Limits at Infinity and Horizontal Asymptotes - AP Calculus AB

1. For which of the following functions are $y=2$ and $y=-2$ the horizontal asymptotes?
   Answer: $f(x) = \frac{2x}{\sqrt{x^2+1}}$ 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 $x \to -\infty$.
2. What is the value of $\lim_{x \to \infty} \frac{3x + \sin(x)}{2x - \cos(x)}$?
   Answer: $\frac{3}{2}$ 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 $\frac{3+0}{2-0} = \frac{3}{2}$.
3. What is the value of the limit $\lim_{x \to \infty} \frac{2^{x+1} - 3^x}{3^{x+1} + 2^x}$?
   Answer: $-\frac{1}{3}$ 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 $\frac{2(0) - 1}{3 + 0} = -\frac{1}{3}$.
4. The function $f$ is given by $f(x) = (ax^2+b)e^{-x}$. The graph of $f$ has a horizontal asymptote at $y=0$ as $x \to \infty$. What must be true about the constants $a$ and $b$?
   Answer: The asymptote exists for any finite constants $a$ and $b$. 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.
5. A population of bacteria is modeled by the logistic function $P(t) = \frac{2000}{1 + 49e^{-0.2t}}$, where $t$ is time in days. What number does the population approach as time increases without bound?
   Answer: $2000$ 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.
6. What are all the horizontal asymptotes of the graph of $f(x) = \frac{2}{\pi} \arctan(x) + 3$?
   Answer: $y=2$ and $y=4$ 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 $y=4$.
7. What is the value of $\lim_{x \to \infty} \frac{\ln(x^2)}{3x}$?
   Answer: $0$ 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 $\lim_{x \to \infty} \frac{d/dx(2\ln(x))}{d/dx(3x)} = \lim_{x \to \infty} \frac{2/x}{3} = \lim_{x \to \infty} \frac{2}{3x} = 0$.
8. What is the value of $\lim_{x \to -\infty} \frac{e^x + 5}{e^{2x} - 3}$?
   Answer: $-\frac{5}{3}$ 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 $\lim_{x \to -\infty} \frac{e^x + 5}{e^{2x} - 3} = \frac{0 + 5}{0 - 3} = -\frac{5}{3}$.
9. Which of the following describes all horizontal asymptotes of the graph of $f(x) = \frac{\sqrt{4x^2 + 1}}{3x - 5}$?
   Answer: $y = \frac{2}{3}$ and $y = -\frac{2}{3}$ 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 $y = -\frac{2}{3}$.
10. What is the value of $\lim_{x \to \infty} \frac{(2x-1)^3}{x(4x+1)^2}$?
    Answer: $\frac{1}{2}$ 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 $\frac{8}{16} = \frac{1}{2}$.
11. The graph of the function $y = f(x)$ has a horizontal asymptote at $y=5$. Which of the following statements must be true?
    Answer: $\lim_{x \to \infty} f(x) = 5$ or $\lim_{x \to -\infty} f(x) = 5$ 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.
12. Let $f(x) = e^{-x}$ and $g(x) = \frac{1}{x^2+1}$. What is $\lim_{x \to \infty} f(g(x))$?
    Answer: $1$ 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: $\lim_{x \to \infty} f(g(x)) = f(\lim_{x \to \infty} g(x)) = f(0) = e^{-0} = 1$.
13. What is the value of the limit $\lim_{x \to -\infty} \frac{x + 2}{\sqrt{9x^2 - 5x}}$?
    Answer: $-\frac{1}{3}$ 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 $-x = \sqrt{x^2}$. $\lim_{x \to -\infty} \frac{(x+2)/x}{(\sqrt{9x^2-5x})/(-x)} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{(9x^2-5x)/x^2}} = \lim_{x \to -\infty} \frac{1+2/x}{-\sqrt{9-5/x}} = \frac{1+0}{-\sqrt{9-0}} = -\frac{1}{3}$.
14. Let $g(x) = \frac{f(x)}{x^2}$, where $f$ is a function such that $\lim_{x \to \infty} f(x) = \infty$. What is the value of $\lim_{x \to \infty} g(x)$?
    Answer: The limit cannot be determined from the information given. 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.
15. What is the value of $\lim_{x \to \infty} \left(5 - \frac{\sin(3x)}{x^2}\right)$?
    Answer: $5$ 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 $5 - 0 = 5$.
16. A function $f$ has the property that $\lim_{x \to \infty} f(x) = L$ and $\lim_{x \to -\infty} f(x) = M$, where $L$ and $M$ are finite constants and $L \neq M$. Which of the following must be true about the graph of $f$?
    Answer: The graph of $f$ has two distinct horizontal asymptotes. 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.
17. Let $f$ be a function such that $\lim_{x \to \infty} f(x) = 4$. Which of the following statements must be true?
    Answer: The graph of $f$ has a horizontal asymptote at $y=4$. 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.
18. Which of the following limits is equal to 0?
    Answer: $\lim_{x \to \infty} \frac{x^{100}}{e^x}$ 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 $\lim_{x \to \infty} (\frac{5}{4})^x = \infty$.
19. The concentration of a drug in a patient's bloodstream, in mg/L, $t$ hours after administration is given by the function $C(t) = \frac{150t}{t^2 + 4}$. What does the concentration of the drug in the bloodstream approach as time increases without bound?
    Answer: $0$ mg/L 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.
20. What is the value of the limit $\lim_{x \to \infty} \frac{3x^2 - 5x + 1}{7 - 2x^2}$?
    Answer: $-\frac{3}{2}$ 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 $\frac{3}{-2} = -\frac{3}{2}$.

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