Limits at Infinity & Horizontal Asymptotes - Calculus 1

Let $f(x) = \sqrt{9x^2+kx} - 3x$ , where $k$ is a positive constant. It is given that $\lim_{x \to \infty} f(x) = L$ , where $L$ is a finite, non-zero number. What is the value of the ratio $k/L$ ?
Answer: 6 To evaluate the limit of the indeterminate form $\infty - \infty$ , we multiply by the conjugate: $L = \lim_{x \to \infty} (\sqrt{9x^2+kx} - 3x) \frac{\sqrt{9x^2+kx} + 3x}{\sqrt{9x^2+kx} + 3x}$ $L = \lim_{x \to \infty} \frac{(9x^2+kx) - (3x)^2}{\sqrt{9x^2+kx} + 3x} = \lim_{x \to \infty} \frac{kx}{\sqrt{9x^2+kx} + 3x}$ Now, we divide the numerator and denominator by $x$ . For the term in the square root, we divide by $\sqrt{x^2}=x$ since $x>0$ . $L = \lim_{x \to \infty} \frac{k}{\sqrt{9+k/x} + 3} = \frac{k}{\sqrt{9+0} + 3} = \frac{k}{3+3} = \frac{k}{6}$ We are given that $L = k/6$ . The question asks for the value of $k/L$ . $\frac{k}{L} = \frac{k}{k/6} = 6$
Let $f(x) = \sqrt{4x^2+8x} - 2x$ . Find the horizontal asymptote of the graph of $f(x)$ as $x \to \infty$ .
Answer: $y=2$ The limit as $x \to \infty$ is of the indeterminate form $\infty - \infty$ . To resolve this, we multiply by the conjugate: $f(x) = (\sqrt{4x^2+8x} - 2x) \cdot \frac{\sqrt{4x^2+8x} + 2x}{\sqrt{4x^2+8x} + 2x} = \frac{(4x^2+8x) - (4x^2)}{\sqrt{4x^2+8x} + 2x} = \frac{8x}{\sqrt{4x^2+8x} + 2x}$ . Now, we find the limit by dividing the numerator and denominator by $x$ : $\lim_{x \to \infty} \frac{8}{\sqrt{4+8/x} + 2} = \frac{8}{\sqrt{4+0} + 2} = \frac{8}{2+2} = 2$ . So, the horizontal asymptote is $y=2$ .
Consider the piecewise function $f(x) = \begin{cases} \frac{10x+7}{5x-2} & x \ge 1000 \\ \frac{ax^2-3x}{2x^2+1} & x < 1000 \end{cases}$ The graph of $y=f(x)$ has two distinct horizontal asymptotes. Which of the following must be true about the constant $a$ ?
Answer: $a \neq 4$ A function has horizontal asymptotes determined by its limits as $x \to \infty$ and $x \to -\infty$ . We need to calculate both limits for the given piecewise function. For $x \to \infty$ , we use the first piece: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{10x+7}{5x-2} = \frac{10}{5} = 2$ So, $y=2$ is one horizontal asymptote. For $x \to -\infty$ , we use the second piece: $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{ax^2-3x}{2x^2+1} = \frac{a}{2}$ So, $y=a/2$ is the other horizontal asymptote. The problem states that the function has two distinct horizontal asymptotes. This means the two $y$ -values must be different: $2 \neq \frac{a}{2}$ Multiplying by 2 gives $4 \neq a$ . Therefore, $a$ can be any real number except 4.
Let $g(x)$ be a function such that $\lim_{x \to \infty} g(x) = 2$ . Let $f(x) = \frac{3(g(x))^2 - g(x)}{g(x) + 1}$ . The graph of $y=f(x)$ has a horizontal asymptote as $x \to \infty$ . What is the equation of this asymptote?
Answer: $y = 10/3$ The horizontal asymptote of $f(x)$ is given by the limit $\lim_{x \to \infty} f(x)$ . We are given that $\lim_{x \to \infty} g(x) = 2$ . Since the expression for $f(x)$ is a rational function of $g(x)$ and the denominator's limit is not zero, we can substitute the limit of $g(x)$ into the expression: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3(g(x))^2 - g(x)}{g(x) + 1}$ $= \frac{3(\lim_{x \to \infty} g(x))^2 - (\lim_{x \to \infty} g(x))}{(\lim_{x \to \infty} g(x)) + 1} = \frac{3(2)^2 - 2}{2 + 1}$ $= \frac{3(4) - 2}{3} = \frac{12 - 2}{3} = \frac{10}{3}$ Thus, the equation of the horizontal asymptote is $y = 10/3$ .
A continuous function $h(x)$ is defined for all real numbers. If $\lim_{x \to \infty} h(x) = -\infty$ and $\lim_{x \to -\infty} h(x) = 5$ , which of the following statements must be true about $h(x)$ ?
Answer: The graph of $h(x)$ has at least one x-intercept. When you encounter a problem involving limits at infinity and continuity, think about how the function's behavior at the extremes constrains what must happen in between. Given that $\lim_{x \to \infty} h(x) = -\infty$ and $\lim_{x \to -\infty} h(x) = 5$ , we know that as $x$ moves far to the right, $h(x)$ becomes arbitrarily negative, while as $x$ moves far to the left, $h(x)$ approaches 5. Since $h(x)$ is continuous everywhere, we can apply the Intermediate Value Theorem. For any sufficiently large positive value of $x$ , we have $h(x) < 0$ (since the function approaches $-\infty$ ). For any sufficiently large negative value of $x$ , we have $h(x)$ close to 5, so $h(x) > 0$ . Since $h(x)$ is continuous and changes from positive to negative values, it must cross the x-axis at least once, making choice D correct. Choice A is wrong because the function doesn't need to be monotonically decreasing everywhere—it could oscillate while still satisfying the given limits. Choice B is incorrect because $h(x)$ approaches $-\infty$ as $x \to \infty$ , meaning it can become arbitrarily large in the positive direction without having a maximum. Choice C is false because while $h(x)$ approaches 5 as $x \to -\infty$ , continuity doesn't guarantee the function ever actually equals 5. Remember: when a continuous function has different signs in different regions, the Intermediate Value Theorem guarantees it must cross zero somewhere between those regions.
If $y=L$ is a horizontal asymptote for the graph of a differentiable function $f(x)$ , which of the following statements about the derivative $f'(x)$ is necessarily true?
Answer: If $\lim_{x \to \infty} f'(x)$ exists, then its value must be 0. If a function $f(x)$ approaches a finite limit $L$ , it cannot continue to have a non-zero slope in the limit. If $\lim_{x \to \infty} f'(x) = C$ where $C \neq 0$ , then for large $x$ , $f(x)$ would behave like a line with slope $C$ , and would go to $\infty$ or $-\infty$ , contradicting that $\lim_{x \to \infty} f(x) = L$ . Therefore, if the limit of the derivative exists, it must be 0. However, the limit of the derivative does not have to exist (e.g., $f(x) = L + \sin(x^2)/x$ ), which makes choice A incorrect.
The function $f(x)$ is differentiable for all real numbers and has a horizontal asymptote at $y=10$ . Which of the following statements must be false?
Answer: The graph of $f(x)$ has a global minimum at $x=d$ where $f(d)=12$ . If the graph has a global minimum at $f(d)=12$ , it means that for all $x$ , $f(x) \ge 12$ . However, the statement that $y=10$ is a horizontal asymptote means that $\lim_{x \to \infty} f(x) = 10$ (or as $x \to -\infty$ ). By the definition of a limit, the function's values must get arbitrarily close to 10. This means for any $\epsilon > 0$ , there is an $N$ such that for $x>N$ , $|f(x)-10| < \epsilon$ . If we choose $\epsilon = 1$ , then there must be values of $f(x)$ between 9 and 11. This contradicts the statement that $f(x) \ge 12$ for all $x$ . Therefore, the function cannot have a global minimum at 12.
Consider the function $f(x) = 5 + \frac{\cos(\pi x)}{x^2+1}$ . Which statement best describes the relationship between the graph of $f(x)$ and the line $y=5$ ?
Answer: $y=5$ is a horizontal asymptote, and the graph intersects it infinitely many times. As $x \to \pm\infty$ , the term $\frac{\cos(\pi x)}{x^2+1}$ approaches 0 by the Squeeze Theorem, since $-1 \le \cos(\pi x) \le 1$ and $\frac{1}{x^2+1} \to 0$ . Thus, $\lim_{x \to \pm\infty} f(x) = 5$ , and $y=5$ is a horizontal asymptote. The graph intersects this line when $f(x)=5$ , which requires $\frac{\cos(\pi x)}{x^2+1} = 0$ . This occurs when $\cos(\pi x) = 0$ , which happens at $x = n + 1/2$ for any integer $n$ . This provides infinitely many intersection points.
What is the value of $\lim_{x \to \infty} \frac{4x - \cos(2x)}{x + \sin(x^2)}$ ?
Answer: 4 To evaluate this limit, we can divide both the numerator and the denominator by the highest power of $x$ , which is $x$ . $\lim_{x \to \infty} \frac{4x - \cos(2x)}{x + \sin(x^2)} = \lim_{x \to \infty} \frac{\frac{4x}{x} - \frac{\cos(2x)}{x}}{\frac{x}{x} + \frac{\sin(x^2)}{x}} = \lim_{x \to \infty} \frac{4 - \frac{\cos(2x)}{x}}{1 + \frac{\sin(x^2)}{x}}$ We know that $-1 \le \cos(2x) \le 1$ and $-1 \le \sin(x^2) \le 1$ . Therefore, as $x \to \infty$ , the terms $\frac{\cos(2x)}{x}$ and $\frac{\sin(x^2)}{x}$ both approach 0 by the Squeeze Theorem. $\frac{4 - 0}{1 + 0} = 4$ The oscillation of the sine and cosine functions does not cause the limit to not exist because their effect is dampened by the division by $x$ .
What is the value of the limit $\lim_{x \to \infty} \frac{\ln(x^4 + 1)}{\ln(x^2 + 1)}$ ?
Answer: 2 This limit is of the indeterminate form $\infty/\infty$ . We can use L'Hôpital's Rule, but a more direct method uses properties of logarithms. For large $x$ , $x^4+1 \approx x^4$ and $x^2+1 \approx x^2$ . So, we can analyze the dominant terms inside the logarithms. $\lim_{x \to \infty} \frac{\ln(x^4 + 1)}{\ln(x^2 + 1)} = \lim_{x \to \infty} \frac{\ln(x^4(1 + 1/x^4))}{\ln(x^2(1 + 1/x^2))}$ $= \lim_{x \to \infty} \frac{\ln(x^4) + \ln(1 + 1/x^4)}{\ln(x^2) + \ln(1 + 1/x^2)} = \lim_{x \to \infty} \frac{4\ln(x) + \ln(1 + 1/x^4)}{2\ln(x) + \ln(1 + 1/x^2)}$ As $x \to \infty$ , $1/x^4 \to 0$ and $1/x^2 \to 0$ , so $\ln(1 + 1/x^4) \to \ln(1) = 0$ and $\ln(1 + 1/x^2) \to \ln(1) = 0$ . The limit becomes: $\lim_{x \to \infty} \frac{4\ln(x)}{2\ln(x)} = \frac{4}{2} = 2$
Consider the family of functions $f_c(x) = \frac{4x}{\sqrt{x^2+c}}$ where $c$ is a positive constant. Which of the following statements is true?
Answer: The horizontal asymptotes are $y = 4$ and $y = -4$ , regardless of the value of $c$ . When analyzing rational functions for horizontal asymptotes, you need to examine the behavior as x approaches both positive and negative infinity. The key is to focus on the highest-degree terms in the numerator and denominator. For $f_c(x) = \frac{4x}{\sqrt{x^2+c}}$ , let's find the limits as x approaches ±∞. When x is very large (positive or negative), the constant c becomes negligible compared to x², so $\sqrt{x^2+c} \approx \sqrt{x^2} = |x|$ . As $x \to +\infty$ : $\lim_{x \to +\infty} \frac{4x}{\sqrt{x^2+c}} = \lim_{x \to +\infty} \frac{4x}{|x|} = \lim_{x \to +\infty} \frac{4x}{x} = 4$ As $x \to -\infty$ : $\lim_{x \to -\infty} \frac{4x}{\sqrt{x^2+c}} = \lim_{x \to -\infty} \frac{4x}{|x|} = \lim_{x \to -\infty} \frac{4x}{-x} = -4$ The horizontal asymptotes are y = 4 and y = -4, regardless of c's value. Choice A is wrong because there are two horizontal asymptotes, not one. Choice B incorrectly suggests the asymptotes depend on c by including $\sqrt{c}$ terms. Choice C is wrong because the asymptotes don't change with c – they remain at ±4 for any positive c. Study tip: For rational functions involving square roots, remember that $\sqrt{x^2} = |x|$ , which equals x when x > 0 and -x when x < 0. This sign consideration is crucial for finding different horizontal asymptotes as x approaches positive versus negative infinity.
Find all horizontal asymptote(s) of the function $f(x) = \frac{|x-2|}{3x+1}$ .
Answer: $y=1/3$ and $y=-1/3$ . When finding horizontal asymptotes of rational functions involving absolute values, you need to analyze the behavior as x approaches both positive and negative infinity, since the absolute value expression behaves differently in each direction. For $f(x) = \frac{|x-2|}{3x+1}$ , you must consider two cases based on the absolute value definition. When $x > 2$ , we have $|x-2| = x-2$ , so $f(x) = \frac{x-2}{3x+1}$ . As $x \to +\infty$ , divide numerator and denominator by the highest power of x: $\lim_{x \to +\infty} \frac{x-2}{3x+1} = \lim_{x \to +\infty} \frac{1-\frac{2}{x}}{3+\frac{1}{x}} = \frac{1}{3}$ When $x < 2$ , we have $|x-2| = -(x-2) = 2-x$ , so $f(x) = \frac{2-x}{3x+1}$ . As $x \to -\infty$ : $\lim_{x \to -\infty} \frac{2-x}{3x+1} = \lim_{x \to -\infty} \frac{\frac{2}{x}-1}{3+\frac{1}{x}} = \frac{-1}{3}$ Therefore, this function has two horizontal asymptotes: $y = \frac{1}{3}$ and $y = -\frac{1}{3}$ . Choice A gives only $y = 1/3$ , missing the negative infinity behavior. Choice C gives only $y = -1/3$ , missing the positive infinity behavior. Choice D incorrectly includes $y = 0$ , which would only occur if the numerator's degree were less than the denominator's degree. Choice B correctly identifies both asymptotes. Remember: when rational functions contain absolute values, always check the limit behavior in both directions separately, as the absolute value creates different expressions for positive and negative inputs.
A function $f(x)$ satisfies the inequality $\frac{4x^2 - \sin(x)}{x^2+1} \le f(x) \le \frac{4x^2 + e^{-x}}{x^2}$ for all $x > 1$ . What can be concluded about the horizontal asymptotes of $f(x)$ ?
Answer: The graph of $f(x)$ has a horizontal asymptote at $y=4$ . We use the Squeeze Theorem. We evaluate the limit of the bounding functions as $x \to \infty$ . For the left side: $\lim_{x \to \infty} \frac{4x^2 - \sin(x)}{x^2+1} = \lim_{x \to \infty} \frac{4 - \sin(x)/x^2}{1+1/x^2} = \frac{4-0}{1+0} = 4$ . For the right side: $\lim_{x \to \infty} \frac{4x^2 + e^{-x}}{x^2} = \lim_{x \to \infty} (4 + \frac{e^{-x}}{x^2}) = 4+0=4$ . Since $f(x)$ is between two functions that both approach 4, by the Squeeze Theorem, $\lim_{x \to \infty} f(x) = 4$ . So, $y=4$ is a horizontal asymptote.
Let $f(x)$ be a function defined for all real numbers. What is the set of all possible numbers of distinct horizontal asymptotes that the graph of $y=f(x)$ can have?
Answer: {0, 1, 2} A horizontal asymptote is determined by the limit of the function as $x \to \infty$ or as $x \to -\infty$ . Since there are only two such limits to consider, a function can have at most two horizontal asymptotes. - 0 horizontal asymptotes: $f(x) = x$ has no horizontal asymptotes as $\lim_{x \to \pm\infty} x = \pm\infty$ . - 1 horizontal asymptote: $f(x) = 1/x$ has one horizontal asymptote $y=0$ since $\lim_{x \to \pm\infty} (1/x) = 0$ . Also, $f(x)=e^x$ has one HA at $y=0$ as $x \to -\infty$ . - 2 horizontal asymptotes: $f(x) = \arctan(x)$ has two horizontal asymptotes, $y=\pi/2$ as $x \to \infty$ and $y=-\pi/2$ as $x \to -\infty$ . It is not possible to have more than two horizontal asymptotes. Therefore, the set of all possible numbers of horizontal asymptotes is {0, 1, 2}.
Let $f(x) = \int_{0}^{x} \frac{1}{1+t^2} dt$ . What is the horizontal asymptote of the graph of $g(x) = 2f(x) + 3$ as $x \to \infty$ ?
Answer: $y = \pi + 3$ The integral $\int \frac{1}{1+t^2} dt = \arctan(t) + C$ . Therefore, $f(x) = [\arctan(t)]_{0}^{x} = \arctan(x) - \arctan(0) = \arctan(x)$ . We need to find the limit of $g(x)$ as $x \to \infty$ : $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} (2\arctan(x) + 3)$ . Since $\lim_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ , the limit is $2(\frac{\pi}{2}) + 3 = \pi+3$ . Thus, the horizontal asymptote is $y = \pi+3$ .
Evaluate the limit: $\lim_{x \to \infty} \frac{x^{10} + 2^x}{x^2 + 3^x}$
Answer: 0 This limit involves comparing the growth rates of polynomial and exponential functions. As $x \to \infty$ , exponential functions grow much faster than polynomial functions. In the numerator, $2^x$ is the dominant term because it grows faster than $x^{10}$ . In the denominator, $3^x$ is the dominant term because it grows faster than $x^2$ . Therefore, the limit is governed by the ratio of the dominant terms: $\lim_{x \to \infty} \frac{x^{10} + 2^x}{x^2 + 3^x} = \lim_{x \to \infty} \frac{2^x}{3^x}$ This can be rewritten as: $\lim_{x \to \infty} \left(\frac{2}{3}\right)^x$ Since the base $2/3$ is between 0 and 1, the limit of this exponential function as $x \to \infty$ is 0.
The function $f(x) = \frac{2x-1}{x-3}$ has a horizontal asymptote at $y=2$ and a vertical asymptote at $x=3$ . What are the horizontal and vertical asymptotes of its inverse function, $f^{-1}(x)$ ?
Answer: Horizontal asymptote at $y=3$ , vertical asymptote at $x=2$ . When you encounter questions about asymptotes of inverse functions, remember that finding an inverse function essentially swaps the roles of x and y coordinates. This means the asymptotes will also swap their orientations. To find the asymptotes of $f^{-1}(x)$ , you need to understand what happens when we reflect the original function across the line $y = x$ . The horizontal asymptote of the original function becomes the vertical asymptote of the inverse, and the vertical asymptote of the original becomes the horizontal asymptote of the inverse. Given that $f(x) = \frac{2x-1}{x-3}$ has a horizontal asymptote at $y = 2$ and a vertical asymptote at $x = 3$ , the inverse function $f^{-1}(x)$ will have: - A vertical asymptote at $x = 2$ (the y-value of the original's horizontal asymptote) - A horizontal asymptote at $y = 3$ (the x-value of the original's vertical asymptote) Looking at the answer choices: Choice A incorrectly uses values that don't relate to the original asymptotes. Choice B keeps the same asymptotes as the original function, ignoring the coordinate swap that occurs with inverse functions. Choice C appears to use some manipulation of the rational function's coefficients but misses the fundamental relationship between a function and its inverse. Only choice D correctly identifies that the horizontal asymptote is at $y = 3$ and the vertical asymptote is at $x = 2$ . Study tip: Always remember the asymptote swap rule for inverse functions - horizontal becomes vertical, and vertical becomes horizontal, with the same numerical values.
For the function $f(x) = \frac{2x^3 - x + 5}{x^2+3}$ , what is the end behavior as $x \to \infty$ ?
Answer: The graph's behavior is approximated by the line $y=2x$ . Since the degree of the numerator (3) is one greater than the degree of the denominator (2), the function has a slant (oblique) asymptote, not a horizontal one. To find it, we perform polynomial long division: $(2x^3 - x + 5) \div (x^2+3) = 2x$ with a remainder of $-7x+5$ . So, $f(x) = 2x + \frac{-7x+5}{x^2+3}$ . As $x \to \infty$ , the remainder term $\frac{-7x+5}{x^2+3}$ approaches 0. Therefore, the graph of $f(x)$ approaches the line $y=2x$ .
Let $k$ be a positive integer. For which function $g(x)$ is it guaranteed that the graph of $f(x) = \frac{e^{2x}}{g(x)}$ has a horizontal asymptote at $y=0$ ?
Answer: $g(x) = e^{x^2}$ For the horizontal asymptote to be at $y=0$ , we need $\lim_{x \to \infty} f(x) = 0$ . This occurs when the denominator $g(x)$ grows significantly faster than the numerator $e^{2x}$ . Comparing the growth rates: (A) $e^{2x}$ grows faster than any polynomial $x^k$ . (B) $(e^x-1)^2 = e^{2x} - 2e^x + 1$ , so the limit would be 1. (D) $100e^{2x}$ grows at the same rate as $e^{2x}$ , and the limit would be 1/100. (C) $e^{x^2}$ grows faster than $e^{2x}$ because its exponent $x^2$ grows faster than $2x$ . Therefore, $\lim_{x \to \infty} \frac{e^{2x}}{e^{x^2}} = 0$ .
For the function $f(x) = 3 + \frac{1}{x^2 + e^{-x}}$ , which statement is true about its horizontal asymptote $y=3$ ?
Answer: The graph approaches $y=3$ from above as $x \to \infty$ and as $x \to -\infty$ . First, we confirm the horizontal asymptote. As $x \to \infty$ , $e^{-x} \to 0$ and $x^2 \to \infty$ , so the denominator goes to $\infty$ and the fraction goes to 0. Thus $\lim_{x \to \infty} f(x) = 3$ . As $x \to -\infty$ , $e^{-x} \to \infty$ , so the denominator goes to $\infty$ and the fraction goes to 0. Thus $\lim_{x \to -\infty} f(x) = 3$ . To determine the direction of approach, we analyze the sign of the fraction $\frac{1}{x^2 + e^{-x}}$ . For any real $x$ , $x^2 \ge 0$ and $e^{-x} > 0$ . Therefore, the denominator $x^2 + e^{-x}$ is always positive. This means the fraction is always positive, so $f(x)$ is always greater than 3. Thus, the graph approaches $y=3$ from above in both directions.

Start this quiz