Answer: Yes, because $\delta = \varepsilon / 3 \leq \varepsilon / 2$, and any δ that is at most $\varepsilon / 2$ satisfies the definition.

Find the standard δ: $|f(x) - 13| = |2x + 7 - 13| = 2|x - 3|$. For this to be less than $\varepsilon$, we need $|x - 3| < \varepsilon / 2$. So $\delta = \varepsilon / 2$ is the natural choice. Any smaller δ also works: If $\delta' \leq \delta$, then $0 < |x - 3| < \delta'$ implies $0 < |x - 3| < \delta$, so the implication still holds. Since $\varepsilon / 3 < \varepsilon / 2$, the student's δ is valid. Why distractors fail: Option A gives a correct conclusion but with incomplete reasoning (it doesn't explain why being smaller works logically). Option B is wrong because δ need not be exactly $\varepsilon / 2$; any positive value $\leq \varepsilon / 2$ works. Option D reverses the inequality — δ must be at most $\varepsilon / |m|$, not larger.

Answer: $\delta = \min(1, \varepsilon / 2)$

Factor the expression: $|x^2 - 3x - (-2)| = |x^2 - 3x + 2| = |(x - 1)(x - 2)| = |x - 2||x - 1|$. Bound the auxiliary factor: Restrict $\delta \leq 1$: if $|x - 2| < 1$, then $1 < x < 3$, so $0 < x - 1 < 2$ and $|x - 1| < 2$. Choose δ: $|x - 2||x - 1| < 2|x - 2| < 2\delta$. For this to be less than $\varepsilon$, we need $\delta \leq \varepsilon / 2$. So $\delta = \min(1, \varepsilon / 2)$ satisfies both constraints. Why distractors fail: Option A: $\delta = \min(1, \varepsilon)$ gives $2\delta \leq 2\varepsilon$, which does not guarantee $|f(x) - L| < \varepsilon$. Option C: $\delta = \varepsilon/5$ has no $\min(1, \cdot)$ restriction, so for large $\varepsilon$ (e.g., $\varepsilon = 100$), $\delta = 20$, and the bound $|x - 1| < 2$ is no longer valid since $x$ could be far from 2. Option D: $\delta = \varepsilon/2$ without the minimum restriction has the same problem — when $\varepsilon$ is large (e.g., $\varepsilon = 10$, so $\delta = 5$), the assumption $|x - 2| < 1$ that gave us $|x - 1| < 2$ is violated, so the bound breaks down.

Answer: For every $\varepsilon > 0$ there exists $\delta > 0$ such that if $0 < |x - a| < \delta$ then $|f(x) - L| < \varepsilon$.

Recall the precise statement: The definition requires: for every $\varepsilon > 0$, there exists $\delta > 0$ such that $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon$. Why Option B is correct: It matches the standard definition exactly, with the correct order of quantifiers and the strict inequality $0 < |x - a|$ excluding $x = a$. Why distractors fail: Option A swaps ε and δ in the quantifiers. Option C omits the condition $0 < |x - a|$, which is essential to exclude the point $x = a$ itself. Option D reverses the quantifier order, making ε fixed rather than universally quantified.

Answer: If $0 < |x - 1| < 0.01$, then $|4x - 4| = 4|x - 1| < 4(0.01) = 0.04 = \varepsilon$. ✓

Set up the verification: We need: if $0 < |x - 1| < \delta = 0.01$, then $|(4x - 1) - 3| = |4x - 4| = 4|x - 1| < \varepsilon = 0.04$. Why Option B is correct: Substituting the bound: $4|x - 1| < 4 \cdot 0.01 = 0.04 = \varepsilon$. The inequality holds (we accept $\leq \varepsilon$ or $< \varepsilon$ depending on convention, and $<$ is satisfied for all $|x-1| < 0.01$). Why distractors fail: Option A incorrectly substitutes 0.04 (ε) instead of 0.01 (δ) into the bound on $|x - 1|$, then falsely claims 0.16 < 0.04. Option C drops the factor of 4. Option D simplifies $(4x - 1) - 3$ incorrectly as $|4x + 2|$ instead of $|4x - 4|$.

Answer: Both proofs are valid; any $\delta \leq \varepsilon / 3$ satisfies the definition.

Verify the limit: $3(5) - 7 = 8$, confirming the limit is indeed 8. Check both δ choices: $|(3x - 7) - 8| = 3|x - 5|$. We need $3|x - 5| < \varepsilon$, so $|x - 5| < \varepsilon/3$. Any $\delta \leq \varepsilon/3$ works. Since $\varepsilon/4 < \varepsilon/3$, both choices are valid. Why distractors fail: Option A incorrectly claims uniqueness. Option B incorrectly claims only the smaller δ works. Option D is wrong because the limit computation is correct.

Answer: Yes, because $|x^2 - 1| = |x - 1||x + 1| < \delta \cdot 3 \leq (\varepsilon/3) \cdot 3 = \varepsilon$.

Check the bound on |x + 1|: If $|x - 1| < 1$, then $0 < x < 2$, so $1 < x + 1 < 3$. Thus $|x + 1| < 3$ is correct. Verify the chain of inequalities: $|x^2 - 1| = |x-1||x+1| < \delta \cdot 3$. With $\delta = \min(1, \varepsilon/3)$, we have $\delta \leq \varepsilon/3$, so $\delta \cdot 3 \leq \varepsilon$. Why distractors fail: Option A is wrong because the bound is valid. Option B is wrong because there is no such rule. Option D is wrong because when $\varepsilon \geq 1$, $\delta = \min(1, \varepsilon/3) = 1$ (if $\varepsilon \geq 3$) or $\varepsilon/3$ (if $1 \leq \varepsilon < 3$), and the proof still works.

Answer: The proof is missing the condition $0 < |x - 0|$; it should say $0 < |x| < \delta$ instead of just $|x| < \delta$.

Check the δ computation: The algebra and δ choice are correct: $|(3x+2) - 2| = 3|x|$, and choosing $\delta = \varepsilon/3$ yields $3|x| < 3 \cdot \varepsilon/3 = \varepsilon$. The limit value $3(0) + 2 = 2$ is also correct. Identify the missing condition: The formal definition requires $0 < |x - a| < \delta$, not just $|x - a| < \delta$. The strict lower bound $0 <$ ensures $x \neq a$. While this omission doesn't affect the final inequality here, it is a formal error in stating the definition. Why distractors fail: Option A is wrong because $\varepsilon / 3$ is a correct and sufficient choice. Option C is wrong because $|3x| = 3|x|$ is a standard absolute value property. Option D is wrong because $\lim_{x \to 0}(3x + 2) = 2$ is correct.

Answer: $\delta = \varepsilon / 5$

Set up the inequality: We need $|f(x) - L| < \varepsilon$, i.e., $|(5x+3) - 13| = |5x - 10| = 5|x - 2| < \varepsilon$. Solve for the required bound on |x − 2|: From $5|x - 2| < \varepsilon$ we get $|x - 2| < \varepsilon / 5$. So choosing $\delta = \varepsilon / 5$ works. Why distractors fail: Option A uses $5\varepsilon$, which is too large and would not guarantee $|f(x) - L| < \varepsilon$. Option C uses subtraction, which is dimensionally inappropriate and could be negative. Option D divides by 3 instead of the correct slope magnitude 5.

Answer: δ depends on ε (and sometimes on the point $a$), and any δ that satisfies the implication is valid — it need not equal ε.

Relationship between δ and ε: In any ε-δ proof, $\delta$ is chosen as a function of $\varepsilon$ (and possibly the point $a$). The proof need only show that the chosen $\delta$ makes the implication true. Why Option C is correct: It accurately states that δ depends on ε but need not equal it. For instance, for $f(x) = 3x$, one might choose $\delta = \varepsilon / 3$. Why distractors fail: Option A perpetuates the misconception. Option B adds a false restriction — δ can be larger than ε in some cases. Option D reverses the logical dependency; δ is chosen after ε is given.

Answer: It leads to $\delta = \min(2, \varepsilon/8)$, which is a valid but slightly less tight choice than $\min(1, \varepsilon/7)$.

Compare the two approaches: With $\delta \leq 1$: $|x+3| < 7$, so $\delta = \min(1, \varepsilon/7)$. With $\delta \leq 2$: $|x+3| < 8$, so $\delta = \min(2, \varepsilon/8)$. Both are valid proofs: In either case, $|x^2 - 9| = |x-3||x+3|$ is bounded by $\varepsilon$. The initial restriction is a proof technique, and many valid choices exist. Why distractors fail: Option A is wrong because the proof is still valid with $\delta \leq 2$. Option C reverses the comparison — a wider restriction gives a weaker bound on $|x+3|$, not tighter. Option D is wrong because $\min(1, \varepsilon/7) \neq \min(2, \varepsilon/8)$ in general.

Answer: $\delta = \min(1, \sqrt{\varepsilon})$

Simplify the expression: $|x^2 + 2x + 1| = |(x+1)^2| = |x+1|^2$. We need $|x+1|^2 < \varepsilon$, i.e., $|x+1| < \sqrt{\varepsilon}$. Choose δ carefully: Setting $\delta = \sqrt{\varepsilon}$ would work if $\sqrt{\varepsilon}$ is not too large. To be safe (and follow standard practice), we use $\delta = \min(1, \sqrt{\varepsilon})$ to keep δ bounded. Why distractors fail: Option A: $\delta = \varepsilon^2$ gives $|x+1|^2 < \varepsilon^4$, not $\varepsilon$. Option B: $\delta = \sqrt{\varepsilon}$ works for small ε but lacks the standard bounding restriction. However, it is technically valid — but Option D is the more rigorous standard choice that also keeps δ bounded. Option C: $\delta = \min(1, \varepsilon)$ gives $|x+1|^2 < \varepsilon^2 \neq \varepsilon$ when $\varepsilon < 1$.

Answer: δ may depend on the given ε, and a valid δ must be produced for each possible ε.

Quantifier order matters: In logic, $\forall \varepsilon > 0,\; \exists \delta > 0$ means the choice of $\delta$ comes after and may depend on $\varepsilon$. Swapping the order would change the meaning entirely. Why Option B is correct: Because ε is universally quantified first, δ must be found for every possible ε. This means δ is typically a function of ε. Why distractors fail: Option A reverses the dependency. Option C is wrong because they are not independent — δ depends on ε. Option D is wrong because the definition demands success for all ε, not just one pair.

Answer: To exclude the point $x = a$ so that the value $f(a)$ (or its existence) does not affect whether the limit exists.

Role of the strict inequality: The condition $0 < |x - a|$ means $x \neq a$. The limit describes the behavior of $f(x)$ as $x$ approaches $a$, not the value at $a$ itself. Why Option B is correct: Limits are about what happens near $a$, regardless of whether $f(a)$ is defined or equals $L$. Excluding $x = a$ captures this idea. Why distractors fail: Option A is irrelevant — $\delta > 0$ is guaranteed by its own quantifier. Option C confuses limits with continuity; continuity additionally requires $f(a) = L$. Option D is wrong because $\varepsilon > 0$ is already ensured by the universal quantifier.

Answer: The algebraic limit laws (sum, product, quotient rules for limits) are themselves theorems that require the ε-δ definition for their proofs; removing it would leave these laws unproven.

Foundation vs. computation: The algebraic rules we use daily (e.g., the limit of a sum is the sum of the limits) are not axioms — they are theorems derived from the ε-δ definition. Why Option B is the strongest counterargument: Removing the formal definition would undermine the logical foundation of all limit laws, leaving calculus without rigorous justification. This directly refutes the claim that the definition is unnecessary. Why distractors fail: Option A concedes the classmate's point rather than refuting it. Option C is far too narrow — the definition underpins all of calculus, not just trigonometric limits. Option D is wrong because numerical approximation doesn't require ε-δ.

Answer: $\delta = 0.05$

Compute the required bound: $|(2x - 1) - 5| = |2x - 6| = 2|x - 3|$. For this to be less than $\varepsilon = 0.1$, we need $|x - 3| < 0.05$. Why Option B is correct: $\delta = 0.05$ gives $2|x - 3| < 2(0.05) = 0.1 = \varepsilon$. This is the exact threshold, and any $\delta \leq 0.05$ works. Why distractors fail: Option A: $\delta = 0.1$ gives $2(0.1) = 0.2 > 0.1$. Option C: $\delta = 0.5$ gives $2(0.5) = 1.0 > 0.1$. Option D: $\delta = 0.2$ gives $2(0.2) = 0.4 > 0.1$. All exceed ε.

Answer: $\delta = \min(1, \varepsilon / 9)$

Combine the two constraints: We need $\delta \leq 1$ (to bound $|x+4|$) and $9|x - 4| < \varepsilon$, i.e., $|x - 4| < \varepsilon / 9$. So $\delta$ must satisfy both: $\delta \leq 1$ and $\delta \leq \varepsilon / 9$. Why Option B is correct: $\delta = \min(1, \varepsilon/9)$ satisfies both constraints simultaneously. Then $|x^2 - 16| = |x-4||x+4| < \delta \cdot 9 \leq (\varepsilon/9) \cdot 9 = \varepsilon$. Why distractors fail: Option A ignores the restriction $\delta \leq 1$; for large ε, $\varepsilon/9 > 1$ would invalidate the bound on $|x+4|$. Option C multiplies by 9 instead of dividing — this gives too large a δ. Option D uses 16 instead of 9, which doesn't match the bound derived.

Answer: The ε-δ definition provides the rigorous logical foundation that justifies the algebraic limit laws and ensures results hold without relying on intuition alone.

Purpose of rigor: Algebraic limit laws (sum, product, quotient rules) are themselves theorems that must be proved. The ε-δ definition is the foundation from which these laws are derived. Why Option B is correct: Without the formal definition, we cannot prove the limit laws or resolve subtle cases; it provides the logical bedrock of calculus. Why distractors fail: Option A dismisses the definition's ongoing importance. Option C is too narrow — the definition applies to all limit statements, not just discontinuous functions. Option D is false because algebraic techniques handle polynomials easily via direct substitution.

Answer: It is valid: if $0 < |x| < \varepsilon^{1/3}$, then $|x^3 - 0| = |x|^3 < \varepsilon$, and there is no requirement that $\delta \leq 1$.

Verify the algebra: $|x^3 - 0| = |x|^3$. If $|x| < \varepsilon^{1/3}$, then $|x|^3 < (\varepsilon^{1/3})^3 = \varepsilon$. The inequality holds for all $\varepsilon > 0$. Is δ ≤ 1 required?: No. Restricting $\delta \leq 1$ is a common technique to bound auxiliary factors in polynomial proofs, but it is not part of the definition. If δ works without the restriction, the proof is valid. Why distractors fail: Option A correctly identifies the algebra but wrongly claims δ > 1 is problematic. Option C states a false rule. Option D is wrong because the proof works for all ε > 0; when $\varepsilon \geq 1$, $\delta = \varepsilon^{1/3} \geq 1$ is perfectly valid.

Answer: No matter how small a tolerance ε is chosen around the limit value, a corresponding neighborhood δ around the input can be found to keep the function within that tolerance.

Core meaning of the quantifiers: The phrase 'for every $\varepsilon > 0$' means the challenger can pick any positive tolerance, no matter how tiny. 'There exists $\delta > 0$' means the prover must respond with a suitable neighborhood size. Why the correct answer works: Option B captures this challenge-response structure: any tolerance around $L$ can be matched by an appropriate neighborhood around $a$. Why distractors fail: Option A is wrong because δ generally depends on ε — a single δ need not work for all ε. Option C reverses the goal; we need δ to satisfy the condition, not fail it. Option D is wrong because there is no requirement that ε < δ.

Answer: 6

Determine the range of x: If $|x - 2| < 1$, then $1 < x < 3$, so $4 < x + 3 < 6$. Find the tightest bound: The maximum value of $|x + 3|$ in this interval is strictly less than 6 (approaching 6 as $x \to 3$). The tightest integer upper bound is 6. Why distractors fail: Option A (5) is too small — at $x = 2.5$, $|x+3| = 5.5 > 5$. Option C (7) is a valid upper bound but not the tightest. Option D (4) is clearly too small since $|x + 3| > 4$ throughout the interval.

Answer: $\lim_{x \to 2} f(x) \neq 5$.

Negate the ε-δ definition: The negation of 'for every $\varepsilon > 0$ there exists $\delta > 0$ such that ...' is 'there exists $\varepsilon > 0$ such that for every $\delta > 0$ there is a counterexample.' Finding one such $\varepsilon$ disproves the claimed limit value. Why Option B is correct: Since $\varepsilon = 0.01$ is a witness to the failure of the definition, we can conclude the limit is not 5. However, this says nothing about whether the limit exists with a different value. Why distractors fail: Option A is contradicted by the evidence. Option C goes too far — the limit might exist but equal something other than 5. Option D is wrong because finding even one ε that fails for all δ is sufficient to disprove the specific limit claim.

Answer: It provides a bound on the factor $|x + 4|$ in the expression $|x^2 - 16| = |x - 4||x + 4|$, so that the product can be controlled by $|x - 4|$ alone.

Factor the key expression: $|x^2 - 16| = |x - 4||x + 4|$. To bound this by $\varepsilon$, we need to control both factors. Why we restrict δ: If $|x - 4| < 1$, then $3 < x < 5$, so $7 < x + 4 < 9$, giving $|x + 4| < 9$. This yields $|x^2 - 16| < 9|x - 4|$, reducing the problem to a linear bound. Why Option B is correct and distractors fail: Option B identifies the strategic purpose. Option A is irrelevant — squaring doesn't require positivity. Option C is wrong because δ is ultimately $\min(1, \varepsilon/9)$, not always 1. Option D dismisses the mathematical necessity of the bound.