to -\infty$ could be different or not exist. Option D is incorrect because a function can cross its horizontal asymptote.

Question 12

1. $-\frac{1}{3}$ (correct answer)
2. $-\frac{1}{9}$
3. $\frac{1}{3}$
4. $3$

Explanation: To evaluate the limit as $x \to -\infty$, we divide the numerator and denominator by the highest power of $x$ in the denominator, which is $\sqrt{x^2} = |x|$. Since $x \to -\infty$, $x$ is negative, so $|x| = -x$. Divide the numerator by $x$ and the denominator by $-x = \sqrt{x^2}$. $\lim_{x \to -\infty} \frac{(x+2)/x}{(\sqrt{9x^2-5x})/(-x)} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{(9x^2-5x)/x^2}} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{9-5/x}} = \frac{1+0}{-\sqrt{9-0}} = -\frac{1}{3}$.

Question 13

1. $0$ (correct answer)
2. $\frac{2}{3}$
3. $1$
4. The limit does not exist.

Explanation: We can rewrite $\ln(x^2)$ as $2\ln(x)$. The limit becomes $\lim_{x \to \infty} \frac{2\ln(x)}{3x}$. Since polynomial functions (like $3x$) grow faster than logarithmic functions (like $2\ln(x)$), the limit is 0. Alternatively, applying L'Hôpital's Rule since the limit is of the form $\frac{\infty}{\infty}$, we get $\lim_{x \to \infty} \frac{d/dx(2\ln(x))}{d/dx(3x)} = \lim_{x \to \infty} \frac{2/x}{3} = \lim_{x \to \infty} \frac{2}{3x} = 0$.

Question 14

1. $y=3$ only
2. $y=4$ only
3. $y=2$ and $y=4$ (correct answer)
4. $y=3-\frac{2}{\pi}$ and $y=3+\frac{2}{\pi}$

Explanation: To find horizontal asymptotes, we evaluate the limits as $x \to \infty$ and $x \to -\infty$. We know that $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$. So, $\lim_{x \to \infty} f(x) = \frac{2}{\pi} \left(\frac{\pi}{2}\right) + 3 = 1 + 3 = 4$. And $\lim_{x \to -\infty} f(x) = \frac{2}{\pi} \left(-\frac{\pi}{2}\right) + 3 = -1 + 3 = 2$. Therefore, the horizontal asymptotes are $y=2$ and $y=4$.

Question 15

1. $\frac{2}{3}$
2. $-\frac{1}{3}$ (correct answer)
3. $1$
4. The limit does not exist.

Explanation: The term with the largest base, $3^x$, dominates as $x \to \infty$. Divide the numerator and denominator by $3^x$: $\lim_{x \to \infty} \frac{\frac{2 \cdot 2^x}{3^x} - \frac{3^x}{3^x}}{\frac{3 \cdot 3^x}{3^x} + \frac{2^x}{3^x}} = \lim_{x \to \infty} \frac{2(\frac{2}{3})^x - 1}{3 + (\frac{2}{3})^x}$. Since $\lim_{x \to \infty} (\frac{2}{3})^x = 0$, the limit is $\frac{2(0) - 1}{3 + 0} = -\frac{1}{3}$.

Question 16

1. The asymptote exists for any finite constants $a$ and $b$. (correct answer)
2. Both $a=0$ and $b=0$ must be true.
3. The constants must be positive, $a>0$ and $b>0$.
4. At least one of the constants, $a$ or $b$, must be zero.

Explanation: The limit is $\lim_{x \to \infty} \frac{ax^2+b}{e^x}$. This is of the indeterminate form $\frac{\infty}{\infty}$ (if $a \neq 0$). The exponential function $e^x$ grows faster than any polynomial. Therefore, the limit is 0 for any finite values of $a$ and $b$. If $a=0$, the limit is $\lim_{x \to \infty} \frac{b}{e^x} = 0$. No specific conditions on $a$ and $b$ are needed, other than that they are finite.

Question 17

1. $f(x) = \frac{2x}{\sqrt{x^2+1}}$ (correct answer)
2. $f(x) = \frac{2x^2}{x^2+1}$
3. $f(x) = \frac{4x}{x+1}$
4. $f(x) = 2e^{-x} + 2$

Explanation: We check the limits as $x \to \infty$ and $x \to -\infty$. For A: $\lim_{x \to \infty} \frac{2x}{\sqrt{x^2+1}} = \lim_{x \to \infty} \frac{2x}{|x|} = 2$. And $\lim_{x \to -\infty} \frac{2x}{\sqrt{x^2+1}} = \lim_{x \to -\infty} \frac{2x}{|x|} = \lim_{x \to -\infty} \frac{2x}{-x} = -2$. So A is correct. For B, the limit is 2 in both directions. For C, the limit is 4 in both directions. For D, the limit is 2 as $x \to \infty$ but $\infty$ as $x \to -\infty$.

Question 18

1. $0$
2. $1$
3. The limit does not exist and is not infinite.
4. The limit cannot be determined from the information given. (correct answer)

Explanation: The value of the limit depends on the relative growth rates of $f(x)$ and $x^2$. If $f(x)$ grows slower than $x^2$ (e.g., $f(x) = x$), the limit is 0. If $f(x)$ grows at the same rate as $x^2$ (e.g., $f(x)=5x^2$), the limit is a nonzero constant. If $f(x)$ grows faster than $x^2$ (e.g., $f(x)=x^3$), the limit is $\infty$. Since the growth rate of $f(x)$ is unknown, the limit cannot be determined.

Question 19

1. $2$
2. $5$ (correct answer)
3. $0$
4. The limit does not exist.

Explanation: We can evaluate the limit of each term separately. The limit of the constant 5 is 5. For the second term, we use the Squeeze Theorem. Since $-1 \le \sin(3x) \le 1$, it follows that $-\frac{1}{x^2} \le \frac{\sin(3x)}{x^2} \le \frac{1}{x^2}$ for $x > 0$. As $x \to \infty$, both $-\frac{1}{x^2}$ and $\frac{1}{x^2}$ approach 0. Thus, $\lim_{x \to \infty} \frac{\sin(3x)}{x^2} = 0$. The overall limit is $5 - 0 = 5$.

Question 20

1. $0$
2. $1$ (correct answer)
3. $e$
4. The limit does not exist.

Explanation: To find the limit of the composite function $f(g(x))$, we first find the limit of the inner function $g(x)$ as $x \to \infty$. $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} \frac{1}{x^2+1} = 0$. Since $f(x) = e^{-x}$ is continuous everywhere, we can find the limit by substituting the limit of the inner function into the outer function: $\lim_{x \to \infty} f(g(x)) = f(\lim_{x \to \infty} g(x)) = f(0) = e^{-0} = 1$.