← Back to 2. Limits and Continuity an Introduction

📝 Epsilon N definition of limits at infinity (18 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 18 questions available

←Previous Topic:Epsilon delta definition of limit📝 Epsilon N definition of limits at infinityM delta definition of infinite limits:Next Topic→

What is Epsilon N definition of limits at infinity?

Definition:
The epsilon-N definition for $\lim_{x \to \infty} f(x) = L$ states that for every $\epsilon > 0$ , there exists a positive number $N$ such that if $x > N$ , then $|f(x) - L| < \epsilon$ . This formalizes the idea that $f(x)$ gets arbitrarily close to $L$ for all sufficiently large $x$ , providing a rigorous foundation for end-behavior analysis and horizontal asymptotes.

Example:
Prove $\lim_{x \to \infty} \frac{1}{x} = 0$ .
Solution: For $\epsilon > 0$ , choose $N = 1/\epsilon$ . If $x > N$ , then $|1/x - 0| = 1/x < 1/N = \epsilon$ .

Reason:
This definition is crucial for verifying limits at infinity rigorously, which is needed in real analysis and for proving convergence of sequences and series in higher mathematics.

Easy

Medium

Hard

📝 All Epsilon N definition of limits at infinity MCQs

Q1. Given that \ $\\lim_{x\\to 2} (3x-5)=1\$ , which of the following statements correctly describes the relationship between \ $\\epsilon\$ and \ $\\delta\$ in the proof?

A.\

\\delta = \\epsilon/3\

✅

B.\

\\delta = 3\\epsilon\

C.\

\\delta = \\epsilon\

D.\

\\delta = \\epsilon^{2}\

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Because the limit proof reduces to \ $|3x-5-1|<\\epsilon\$ which simplifies to \ $3|x-2|<\\epsilon\$ , choosing \ $\\delta = \\epsilon/3\$ guarantees that whenever \ $0<|x-2|<\\delta\$ the inequality \ $|3x-5-1|<\\epsilon\$ holds, satisfying the epsilon‑delta condition.

Q2. What does the symbol \ $\\lim_{x\\to a} f(x)=L\$ signify in the epsilon‑delta definition?

A.The function equals L at x=a.

B.For every \

\\epsilon>0\

there exists \

\\delta>0\

such that if \

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

then \

|f(x)-L|<\\epsilon\

. ✅

C.The derivative at a equals L.

D.The function is continuous at a.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The notation means that for each positive tolerance \ $\\epsilon\$ we can find a positive distance \ $\\delta\$ so that whenever \ $x\$ lies within that distance of \ $a\$ (but not at \ $a\$ itself) the function values stay within \ $\\epsilon\$ of \ $L\$ . It does not require the function to be defined at \ $a\$ .

Q3. If for a function \ $f\$ we have \ $|f(x)-L|<\\epsilon\$ whenever \ $0<|x-a|<\\delta\$ , which of the following statements must be true?

A.The same \

\\delta\

works for all \

\\epsilon\

B.For each \

\\epsilon\

there exists a possibly different \

\\delta\

. ✅

C.\

\\delta\

can be chosen independent of \

\\epsilon\

D.The limit does not exist.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The epsilon‑delta definition requires that for every positive \ $\\epsilon\$ we can find a (potentially different) \ $\\delta\$ that makes the implication true. It does not demand a single \ $\\delta\$ to work for all \ $\\epsilon\$ ; the relationship is quantified separately for each tolerance.

Q4. When proving \ $\\lim_{x\\to2}(3x-5)=1\$ , two possible choices for \ $\\delta\$ are \ $\\delta=\\epsilon/3\$ and \ $\\delta=\\sqrt{\\epsilon}\$ . Which choice yields the most efficient proof?

A.\

\\delta=\\epsilon\

B.\

\\delta=\\epsilon/3\

✅

C.\

\\delta=\\sqrt{\\epsilon}\

D.Both are equally efficient.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The inequality derived from the limit is \ $3|x-2|<\\epsilon\$ . Directly solving for \ $|x-2|\$ gives \ $\\delta=\\epsilon/3\$ , which matches the algebraic requirement without extra bounding steps. The square‑root choice introduces unnecessary complexity, making the former the most straightforward option.

Q5. Apply the epsilon‑delta definition to prove \ $\\lim_{x\\to3} x^{2}=9\$ . Which expression correctly relates \ $\\delta\$ to \ $\\epsilon\$ ?

A.\

\\delta = \\epsilon/6\

B.\

\\delta = \\sqrt{\\epsilon}\

C.\

\\delta = \\min\\{1,\\epsilon/7\\}\

✅

D.\

\\delta = \\epsilon/(2\\cdot3)\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Starting from \ $|x^{2}-9|=|x-3||x+3|\$ , restrict \ $|x-3|<1\$ so that \ $|x+3|<7\$ . Then require \ $|x-3|<\\epsilon/7\$ . Combining the two restrictions gives \ $\\delta=\\min\\{1,\\epsilon/7\\}\$ , which ensures the product stays below \ $\\epsilon\$ .

Q6. For \ $f(x)=\\frac{1}{x}\$ and \ $a=1\$ , the limit is 1. If \ $\\epsilon=0.2\$ , which of the following \ $\\delta\$ values guarantees \ $|f(x)-1|<0.2\$ whenever \ $0<|x-1|<\\delta\$ ?

A.\

\\delta=0.1\

✅

B.\

\\delta=0.05\

C.\

\\delta=0.2\

D.\

\\delta=0.04\

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: We need \ $|1/x-1|<0.2\$ which is \ $|x-1|/|x|<0.2\$ . If we first require \ $|x-1|<0.1\$ , then \ $|x|>0.9\$ . Consequently \ $|x-1|<0.2\\times0.9=0.18\$ . The smaller bound, \ $0.1\$ , satisfies both conditions, so \ $\\delta=0.1\$ works.

Q7. When proving \ $\\lim_{x\\to0}\\sin x =0\$ , a student chooses \ $\\delta=\\epsilon\$ . Which of the following best evaluates this choice?

A.It works for all \

\\epsilon\

because \

|\\sin x|\\le|x|\

. ✅

B.It is valid but not optimal.

C.It fails for \

\\epsilon>1\

D.It is unnecessary since \

\\delta\

must depend on \

\\epsilon\

differently.

💡 Difficulty: hard | ✅ Correct: A

Q8. Which of the following outlines a correct epsilon‑delta proof for \ $\\lim_{x\\to2} x^{2}=4\$ ?

A.Show \

|x^{2}-4|<\\epsilon\

directly.

B.Factor as \

|x-2||x+2|<\\epsilon\

and bound \

|x+2|\

by 5 when \

|x-2|<1\

, then choose \

\\delta=\\min\\{1,\\epsilon/5\\}\

. ✅

C.Use derivative at 2.

D.Assume continuity.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Writing \ $|x^{2}-4|=|x-2||x+2|\$ and restricting \ $|x-2|<1\$ forces \ $|x+2|<5\$ . The inequality then becomes \ $|x-2|<\\epsilon/5\$ . Selecting \ $\\delta=\\min\\{1,\\epsilon/5\\}\$ guarantees both the bound on \ $|x+2|\$ and the desired closeness, completing the proof.

Q9. A student argues that because \ $|x-2|<\\delta\$ implies \ $|3x-6|<3\\delta\$ , the limit proof is complete without linking \ $\\delta\$ to \ $\\epsilon\$ . What is the flaw?

A.The inequality sign is reversed.

B.The factor 3 is incorrect.

C.The proof omits the requirement that \

3\\delta<\\epsilon\

. ✅

D.The argument assumes \

\\delta\

can be any positive number.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The step shows how the difference scales, but the definition demands that the scaled difference be less than \ $\\epsilon\$ . Without explicitly choosing \ $\\delta\$ so that \ $3\\delta<\\epsilon\$ , the argument fails to satisfy the epsilon‑delta condition.

Q10. Which mathematician is credited with formalizing the epsilon‑delta definition of a limit?

A.Karl Weierstrass ✅

B.Augustin-Louis Cauchy

C.Isaac Newton

D.Leonhard Euler

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Karl Weierstrass is widely recognized for introducing the rigorous epsilon‑delta framework that precisely defines limits, continuity, and related concepts in analysis, providing the foundation for modern real analysis.

Q11. How does the definition of a right‑hand limit differ from the two‑sided limit?

A.It replaces \

|x-a|\

with \

x-a\

and requires \

x>a\

. ✅

B.It uses \

x<a\

C.It does not require \

\\epsilon\

D.It requires continuity at a.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A right‑hand limit considers only points to the right of \ $a\$ ; thus the condition becomes \ $0<x-a<\\delta\$ (or \ $0<|x-a|<\\delta\$ with \ $x>a\$ ). This contrasts with the two‑sided definition, which allows approach from both sides.

Q12. Why does the epsilon‑delta definition exclude the point \ $x=a\$ by using \ $0<|x-a|<\\delta\$ ?

A.To avoid division by zero.

B.Because the function may be undefined at \

a\

. ✅

C.To ensure the limit is unique.

D.To simplify calculations.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Excluding \ $x=a\$ allows the definition to apply even when the function is not defined at that point. The limit describes the behavior of \ $f(x)\$ as \ $x\$ approaches \ $a\$ from nearby values, independent of the function's value (or lack thereof) at \ $a\$ .

Q13. If \ $\\epsilon=0.5\$ for the function \ $f(x)=3x-5\$ at \ $a=2\$ , what is the appropriate \ $\\delta\$ ?

A.\

\\delta=0.1667\

✅

B.\

\\delta=0.5\

C.\

\\delta=1.5\

D.\

\\delta=0.3333\

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: From the proof we have \ $3|x-2|<\\epsilon\$ , so \ $|x-2|<\\epsilon/3\$ . Substituting \ $\\epsilon=0.5\$ gives \ $\\delta=0.5/3\\approx0.1667\$ , which guarantees the required inequality.

Q14. For \ $f(x)=\\frac{x^{2}-4}{x-2}\$ (with \ $x\\neq2\$ ), determine \ $\\lim_{x\\to2} f(x)\$ using epsilon‑delta. Which choice of \ $\\delta\$ correctly demonstrates the limit equals 4?

A.Choose \

\\delta = \\min\\{1,\\epsilon/5\\}\

B.Choose \

\\delta = \\epsilon\

. ✅

C.Choose \

\\delta = \\sqrt{\\epsilon}\

D.No \

\\delta\

can be found because the function is undefined at 2.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Since \ $f(x)=x+2\$ for \ $x\\neq2\$ , we have \ $|f(x)-4|=|x-2|\$ . To make this less than \ $\\epsilon\$ it suffices to require \ $|x-2|<\\epsilon\$ . Thus choosing \ $\\delta=\\epsilon\$ satisfies the epsilon‑delta condition.

Q15. Which general formula gives a suitable \ $\\delta\$ for any linear function \ $f(x)=mx+b\$ to prove \ $\\lim_{x\\to a} f(x)=ma+b\$ ?

A.\

\\delta = \\epsilon/|m|\

✅

B.\

\\delta = \\epsilon\

C.\

\\delta = |m|\\epsilon\

D.\

\\delta = \\sqrt{\\epsilon}\

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For a linear function, \ $|f(x)-(ma+b)|=|m||x-a|\$ . To ensure this is less than \ $\\epsilon\$ , we need \ $|x-a|<\\epsilon/|m|\$ . Hence \ $\\delta=\\epsilon/|m|\$ provides a direct and universal choice for any slope \ $m\$ .

Q16. If \ $\\lim_{x\\to a} f(x)=L\$ exists, must \ $f(a)\$ be defined?

A.Yes, the limit implies the function is defined at a. ✅

B.No, the limit can exist even if \

f(a)\

is undefined.

C.Only if the limit equals \

f(a)\

D.Only for continuous functions.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The existence of a limit concerns the behavior of \ $f(x)\$ as \ $x\$ approaches \ $a\$ ; it does not require the function to have a value at \ $a\$ . Therefore the limit can exist even when \ $f(a)\$ is not defined or is defined differently.

Q17. Which statement correctly relates continuity at \ $a\$ to the epsilon‑delta limit definition?

A.Continuity requires \

\\lim_{x\\to a} f(x)=f(a)\

and the limit exists. ✅

B.Continuity only requires the function to be defined at \

a\

C.Continuity is unrelated to limits.

D.Continuity demands \

\\delta=\\epsilon\

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A function is continuous at \ $a\$ precisely when the limit as \ $x\$ approaches \ $a\$ exists and equals the function's value at that point. This ties the epsilon‑delta definition of limits directly to the formal definition of continuity.

Q18. Consider the piecewise function \ $g(x)=\\begin{cases}x^{2}& x<1\\\\ 2x& x\\ge1\\end{cases}\$ . Using the epsilon‑delta definition, does \ $\\lim_{x\\to1} g(x)\$ exist?

A.Yes, because both sides approach 1.

B.Yes, because both sides approach 2.

C.No, because the left‑hand limit is 1 and the right‑hand limit is 2. ✅

D.No, because the function is not defined at 1.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Evaluating each piece at \ $x=1\$ gives \ $\\lim_{x\\to1^-}x^{2}=1\$ and \ $\\lim_{x\to1^+}2x=2\$ . Since the one‑sided limits differ, there is no single \ $L\$ satisfying the epsilon‑delta condition, so the overall limit does not exist.

🔗 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:Epsilon delta definition of limit📝 Epsilon N definition of limits at infinityM delta definition of infinite limits:Next Topic→