← Back to 2. Limits and Continuity an Introduction

📝 Infinite limits and unbounded behavior (15 MCQs)

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

←Previous Topic:One sided vs two sided limits📝 Infinite limits and unbounded behaviorVertical asymptotes from infinite limits:Next Topic→

What is Infinite limits and unbounded behavior?

Definition:
An infinite limit occurs when the values of $f(x)$ increase or decrease without bound as $x$ approaches a finite number $a$ , written as $\lim_{x \to a} f(x) = \infty$ or $-\infty$ . This indicates that the function grows arbitrarily large in magnitude near $a$ , often due to division by zero or vertical asymptotes, and the function's output becomes unbounded.

Example:
Evaluate $\lim_{x \to 1} \frac{1}{(x-1)^2}$ .
Solution: As $x \to 1$ , $(x-1)^2 \to 0^+$ , so $\frac{1}{(x-1)^2} \to \infty$ . Thus, $\lim_{x \to 1} \frac{1}{(x-1)^2} = \infty$ .

Reason:
This concept identifies vertical asymptotes and extreme behavior, which is crucial for graphing functions and understanding singularities in physical models, such as gravitational forces near a point mass.

Easy

Medium

Hard

📝 All Infinite limits and unbounded behavior MCQs

Q1. Consider $f(x)=\frac{1}{x}$ . Which of the following statements correctly describes the limit as $x\to 0^{+}$ ?

A.The limit equals 0.

B.The limit equals

+\infty

. ✅

C.The limit equals

-\infty

D.The limit does not exist because the function oscillates.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because $f(x)=1/x$ grows without bound in the positive direction when $x$ approaches 0 from the right, the limit is expressed as $+\infty$ . The other choices either give a finite value, a negative infinity, or claim oscillation, none of which match the actual behavior.

Q2. If \ $\\displaystyle \\lim_{x\\to a^-} f(x)=+\\infty\$ and \ $\\displaystyle \\lim_{x\\to a^+} f(x)=+\\infty\$ , which conclusion about \ $\\displaystyle \\lim_{x\\to a} f(x)\$ is valid?

A.The limit equals \

+\\infty\

. ✅

B.The limit does not exist because the left and right limits are different.

C.The limit equals \

-\\infty\

D.The limit equals a finite number.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When both one‑sided limits approach the same infinite value, the two‑sided limit exists and is defined as that same infinite value, namely \ $+\\infty\$ . The other options either contradict the given behavior or incorrectly assert finiteness.

Q3. Suppose \ $g(x)=\\frac{1}{(x-2)^2}\$ . Analyze the behavior of \ $g(x)\$ as \ $x\$ approaches 2 from either side and determine which statement is true.

A.\

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

and \

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

B.Both one‑sided limits equal \

+\\infty\

. ✅

C.Both one‑sided limits equal \

-\\infty\

D.The limits are finite but different.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Because the denominator \ $(x-2)^2\$ is always positive, the fraction is always positive and grows without bound as \ $x\$ approaches 2 from either side, giving \ $+\\infty\$ for both limits. The other options incorrectly assign signs or finiteness.

Q4. Compare the graphs of \ $h(x)=\\frac{1}{x}\$ and \ $k(x)=\\frac{-1}{x}\$ near \ $x=0\$ . Which description is accurate?

A.Both functions increase without bound on the right side of zero.

B.\

h(x)\

increases without bound on the right, while \

k(x)\

decreases without bound on the right. ✅

C.Both functions decrease without bound on the left side of zero.

D.\

h(x)\

decreases without bound on the left, and \

k(x)\

increases without bound on the left.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For \ $h(x)=1/x\$ , as \ $x\\to0^{+}\$ the values become arbitrarily large positive, while \ $k(x)=-1/x\$ becomes arbitrarily large negative. Thus the right‑hand behavior differs in sign, making option B correct; the other options mischaracterize the sign changes.

Q5. Given the functions \ $p(x)=\\frac{1}{(x+3)}\$ and \ $q(x)=\\frac{1}{(x+3)^2}\$ , which of the following best characterizes their infinite limit behavior as \ $x\\to -3\$ ?

A.Both have limits \

+\\infty\

from the left and right.

B.\

p(x)\

has opposite signs from each side, while \

q(x)\

approaches \

+\\infty\

from both sides. ✅

C.Both approach \

-\\infty\

from both sides.

D.\

p(x)\

approaches a finite value, while \

q(x)\

diverges to \

-\\infty\

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The denominator of \ $p(x)\$ changes sign at \ $x=-3\$ , giving opposite infinities on each side. The squared denominator in \ $q(x)\$ is always positive, so \ $q(x)\$ grows positively without bound from both sides, yielding \ $+\\infty\$ . Hence option B correctly captures the contrasting behaviors.

Q6. For the function \ $r(x)=\\frac{x}{(x-1)}\$ , determine the relationship between the sign of the infinite limit as \ $x\\to 1^{-}\$ and as \ $x\\to 1^{+}\$ .

A.Both one‑sided limits are \

+\\infty\

B.Both one‑sided limits are \

-\\infty\

C.The left‑hand limit is \

-\\infty\

and the right‑hand limit is \

+\\infty\

. ✅

D.The left‑hand limit is \

+\\infty\

and the right‑hand limit is \

-\\infty\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Near \ $x=1\$ , the numerator is approximately 1, while the denominator changes sign. Approaching from the left gives a negative small denominator, producing a large negative value (\ $-\\infty\$ ). Approaching from the right yields a positive small denominator, giving a large positive value (\ $+\\infty\$ ).

Q7. Consider \ $s(x)=\\frac{(x-4)}{(x-4)^3}\$ . Simplify and deduce the infinite limit behavior as \ $x\\to 4\$ . Which statement is true?

A.The function simplifies to \

\\frac{1}{(x-4)^2}\

and both one‑sided limits are \

+\\infty\

. ✅

B.The function simplifies to \

\\frac{1}{(x-4)^2}\

and both one‑sided limits are \

-\\infty\

C.The function simplifies to \

\\frac{1}{(x-4)^2}\

with opposite signs on each side.

D.The function simplifies to \

\\frac{-1}{(x-4)^2}\

with both limits \

+\\infty\

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Cancelling a factor of \ $(x-4)\$ yields \ $s(x)=1/(x-4)^2\$ . The denominator is squared, so it is always positive, making the whole expression positive and unbounded as \ $x\$ approaches 4 from either side, resulting in \ $+\\infty\$ for both limits.

Q8. Which of the following best defines an infinite limit at a point \ $a\$ ?

A.The function approaches a finite number as \

x\\to a\

B.The function oscillates without approaching any value as \

x\\to a\

C.The function grows without bound (positively or negatively) as \

x\\to a\

. ✅

D.The function has a removable discontinuity at \

a\

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: An infinite limit indicates that as the variable approaches a particular point, the function’s magnitude increases without bound, either to \ $+\\infty\$ or \ $-\\infty\$ . This distinguishes it from finite limits, oscillatory behavior, or removable discontinuities, making option C the accurate definition.

Q9. A function has a vertical asymptote at \ $x = a\$ . Which inference about the limits \ $\\lim_{x\\to a^-} f(x)\$ and \ $\\lim_{x\\to a^+} f(x)\$ must be true?

A.Both one‑sided limits equal the same finite number.

B.At least one of the one‑sided limits is infinite (either \

+\\infty\

or \

-\\infty\

). ✅

C.Both one‑sided limits equal zero.

D.The function is continuous at \

x = a\

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A vertical asymptote occurs when the function’s values become unbounded as the input approaches the asymptote from at least one side. Therefore, at least one one‑sided limit must be infinite. The other options incorrectly claim finiteness, continuity, or zero values.

Q10. Suppose \ $\\displaystyle \\lim_{x\\to a} f(x)=+\\infty\$ . Which of the following statements is always correct?

A.For every real number \

M\

, there exists \

\\delta>0\

such that \

0<|x-a|<\\delta\

implies \

f(x)>M\

. ✅

B.The function must be positive for all \

x\

near \

a\

C.The limit from the left must be \

-\\infty\

D.The function has a hole at \

x=a\

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The formal definition of a limit diverging to \ $+\\infty\$ requires that for any arbitrarily large bound \ $M\$ , we can find a neighborhood around \ $a\$ where the function exceeds \ $M\$ . This captures the notion of unbounded growth. The other statements are not guaranteed by the definition.

Q11. Let \ $t(x)=\\frac{\\sin (1/(x-2))}{(x-2)}\$ . Determine the nature of \ $\\lim_{x\\to 2} t(x)\$ .

A.The limit is \

+\\infty\

B.The limit is \

-\\infty\

C.The limit does not exist because the function oscillates with unbounded amplitude. ✅

D.The limit does not exist because the function oscillates while remaining bounded.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: As \ $x\$ approaches 2, the denominator tends to zero, while the numerator remains bounded between \ $-1\$ and \ $1\$ . The quotient therefore grows without bound, but its sign alternates rapidly, preventing convergence to either \ $+\\infty\$ or \ $-\\infty\$ . Hence the limit does not exist due to unbounded oscillation.

Q12. If a function satisfies \ $\\displaystyle \\lim_{x\\to a^-} f(x)=+\\infty\$ and \ $\\displaystyle \\lim_{x\\to a^+} f(x)=-\\infty\$ , what can be said about the existence of \ $\\displaystyle \\lim_{x\\to a} f(x)\$ ?

A.The two‑sided limit exists and equals \

+\\infty\

B.The two‑sided limit exists and equals \

-\\infty\

C.The two‑sided limit does not exist. ✅

D.The limit equals zero.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a two‑sided limit to exist, the left‑ and right‑hand limits must agree. Here they diverge to opposite infinities, so no single value (finite or infinite) can represent the limit. Consequently, the overall limit does not exist.

Q13. Compare the behavior of \ $u(x)=\\frac{1}{(x-5)}\$ and \ $v(x)=\\frac{-1}{(x-5)^2}\$ as \ $x\$ approaches 5. Which statement correctly describes both one‑sided limits?

A.Both functions approach \

+\\infty\

from the left and \

-\\infty\

from the right.

B.\

u(x)\

approaches opposite infinities from each side, while \

v(x)\

approaches \

-\\infty\

from both sides. ✅

C.Both functions approach \

-\\infty\

from both sides.

D.\

u(x)\

approaches \

+\\infty\

from both sides, while \

v(x)\

approaches \

-\\infty\

from both sides.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The function \ $u(x)=1/(x-5)\$ changes sign at \ $x=5\$ , yielding \ $-\\infty\$ from the left and \ $+\\infty\$ from the right. The function \ $v(x)=-1/(x-5)^2\$ is always negative and its magnitude grows without bound on both sides, giving \ $-\\infty\$ for each one‑sided limit.

Q14. Which of the following transformations will change the sign of an infinite limit at a vertical asymptote?

A.Multiplying the function by a positive constant.

B.Adding a constant to the function.

C.Multiplying the function by \

-1\

. ✅

D.Composing the function with a square root.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Multiplying a function by \ $-1\$ reverses the sign of all its values, so an infinite limit that was previously \ $+\\infty\$ becomes \ $-\\infty\$ and vice versa. Multiplying by a positive constant preserves sign, adding a constant shifts the graph without altering the unbounded direction, and taking a square root is not defined for negative large values.

Q15. What symbol is commonly used to denote that a limit diverges to positive infinity?

A.\

-\\infty\

B.0

C.\

+\\infty\

✅

D.\

\\text{DNE}\

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The standard notation for a limit that grows without bound in the positive direction is the symbol \ $+\\infty\$ . The other symbols represent negative infinity, the number zero, or a generic statement that a limit does not exist, none of which convey positive unbounded growth.

🔗 Related Topics (MCQs)

📝 Tangent line problem and slope MCQs 📝 Area problem and limits MCQs 📝 Limits intuitive understanding MCQs 📝 Left hand and right hand limits MCQs 📝 One sided vs two sided limits MCQs

←Previous Topic:One sided vs two sided limits📝 Infinite limits and unbounded behaviorVertical asymptotes from infinite limits:Next Topic→