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📝 Horizontal asymptotes from limits at infinity (15 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 15 questions available

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What is Horizontal asymptotes from limits at infinity?

Definition:
A horizontal asymptote is a horizontal line $y = L$ such that $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$ , meaning the function approaches $L$ as $x$ tends to positive or negative infinity. A function can have at most two horizontal asymptotes (one for each direction), and they indicate the function's end behavior, often found by comparing degrees of polynomials in rational functions.

Example:
Find horizontal asymptotes of $f(x) = \frac{2x}{x+1}$ .
Solution: $\lim_{x \to \infty} \frac{2x}{x+1} = 2$ , and $\lim_{x \to -\infty} \frac{2x}{x+1} = 2$ , so $y=2$ is the horizontal asymptote.

Reason:
Identifying horizontal asymptotes helps in sketching graphs accurately and in understanding the limiting behavior of functions, which is essential for modeling population growth, radioactive decay, and other processes with saturation.

Easy

Medium

Hard

📝 All Horizontal asymptotes from limits at infinity MCQs

Q1. What is the definition of a horizontal asymptote for a function $f(x)$ ?

A.A line that the graph of

f

approaches as

x\to\infty

x\to-\infty

. ✅

B.A line that the graph of

f

never crosses.

C.A line parallel to the y‑axis that the graph approaches.

D.A line that represents the maximum value of

f

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A horizontal asymptote is a straight line that the function’s graph gets arbitrarily close to as the input grows without bound in either the positive or negative direction. This description matches option A, which explicitly mentions the approach of the graph as $x$ tends to $+\infty$ or $-\infty$ .

Q2. For a rational function $R(x)=\frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials, which degree relationship guarantees a horizontal asymptote at $y=0$ ?

\deg(p)>\deg(q)

\deg(p)=\deg(q)

\deg(p)<\deg(q)

✅

\deg(p)=\deg(q)+1

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When the degree of the numerator is strictly less than the degree of the denominator, the fraction’s value shrinks toward zero as $|x|$ becomes large. Hence the x‑axis ( $y=0$ ) serves as a horizontal asymptote. This condition corresponds to option C, making it the correct choice.

Q3. What is $\displaystyle\lim_{x\to\infty}\frac{2x^{2}+5x-3}{x^{2}-4}$ ?

A.2 ✅

B.-2

C.0

D.The limit does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Both numerator and denominator are degree 2 polynomials, so the limit equals the ratio of the leading coefficients: $\frac{2}{1}=2$ . Therefore the function approaches the constant line $y=2$ as $x$ grows without bound, confirming option A.

Q4. What is $\displaystyle\lim_{x\to\infty}\frac{7x+1}{x^{2}+3}$ ?

A.0 ✅

B.7

\infty

D.The limit does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The numerator grows linearly while the denominator grows quadratically, causing the fraction to shrink toward zero. Formally, $\frac{7x+1}{x^{2}+3}= \frac{7}{x}+O\!\left(\frac{1}{x^{2}}\right)$ , which tends to 0 as $x\to\infty$ . Hence option A is correct.

Q5. Which statement about the function $h(x)=\frac{3x^{3}+2x}{x^{2}-5}$ is true regarding horizontal asymptotes?

y=0

is a horizontal asymptote

y=3x

is a horizontal asymptote

C.The function has no horizontal asymptote ✅

y=3

is a horizontal asymptote

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The numerator’s degree (3) exceeds the denominator’s degree (2) by one, causing the function to grow without bound as $|x|$ increases. Since the limit at infinity is infinite, no finite horizontal line can serve as an asymptote, making option C correct.

Q6. If a function $f(x)$ has a horizontal asymptote $y=1$ , what is the horizontal asymptote of the shifted function $f(x)+4$ ?

y=5

✅

y=1

y=-3

D.The function has no horizontal asymptote

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Adding a constant $k$ to a function translates its graph vertically by $k$ units. Consequently, an existing horizontal asymptote $y=L$ moves to $y=L+k$ . Here $k=4$ and $L=1$ , so the new asymptote is $y=5$ , which is option A.

Q7. Both functions $f(x)=\frac{5x^{2}-3}{2x^{2}+7}$ and $g(x)=\frac{5x^{2}+10}{2x^{2}-1}$ have which horizontal asymptote?

A.Both have the same horizontal asymptote

y=\frac{5}{2}

. ✅

f

has

y=\frac{5}{2}

and

g

has a different asymptote.

f

has a different asymptote while

g

has

y=\frac{5}{2}

D.Neither function has a horizontal asymptote.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When degrees of numerator and denominator are equal, the horizontal asymptote equals the ratio of the leading coefficients. Both functions have leading numerator coefficient 5 and denominator coefficient 2, giving the same asymptote $y=\frac{5}{2}$ . Thus option A is correct.

Q8. If the denominator of $f(x)=\frac{x^{2}+4}{3x^{2}+6}$ is multiplied by 2, producing $h(x)=\frac{x^{2}+4}{6x^{2}+12}$ , what is the new horizontal asymptote?

\frac{1}{3}

\frac{1}{6}

✅

\frac{2}{3}

D.2

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Multiplying the denominator by 2 doubles its leading coefficient from 3 to 6. The horizontal asymptote for a rational function with equal degrees is the ratio of leading coefficients, now $\frac{1}{6}$ . Hence option B correctly reflects the new asymptote.

Q9. For the function $f(x)=\frac{|x|}{x+1}$ , which horizontal asymptote(s) describe its end behavior?

y=1

only

y=-1

only

C.Both

y=1

and

y=-1

✅

D.No horizontal asymptote

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: As $x\to\infty$ , $|x|=x$ and the ratio approaches $\frac{x}{x}=1$ . As $x\to-\infty$ , $|x|=-x$ and the ratio approaches $\frac{-x}{x}=-1$ . Thus the function has two distinct horizontal asymptotes, one for each direction, matching option C.

Q10. Both functions $f(x)=\frac{-2x^{2}+3}{x^{2}-4}$ and $g(x)=\frac{2x^{2}+1}{-x^{2}+5}$ share which horizontal asymptote?

A.Both have the same horizontal asymptote

y=-2

. ✅

f

has

y=-2

and

g

has

y=2

f

has

y=2

and

g

has

y=-2

D.Neither function has a horizontal asymptote.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Each function’s numerator and denominator are degree 2. The asymptote equals the ratio of leading coefficients: for $f$ , $-2/1=-2$ ; for $g$ , $2/(-1)=-2$ . Both share the line $y=-2$ , so option A is correct.

Q11. Does the graph of $f(x)=\frac{x^{2}-1}{x^{2}+1}$ ever cross its horizontal asymptote?

A.It crosses at

x=0

B.It crosses at

x=1

C.It never crosses the horizontal asymptote. ✅

D.It crosses infinitely often.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The horizontal asymptote of this function is $y=1$ because the leading coefficients are both 1. Setting $\frac{x^{2}-1}{x^{2}+1}=1$ leads to $-1=1$ , an impossibility. Therefore the curve never meets the line $y=1$ ; option C is correct.

Q12. If $f(x)$ has a horizontal asymptote $y=3$ , what is the horizontal asymptote of the function $f(2x)$ ?

y=3

✅

y=6

y=1.5

D.The function has no horizontal asymptote

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A horizontal stretch (or compression) changes the input variable but does not affect the limit of the function as $x$ approaches infinity. Consequently, the asymptotic value remains $y=3$ , making option A the correct answer.

Q13. A rational function with horizontal asymptote $y=0$ and a vertical asymptote at $x=2$ is increased by 5. What is the new horizontal asymptote and does the vertical asymptote change?

A.Horizontal asymptote

y=5

; vertical asymptote unchanged ✅

B.Horizontal asymptote

y=5

; vertical asymptote moves

C.Horizontal asymptote

y=0

; vertical asymptote unchanged

D.The function has no horizontal asymptote

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Adding a constant $k$ to a function shifts its entire graph upward by $k$ . The original horizontal asymptote at $y=0$ becomes $y=k$ , here $k=5$ . Vertical asymptotes, determined by zeros of the denominator, are unaffected by such vertical shifts. Thus option A is correct.

Q14. Determine the horizontal asymptote of $f(x)=\frac{x^{2}}{x^{2}+\sin x}$ .

y=0

y=1

✅

y=-1

D.The limit does not exist

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because $|\sin x|\le1$ , the denominator satisfies $x^{2}-1\le x^{2}+\sin x\le x^{2}+1$ . Dividing numerator and denominator by $x^{2}$ yields $\frac{1}{1+\sin x/x^{2}}$ , which approaches 1 as $|x|\to\infty$ . Hence the horizontal asymptote is $y=1$ , option B.

Q15. If $h(x)=f(x)g(x)$ where $f(x)=\frac{x^{2}+1}{x^{2}-1}$ and $g(x)=\frac{2x}{x^{2}+3}$ , what is the horizontal asymptote of $h(x)$ ?

y=0

✅

y=1

y=2

D.The function has no horizontal asymptote

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Function $f$ has equal degrees, giving a horizontal asymptote of $\frac{1}{1}=1$ . Function $g$ has numerator degree 1 and denominator degree 2, so its horizontal asymptote is $0$ . The product’s asymptote is the product of the individual limits: $1 \times 0 = 0$ . Thus option A is correct.

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