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📝 Vertical asymptotes from infinite limits (13 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 13 questions available

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What is Vertical asymptotes from infinite limits?

Definition:
A vertical asymptote is a vertical line $x = a$ where the function $f(x)$ approaches $\pm \infty$ as $x$ approaches $a$ from either the left or right. This occurs when at least one of the one-sided infinite limits is $\infty$ or $-\infty$ , indicating that the function's graph rises or falls indefinitely near that line, often due to factors in the denominator vanishing.

Example:
Find vertical asymptotes for $f(x) = \frac{x}{x-3}$ .
Solution: As $x \to 3^-$ , denominator $\to 0^-$ , so $f \to -\infty$ ; as $x \to 3^+$ , denominator $\to 0^+$ , so $f \to \infty$ . Thus, $x=3$ is a vertical asymptote.

Reason:
Identifying vertical asymptotes helps in sketching accurate graphs and in predicting discontinuities, especially in rational functions, which is essential for solving equations and inequalities in calculus and applied mathematics.

Easy

Medium

Hard

📝 All Vertical asymptotes from infinite limits MCQs

Q1. Given $f(x)=\dfrac{x^{2}-1}{x-1}$ , which of the following describes its vertical asymptote(s)?

A.x=1

B.None ✅

C.x=-1

D.x=0

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Since the factor $x-1$ cancels, the function simplifies to $f(x)=x+1$ which is defined and finite at $x=1$ . Therefore no vertical asymptote exists, making “None” the correct choice.

Q2. Compare $f(x)=\dfrac{1}{x-2}$ and $g(x)=\dfrac{2}{x-2}$ . Which statement about their vertical asymptotes is true?

A.Neither has a vertical asymptote

B.Both have the same vertical asymptote

x=2

✅

f

has a vertical asymptote at

x=2

while

g

does not

g

has a vertical asymptote at

x=2

while

f

does not

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Both functions share the denominator $x-2$ ; when that denominator approaches zero the quotients blow up regardless of the constant numerator. Hence each function possesses the identical vertical asymptote at $x=2$ .

Q3. For $h(x)=\dfrac{x^{2}+3x-4}{x^{2}-4}$ , identify all vertical asymptotes.

x=2

x=-2

x=2

and

x=-2

✅

D.None

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The denominator factors as $(x-2)(x+2)$ . Neither factor cancels with the numerator, so the function becomes unbounded as $x$ approaches $2$ or $-2$ . Consequently both $x=2$ and $x=-2$ are vertical asymptotes.

Q4. Which function has a vertical asymptote at $x=0$ but no horizontal asymptote?

y=\dfrac{1}{x^{2}+1}

y=\dfrac{x}{x^{2}}

y=\dfrac{1}{x}

y=\dfrac{x^{2}+1}{x}

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The expression $\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}$ diverges to $\pm\infty$ as $x\to0$ , giving a vertical asymptote. As $|x|\to\infty$ , the dominant term is $x$ , so the graph grows without bound and does not settle to a horizontal line, eliminating a horizontal asymptote.

Q5. If a rational function has denominator $(x-3)^{2}$ and numerator $(x-3)$ , what is the behavior near $x=3$ ?

A.A vertical asymptote

B.A removable hole ✅

C.Both a hole and an asymptote

D.No special behavior

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: One factor of $(x-3)$ cancels, leaving a reduced function that is continuous at $x=3$ . The original expression is undefined there, but the limit exists and is finite, indicating a removable discontinuity (hole) rather than an infinite blow‑up.

Q6. Which statement is always true about vertical asymptotes of rational functions?

A.They occur only where the denominator is zero ✅

B.They occur only where the numerator is zero

C.They occur where both numerator and denominator are zero

D.They can occur at any

x

-value

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A vertical asymptote arises when the denominator vanishes while the numerator remains non‑zero, causing the function to diverge. The numerator being zero would instead produce a finite value (or a hole), so the only guaranteed condition is a zero denominator with a non‑zero numerator.

Q7. A graph shows a vertical asymptote at $x=-1$ and a horizontal asymptote $y=0$ . Which rational function could produce it?

y=\dfrac{x+1}{x-1}

y=\dfrac{1}{x+1}

✅

y=\dfrac{x^{2}}{x+1}

y=\dfrac{x-2}{(x+1)^{2}}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The function $y=\dfrac{1}{x+1}$ is undefined at $x=-1$ and grows without bound as the input approaches that point, giving a vertical asymptote. As $|x|$ becomes large, the term tends to zero, producing the horizontal asymptote $y=0$ .

Q8. For $p(x)=\dfrac{2x^{3}-x^{2}+5}{x^{3}-4x}$ , find all vertical asymptotes and classify each as simple (multiplicity 1) or multiple (multiplicity > 1).

x=0

(multiple),

x=2

(simple),

x=-2

(simple)

x=0

(simple),

x=2

(multiple),

x=-2

(simple)

x=0

(simple),

x=2

(simple),

x=-2

(simple) ✅

x=0

(multiple),

x=2

(multiple),

x=-2

(multiple)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The denominator factors as $x(x-2)(x+2)$ ; each factor appears once, giving three simple poles. No factor repeats, so each vertical asymptote at $x=0,\;2,\;-2$ is of multiplicity 1, i.e., simple.

Q9. Consider $q(x)=\dfrac{x^{2}-9}{x^{2}-4x+3}$ . Determine $\displaystyle\lim_{x\to3^{-}}q(x)$ and $\displaystyle\lim_{x\to3^{+}}q(x)$ .

A.Both limits are

-\infty

B.Both limits are

+\infty

C.Limits are finite and equal ✅

D.Left limit

-\infty

, right limit

+\infty

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Factorising gives $\dfrac{(x-3)(x+3)}{(x-1)(x-3)}=\dfrac{x+3}{x-1}$ after cancelling $(x-3)$ . The resulting expression is continuous at $x=3$ , yielding $\dfrac{6}{2}=3$ . Hence both one‑sided limits exist and equal the finite value 3.

Q10. The rational function $r(x)=\dfrac{x^{2}+1}{x^{2}-2x+1}$ can be simplified. After simplification, what vertical asymptotes does it have?

A.No vertical asymptotes

x=1

✅

x=-1

x=0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The denominator is $(x-1)^{2}$ ; the numerator never vanishes at $x=1$ . Because the factor does not cancel, the function becomes unbounded as $x$ approaches 1, producing a vertical asymptote at $x=1$ (of multiplicity 2).

Q11. For $f(x)=\dfrac{x^{2}-4}{x^{2}-9}$ , which statement about its vertical asymptotes is false?

A.It has vertical asymptotes at

x=3

and

x=-3

✅

B.The asymptotes are simple

C.The function approaches infinity near

x=3

D.There is a hole at

x=2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The function’s denominator factors as $(x-3)(x+3)$ and the numerator as $(x-2)(x+2)$ ; no common factor exists, so there are no holes. Thus the claim of a hole at $x=2$ is false, while the other statements correctly describe the asymptotic behavior.

Q12. Given $g(x)=\dfrac{x^{2}-5x+6}{x^{2}-3x}$ , determine the x‑values where vertical asymptotes occur and explain why $x=0$ is not an asymptote.

A.Asymptote only at

x=3

;

x=0

not because denominator factor cancels

B.Asymptotes at

x=0

and

x=3

;

x=0

not because numerator also zero

C.Asymptotes at

x=0

and

x=3

;

x=0

not because factor cancels

D.No vertical asymptotes ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The denominator factors to $x(x-3)$ ; the numerator contains $(x-2)(x-3)$ , cancelling the $(x-3)$ factor. The remaining denominator factor $x$ does not cancel, so a vertical asymptote exists at $x=0$ . Thus the correct description is that the only vertical asymptote is at $x=0$ and $x=3$ is a removable hole.

Q13. A vertical asymptote occurs at $x = a$ if ...

A.the function is undefined at

a

and the limit as

x\to a

is infinite ✅

B.the function equals a constant at

a

C.the derivative does not exist at

a

D.the function has a maximum at

a

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By definition, a vertical asymptote at $x=a$ means the function cannot be evaluated at that point (denominator zero) and the values of the function grow without bound as the input approaches $a$ from either side, producing an infinite limit.

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