Calculus 1 Quiz: Limits At Infinity And Horizontal Asymptotes

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Limits At Infinity And Horizontal AsymptotesQuestion 1 of 20

The function $f(x) = \frac{(ax-1)^2}{3x^2 + 5x - 2}$ has a horizontal asymptote at $y=3$ . What is the value of $a^2$ ?

Calculus 1 Quiz

Calculus 1 Quiz: Limits At Infinity And Horizontal Asymptotes

Practice Limits At Infinity And Horizontal Asymptotes in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 Calculus 1.

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

The function $f(x) = \frac{(ax-1)^2}{3x^2 + 5x - 2}$ has a horizontal asymptote at $y=3$ . What is the value of $a^2$ ?

9 (correct answer)
3
81
1

Explanation: To find the horizontal asymptote, we find the limit as

x \to \infty

. First, expand the numerator:

(ax-1)^2 = a^2x^2 - 2ax + 1

. The function is

f(x) = \frac{a^2x^2 - 2ax + 1}{3x^2 + 5x - 2}

. The limit at infinity is the ratio of the leading coefficients, since the degrees of the numerator and denominator are both 2. So,

\lim_{x \to \infty} f(x) = \frac{a^2}{3}

. We are given that the horizontal asymptote is at

y=3

, so we set

\frac{a^2}{3} = 3

, which gives

a^2 = 9

Question 2

The graph of the function $f(x) = \frac{3e^x - 5}{e^x + |e^x - 10|}$ has how many distinct horizontal asymptotes?

Zero
One
Two (correct answer)
Three

Explanation: We check the limits as

x \to \infty

and

x \to -\infty

. For

x \to \infty

e^x

is large, so

e^x - 10 > 0

and

|e^x - 10| = e^x - 10

. The limit is

\lim_{x \to \infty} \frac{3e^x - 5}{e^x + e^x - 10} = \lim_{x \to \infty} \frac{3e^x - 5}{2e^x - 10} = 3/2

. For

x \to -\infty

e^x \to 0

, so

e^x - 10 < 0

and

|e^x - 10| = -(e^x - 10) = 10 - e^x

. The limit is

\lim_{x \to -\infty} \frac{3e^x - 5}{e^x + 10 - e^x} = \lim_{x \to -\infty} \frac{3e^x - 5}{10} = \frac{0 - 5}{10} = -1/2

. Since

3/2 \neq -1/2

, there are two distinct horizontal asymptotes:

y=3/2

and

y=-1/2

Question 3

Which of the following functions has a non-zero horizontal asymptote?

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

Explanation: When analyzing horizontal asymptotes, you need to examine the behavior of each function as x approaches positive and negative infinity. A horizontal asymptote exists when the function approaches a specific finite value in these limits. For choice B,

f(x) = x \sin(\frac{1}{x})

, let's substitute

u = \frac{1}{x}

, so

x = \frac{1}{u}

. As

x \to \infty

, we have

u \to 0

, and the function becomes

\frac{\sin(u)}{u}

. This is the famous limit

\lim_{u \to 0} \frac{\sin(u)}{u} = 1

. Similarly, as

x \to -\infty

, the limit is also 1. Therefore,

y = 1

is a horizontal asymptote. Choice A,

f(x) = \frac{\cos(x)}{e^x}

, has a bounded numerator (cosine oscillates between -1 and 1) but an exponentially growing denominator. As

x \to \infty

, this approaches 0, giving a horizontal asymptote at

y = 0

(which is considered a zero horizontal asymptote). Choice C,

f(x) = x - \sqrt{x^2-4}

, can be rationalized by multiplying by the conjugate. This yields

\frac{4}{x + \sqrt{x^2-4}}

, which approaches 0 as

x \to \infty

. Choice D,

f(x) = \frac{3x^2+1}{x^3-x}

, has a denominator with higher degree than the numerator, so it approaches 0 as

x \to \pm\infty

. Remember: when finding horizontal asymptotes, look for limits that approach non-zero finite values. The key insight here is recognizing the

\frac{\sin(u)}{u}

pattern in disguise.

Question 4

Find all horizontal asymptote(s) of the function $f(x) = \frac{|x-2|}{3x+1}$ .

$y=1/3$ only.
$y=1/3$ and $y=-1/3$ . (correct answer)
$y=-1/3$ only.
$y=0$ and $y=1/3$ .

Explanation: 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.

Question 5

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 $y=0$ ?

$g(x) = x^k$
$g(x) = (e^x - 1)^2$
$g(x) = e^{x^2}$ (correct answer)
$g(x) = 100e^{2x}$

Explanation: 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

Question 6

Which of the following functions has a horizontal asymptote that its graph intersects at least once for $x > 0$ ?

$f(x) = \frac{5x^2}{x^2+1}$
$f(x) = e^{-x} + 2$
$f(x) = \frac{\cos(x)}{x} + 3$ (correct answer)
$f(x) = \frac{1}{x} + 4$

Explanation: A function

f(x)

intersects its horizontal asymptote

y=L

if the equation

f(x)=L

has a solution. We first find the horizontal asymptote for each function by taking the limit as

x \to \infty

. A)

\lim_{x \to \infty} \frac{5x^2}{x^2+1} = 5

. The equation

\frac{5x^2}{x^2+1} = 5

gives

5x^2 = 5x^2+5

, which simplifies to

0=5

, having no solution. B)

\lim_{x \to \infty} (e^{-x} + 2) = 0+2=2

. The equation

e^{-x}+2=2

gives

e^{-x}=0

, which has no solution. C)

\lim_{x \to \infty} (\frac{\cos(x)}{x} + 3) = 0+3=3

, by the Squeeze Theorem. The equation

\frac{\cos(x)}{x} + 3 = 3

gives

\frac{\cos(x)}{x}=0

, which means

\cos(x)=0

. This occurs for

x = \frac{\pi}{2} + n\pi

for any integer

n

. There are infinitely many positive solutions. D)

\lim_{x \to \infty} (\frac{1}{x} + 4) = 0+4=4

. The equation

\frac{1}{x}+4=4

gives

\frac{1}{x}=0

, which has no solution.

Question 7

What is the value of $\lim_{x \to \infty} \frac{4x - \cos(2x)}{x + \sin(x^2)}$ ?

4 (correct answer)
3
1
The limit does not exist.

Explanation: 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

Question 8

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?

$y = 5$
$y = 10/3$ (correct answer)
$y = 2$
$y = 11/3$

Explanation: 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

Question 9

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 $a$ ?

$a = 2$
$a \neq 2$
$a = 4$
$a \neq 4$ (correct answer)

Explanation: 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.

Question 10

A function $f(x)$ has the following properties: $\lim_{x \to \infty} f(x) = -2$ , $\lim_{x \to -\infty} f(x) = -2$ , and $\lim_{x \to 3} f(x) = \infty$ . Which of the following could be an expression for $f(x)$ ?

$f(x) = \frac{-2x^2}{x^2 - 9}$
$f(x) = \frac{20 - 2x^2}{(x-3)^2}$ (correct answer)
$f(x) = \frac{-2x^2 + 1}{(x-3)^2}$
$f(x) = \frac{2x^2 - 1}{(3-x)^2}$

Explanation: We need a function with a horizontal asymptote at

y=-2

and a vertical asymptote at

x=3

where the limit is

\infty

. For a rational function, the HA is

y=-2

if the ratio of leading coefficients is -2 and the degrees of the numerator and denominator are equal. Choices A, B, and C satisfy this. For the VA at

x=3

, the denominator should be zero. All choices except A have a denominator of

(x-3)^2

(3-x)^2

, which ensures the limit is the same from both sides. For the limit to be

+\infty

, the numerator must be positive as

x \to 3

. For choice B, the numerator at

x=3

20 - 2(3^2) = 20 - 18 = 2 > 0

. The denominator

(x-3)^2

approaches

0^+

. So

\lim_{x \to 3} f(x) = \infty

. For choice C, the numerator at

x=3

-2(9)+1 = -17 < 0

, so the limit is

-\infty

Question 11

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 $f(x)$ ?

The graph of $f(x)$ has a horizontal asymptote at $y=4$ . (correct answer)
The graph of $f(x)$ has a horizontal asymptote, but its value cannot be determined.
The graph of $f(x)$ might not have a horizontal asymptote.
The graph of $f(x)$ has horizontal asymptotes at both $y=0$ and $y=4$ .

Explanation: 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.

Question 12

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, $f^{-1}(x)$ ?

Horizontal asymptote at $y=1/2$ , vertical asymptote at $x=-1/3$ .
Horizontal asymptote at $y=2$ , vertical asymptote at $x=3$ .
Horizontal asymptote at $y=-1/3$ , vertical asymptote at $x=1/2$ .
Horizontal asymptote at $y=3$ , vertical asymptote at $x=2$ . (correct answer)

Explanation: 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.

Question 13

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 $h(x)$ ?

The graph of $h(x)$ must be decreasing for all $x$ .
The graph of $h(x)$ has a global maximum value.
The equation $h(x)=5$ has at least one solution.
The graph of $h(x)$ has at least one x-intercept. (correct answer)

Explanation: 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

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.

Question 14

Let $f(x) = \frac{2x^2 + 3}{ (1-ax)(2x) + 5x^2 }$ . If the graph of $y=f(x)$ has a horizontal asymptote at $y = -1$ , what is the value of the constant $a$ ?

$3/2$
$3$
$-7/2$
$7/2$ (correct answer)

Explanation: First, simplify the expression for

f(x)

by expanding the denominator:

f(x) = \frac{2x^2 + 3}{2x - 2ax^2 + 5x^2} = \frac{2x^2 + 3}{(5 - 2a)x^2 + 2x}

The horizontal asymptote is determined by the limit as

x \to \infty

. We look at the ratio of the leading coefficients of the numerator and the denominator.

\lim_{x \to \infty} f(x) = \frac{2}{5 - 2a}

We are given that the horizontal asymptote is at

y=-1

. Therefore, we set the limit equal to -1 and solve for

a

\frac{2}{5 - 2a} = -1

2 = -1(5 - 2a)

2 = -5 + 2a

7 = 2a

a = \frac{7}{2}

Question 15

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 $k/L$ ?

3
6 (correct answer)
9
1/6

Explanation: 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

Question 16

Evaluate the limit: $\lim_{x \to \infty} \frac{x^{10} + 2^x}{x^2 + 3^x}$

0 (correct answer)
1
$+\infty$
$2/3$

Explanation: 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.

Question 17

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?

{0, 1}
{1, 2}
{0, 1, 2} (correct answer)
{0, 1, 2, 3, ...}

Explanation: 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}.

Question 18

What is the value of the limit $\lim_{x \to \infty} \frac{\ln(x^4 + 1)}{\ln(x^2 + 1)}$ ?

$1/2$
1
2 (correct answer)
4

Explanation: 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)}

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

Question 19

Suppose $f(x)$ is a function such that $\lim_{x \to \infty} f(x) = 3$ . Which of the following statements must be true?

The graph of $f(x)$ does not intersect the line $y=3$ for all sufficiently large $x$ .
The derivative $f'(x)$ must approach 0 as $x \to \infty$ .
The graph of $h(x) = f(x+5)$ has a horizontal asymptote at $y=3$ . (correct answer)
The function $f(x)$ must be positive for all sufficiently large $x$ .

Explanation: A horizontal asymptote describes the end behavior of a function. A horizontal shift of the graph, such as

h(x) = f(x+5)

, does not change the end behavior. If

x \to \infty

, then

x+5 \to \infty

, so

\lim_{x \to \infty} h(x) = \lim_{x \to \infty} f(x+5) = 3

. Distractor A is a common misconception; functions can cross their horizontal asymptotes. Distractor B is false; for example,

f(x) = 3 + \frac{\sin(x^2)}{x}

. Distractor D is false; the function approaches 3, so it will be positive for large

x

, but this is a consequence, not a distinct property in this context and less certain than C.

Question 20

Consider the function $f(x) = x - \ln(e^x + 1)$ . Which statement accurately describes its horizontal asymptotes?

The graph has one horizontal asymptote at $y=0$ . (correct answer)
The graph has two horizontal asymptotes at $y=0$ and $y=-1$ .
The graph has one horizontal asymptote at $y=1$ .
The graph has no horizontal asymptotes.

Explanation: We can rewrite

f(x) = x - \ln(e^x(1+e^{-x})) = x - (\ln(e^x) + \ln(1+e^{-x})) = x - (x + \ln(1+e^{-x})) = -\ln(1+e^{-x})

. Now we check the limits. As

x \to \infty

e^{-x} \to 0

, so

f(x) \to -\ln(1+0) = 0

. Thus,

y=0

is a horizontal asymptote. As

x \to -\infty

e^{-x} \to \infty

, so

1+e^{-x} \to \infty

, and

f(x) = -\ln(1+e^{-x}) \to -\infty

. There is no horizontal asymptote as

x \to -\infty

. Therefore, there is only one horizontal asymptote.