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📝 Epsilon delta definition of limit (20 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 20 questions available

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What is Epsilon delta definition of limit?

Definition:
The epsilon-delta definition formalizes the limit $\lim_{x \to a} f(x) = L$ by stating that for every $\epsilon > 0$ , there exists a $\delta > 0$ such that if $0 < |x-a| < \delta$ , then $|f(x) - L| < \epsilon$ . This rigorous definition ensures that $f(x)$ can be made arbitrarily close to $L$ by choosing $x$ sufficiently close to $a$ , providing a precise mathematical foundation for the intuitive notion of a limit.

Example:
Prove $\lim_{x \to 2} (3x-1) = 5$ .
Solution: For $\epsilon > 0$ , choose $\delta = \epsilon/3$ . If $|x-2| < \delta$ , then $|(3x-1)-5| = 3|x-2| < 3\delta = \epsilon$ .

Reason:
This definition is the cornerstone of rigorous calculus, ensuring that limits are not just intuitive but mathematically provable, which is essential for advanced analysis and for proving theorems like continuity and differentiability.

Easy

Medium

Hard

📝 All Epsilon delta definition of limit MCQs

Q1. What does the notation $\lim_{x\to a}f(x)=L$ represent?

A.The function value at a equals L ✅

B.The values of f(x) approach L as x approaches a

C.f is continuous at a

D.The limit does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The expression states that as x gets arbitrarily close to a (but not necessarily equal to a), the function values f(x) become arbitrarily close to the number L. It does not require f(a) to be defined, nor does it guarantee continuity; it merely describes the limiting behavior.

Q2. In the epsilon‑delta definition of a two‑sided limit, what role does the number $\delta$ play?

A.It bounds how close x must be to a to make f(x) within ε of L ✅

B.It measures the distance between L and f(x)

C.It is the limit value

D.It is unrelated to ε

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The δ is chosen after ε is given; it specifies a radius around a such that any x inside that radius (except possibly a itself) forces f(x) to stay within ε of the limit L. This is the core of the ε‑δ formulation, linking closeness in the domain to closeness in the range.

Q3. If for every $\epsilon>0$ there exists a $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\epsilon$ , which of the following must be true?

A.The limit

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

exists and equals L ✅

B.f is discontinuous at a

C.The function is undefined at a

D.No conclusion can be drawn

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The given condition is precisely the formal definition of the limit of f at a being L. It guarantees that no matter how tightly we require f(x) to be near L (by choosing ε), we can find a neighborhood around a (determined by δ) where the condition holds, establishing the limit.

Q4. Suppose we find an interval $(a-\delta, a+\delta)$ that works for $\epsilon=0.1$ but a smaller interval is needed for $\epsilon=0.01$ . What does this indicate about the behavior of f near a?

A.f gets arbitrarily close to L as x approaches a ✅

B.f oscillates wildly near a

C.The limit does not exist

D.f is constant near a

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Needing a smaller δ for a smaller ε reflects the idea that as we demand f(x) be closer to L, we must restrict x to lie nearer to a. This is exactly how limits behave: the function can be made as close as desired to L by taking x sufficiently close to a.

Q5. Given that for each $\epsilon>0$ we can choose $\delta = \min\{1,\epsilon\}$ to satisfy the limit condition, which of the following statements is logically equivalent?

A.The limit exists and the choice of δ can be made uniformly for all ε

B.The limit exists, but δ must depend on ε ✅

C.The limit does not exist because δ is bounded by 1

D.The function is bounded near a

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The definition requires δ to be a function of ε; here δ is explicitly chosen as the smaller of 1 and ε, showing that δ indeed depends on ε. This dependence is necessary for the limit to exist, confirming the second statement.

Q6. If a function f has a hole at x = a but the surrounding points satisfy the ε‑δ condition for a limit L, which of the following is true?

A.The limit exists even though f(a) is undefined ✅

B.The limit cannot exist without a defined value at a

C.f must be continuous at a

D.The limit equals the value of f at a

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A hole (removable discontinuity) does not affect the limit because the ε‑δ condition only concerns points arbitrarily close to a, not the point itself. As long as the condition holds, the limit exists regardless of whether f(a) is defined.

Q7. Compare the intervals $(x_0,x_1)$ and $(a-\delta, a+\delta)$ described in the text. Which statement correctly describes their relationship?

A.The second interval is always larger than the first

B.The second interval is a symmetric subinterval of the first ✅

C.The first interval is symmetric about a

D.They are unrelated

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The interval $(a-\delta, a+\delta)$ is deliberately chosen to be symmetric about a and to lie inside the possibly asymmetric interval $(x_0,x_1)$ . Hence it is a subinterval that extends the same distance on both sides of a.

Q8. Evaluate which of the following choices correctly ensures that the interval $(a-\delta, a+\delta)$ lies entirely inside $(x_0,x_1)$ .

A.Choose

\delta = \min\{a-x_0, x_1-a\}

✅

B.Choose

\delta = (x_1-x_0)/2

C.Choose

\delta = |x_1-x_0|

D.Choose any positive δ

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Taking the minimum of the distances from a to each endpoint guarantees that the symmetric interval does not exceed either side of the original interval, ensuring it stays wholly within $(x_0,x_1)$ .

Q9. Differentiate between the statements: (i) ‘For every ε there exists δ such that…’ and (ii) ‘There exists a δ that works for all ε.’ Which of the following correctly captures the distinction?

A.Both statements are equivalent

B.Statement (i) defines a limit, while (ii) would imply uniform continuity

C.Statement (ii) is stronger and rarely true for limits ✅

D.Statement (i) is false for limits

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Statement (i) is the standard limit definition, allowing δ to depend on ε. Statement (ii) would require a single δ to satisfy every ε, which is an unrealistically strong condition and would only hold in trivial cases; thus (ii) is much stronger.

Q10. Which of the following functions fails to satisfy the two‑sided limit definition at $a=0$ despite being bounded near 0?

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

✅

f(x)=x

f(x)=|x|

f(x)=0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The sine of 1/x oscillates infinitely as x approaches 0, never settling near a single value, so no single L can satisfy the ε‑δ condition. The other functions approach 0 (or are constant) and thus meet the definition.

Q11. If the left‑hand limit and right‑hand limit at a point a are equal, which conclusion follows regarding the two‑sided limit?

A.The two‑sided limit does not exist

B.The two‑sided limit equals the common value ✅

C.The function must be continuous at a

D.Only one‑sided limits are defined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When both one‑sided limits exist and are equal, the definition of a two‑sided limit is satisfied, and the limit equals that common value. Continuity further requires the function to be defined at a and equal to this limit, which is not assumed here.

Q12. Apply the definition of a two‑sided limit to explain why $\lim_{x\to 2}\frac{x^2-4}{x-2}=4$ even though the function is undefined at x=2.

A.Because the simplified expression

x+2

approaches 4 as x→2 ✅

B.Because the limit equals the value at x=2

C.Because the function is continuous everywhere

D.Because limits ignore holes

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Algebraic cancellation yields $\frac{x^2-4}{x-2}=x+2$ for all x≠2. As x approaches 2, the simplified expression approaches 4, satisfying the ε‑δ condition despite the original function’s hole at 2.

Q13. Synthesize the geometric interpretation given in the text: how does shrinking ε affect the corresponding δ‑interval around a?

A.A smaller ε forces a larger δ

B.A smaller ε forces a smaller δ, narrowing the interval around a ✅

C.ε and δ are independent

D.δ becomes negative

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Reducing ε narrows the horizontal band around L; to keep the graph of f inside this band, the vertical strip around a must also shrink, requiring a smaller δ. This visual shrinking reflects the formal ε‑δ relationship.

Q14. Explain why the interval $(a-\delta, a+\delta)$ must be chosen so that $\delta < \min\{x_1-a, a-x_0\}$ . Which principle does this reflect?

A.Ensuring the interval stays within the region where f(x) stays within ε of L ✅

B.Guaranteeing symmetry of the function

C.Making the limit equal to f(a)

D.Maximizing the distance from a

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Choosing δ smaller than both distances to the endpoints guarantees the symmetric interval remains entirely inside the original interval where the function values are already known to lie within ε of L. This safeguards the limit condition on both sides.

Q15. If a function satisfies the ε‑δ condition at a point a, what can be said about the continuity of the function at a?

A.The function is automatically continuous at a ✅

B.Continuity depends on whether f(a) is defined and equals L

C.The function must be discontinuous at a

D.Continuity is unrelated to limits

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The ε‑δ condition guarantees the limit exists and equals some L. For continuity, we additionally require that f(a) be defined and that f(a)=L. If those extra requirements hold, the function is continuous; otherwise, it may have a removable discontinuity.

Q16. Using the definition, which of the following statements correctly describes the relationship between ε and δ?

A.For a fixed ε, many δ may work ✅

B.For a fixed δ, only one ε works

C.ε and δ are always equal

D.δ must be larger than ε

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Given an ε, any δ that is small enough to keep x within the ε‑band will satisfy the condition; often there are infinitely many such δ values, illustrating that δ is not unique for a given ε.

Q17. Given that for ε=0.5 we found δ=0.2, and for ε=0.1 we found δ=0.04, what logical conclusion follows about the limit behavior?

A.The limit does not exist because δ decreases

B.The limit exists and the function approaches L more closely as ε decreases ✅

C.δ must always be half of ε

D.The function is constant

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The decreasing δ values correspond to tighter ε tolerances, showing that as we demand f(x) be closer to L (smaller ε), we must restrict x to be nearer to a (smaller δ). This pattern is exactly what a valid limit exhibits.

Q18. Consider the function $f(x)=\begin{cases}x^2 & x\neq 0\\ 1 & x=0\end{cases}$ . Which statement about $\lim_{x\to 0}f(x)$ is correct?

A.The limit equals 1 because f(0)=1

B.The limit equals 0 because

x^2

approaches 0 ✅

C.The limit does not exist because the function is defined differently at 0

D.The limit equals both 0 and 1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For all x≠0, f(x)=x^2, and as x approaches 0, x^2 tends to 0. The isolated value at x=0 does not affect the limiting process, so the limit is 0 despite f(0)=1.

Q19. How does the concept of a two‑sided limit relate to the idea of a function being ‘approachable’ from both directions on the real line?

A.It requires the function to have the same value from left and right ✅

B.It only considers the right‑hand side

C.It ignores the left‑hand side

D.It demands the function be defined at the point

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A two‑sided limit exists only when the values of the function approach the same number whether x approaches the point from the left or from the right. This bidirectional approach ensures the function is ‘approachable’ from both sides.

Q20. If a function fails to satisfy the ε‑δ condition for some ε>0, which of the following statements is necessarily true?

A.The limit exists but is not equal to L

B.The limit does not exist at that point ✅

C.The function is unbounded near a

D.The function is continuous at a

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Failure to meet the ε‑δ requirement for even a single positive ε means we cannot guarantee that f(x) stays within any prescribed distance of a candidate limit L as x approaches a. Consequently, the limit at that point cannot exist.

🔗 Related Topics (MCQs)

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