Answer: $0$

Bound the oscillating factor: For all $x \neq 0$, $-1 \leq \cos(2/x^3) \leq 1$. Multiplying by $x^4 \geq 0$ gives $-x^4 \leq x^4 \cos(2/x^3) \leq x^4$. Apply the Squeeze Theorem: Since $\lim_{x \to 0}(-x^4) = 0$ and $\lim_{x \to 0} x^4 = 0$, the Squeeze Theorem gives $\lim_{x \to 0} x^4 \cos(2/x^3) = 0$. Why distractors fail: Options A and D ($1$ and $-1$) are the extreme values of cosine, but the $x^4$ factor drives the product to $0$. Option C incorrectly concludes the limit does not exist; the oscillation is neutralized by the vanishing polynomial factor.

Answer: The Squeeze Theorem cannot be applied because the bounding limits are not equal, so no conclusion about $\lim_{x \to a} f(x)$ follows from this information alone.

Recall the equal-limits hypothesis: The Squeeze Theorem requires $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$. When the two bounding limits differ, the theorem's hypotheses are not satisfied. Why Option C is correct: With bounding limits of $2$ and $5$, the Squeeze Theorem is inapplicable. We know $f(x)$ values are between the bounds, but the limit of $f$ could be any value in $[2,5]$, or it might not exist — additional information is needed. Why distractors fail: Option A invents an "average" rule that does not exist. Option B sounds plausible but is not guaranteed — $f$ could oscillate between the bounds, and the limit might not exist. Option D arbitrarily picks the upper bound with no justification.

Answer: $\lim_{x \to 0} x \cos\!\left(\frac{1}{x^2}\right)$

Classify each limit technique: Option A yields $5$ by direct substitution. Option B is a $0/0$ form resolved by factoring: $\frac{(x-1)(x+1)}{x-1} = x+1 \to 2$. Option D is also a $0/0$ form resolved by rationalizing. Only Option C involves an oscillating factor $\cos(1/x^2)$ multiplied by a vanishing factor. Why Option C is correct: Direct substitution fails because $\cos(1/x^2)$ does not approach a single value as $x \to 0$. However, $-|x| \leq x\cos(1/x^2) \leq |x|$, and both bounds go to $0$, making this a textbook Squeeze Theorem application. Why distractors fail: Options A, B, and D are all resolvable by algebraic techniques (substitution, factoring, or rationalization) without needing the Squeeze Theorem.

Answer: Yes, because both $x^2 e^{-1}$ and $x^2 e$ approach $0$ as $x \to 0$, so the limit is $0$.

Verify the bounding inequality: Since $-1 \leq \sin(1/x) \leq 1$ and $e^t$ is increasing, $e^{-1} \leq e^{\sin(1/x)} \leq e$. Multiplying by $x^2 \geq 0$ preserves the inequality direction. Check the bounding limits: $\lim_{x \to 0} x^2 e^{-1} = 0$ and $\lim_{x \to 0} x^2 e = 0$. Both bounding limits equal $0$, so the Squeeze Theorem applies and the limit is $0$. Why distractors fail: Option A is wrong because $e^{\sin(1/x)}$ is bounded between $e^{-1}$ and $e$. Option B is wrong because there is no restriction to polynomial bounds. Option D is wrong because the $x^2$ factor drives both bounds to $0$.

Answer: Two bounding functions $g(x)$ and $h(x)$ must satisfy $g(x) \leq f(x) \leq h(x)$ near $a$ and share the same limit at $a$.

Identify the hypotheses: The Squeeze Theorem has two hypotheses: (1) an inequality $g(x) \leq f(x) \leq h(x)$ holds near $a$, and (2) the bounding functions share the same limit, $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$. Why Option C is correct: Option C captures both required conditions: the bounding inequality and the equal limits of the outer functions. Why distractors fail: Option A is wrong because the theorem works even if $f$ is not continuous at $a$; it only guarantees the limit exists. Option B is wrong because differentiability is not required. Option D is wrong because $f(a)$ need not be defined for the limit to exist.

Answer: Because $|f(x)| \leq g(x)$ is equivalent to $-g(x) \leq f(x) \leq g(x)$, and if $g(x) \to 0$ then both bounds approach $0$.

Unpack the absolute value inequality: The inequality $|f(x)| \leq g(x)$ is equivalent to $-g(x) \leq f(x) \leq g(x)$. This is precisely the setup needed for the Squeeze Theorem. Why Option B is correct: If $g(x) \to 0$ as $x \to a$, then $-g(x) \to 0$ as well. With both outer bounds approaching $0$, the Squeeze Theorem yields $\lim_{x \to a} f(x) = 0$. Why distractors fail: Option A is wrong because $|f(x)| \leq g(x)$ does not mean $f = g$. Option C is wrong because absolute value bounding does not imply continuity. Option D is tempting — it is true that $|f(x)| \to 0 \Leftrightarrow f(x) \to 0$ — but the question asks why the bounding technique is sufficient; the answer is that it sets up the Squeeze Theorem, making Option B the more complete and precise explanation.

Answer: The bounding functions $\pm\frac{1}{|x|}$ both diverge as $x \to 0$, so they do not share a common finite limit and the Squeeze Theorem cannot be applied.

Check the bounding limits: While $|\sin(1/x)| \leq 1$ gives $\left|\frac{\sin(1/x)}{x}\right| \leq \frac{1}{|x|}$, the bounds $\pm\frac{1}{|x|}$ diverge to $\pm\infty$ as $x \to 0$. Why Option B is correct: The Squeeze Theorem requires the two outer functions to converge to the same finite limit $L$. Since $\frac{1}{|x|} \to \infty$, the hypotheses are not met and no conclusion can be drawn. Why distractors fail: Option A is wrong because the inequality is algebraically valid. Option C is wrong because the Squeeze Theorem works for any common finite limit $L$, not just $0$. Option D is wrong because $\frac{\sin(1/x)}{x}$ is not a $0/0$ form (the numerator does not approach a fixed value), and in fact this limit does not exist.

Answer: If $g(x) \leq f(x) \leq h(x)$ near $a$ and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

Recall the theorem statement: The Squeeze Theorem requires three functions where $g(x) \leq f(x) \leq h(x)$ in a neighborhood of $a$ (except possibly at $a$ itself), and the outer two functions share the same limit $L$ as $x \to a$. Why Option A is correct: Option A exactly matches the formal statement: the bounding functions share a common limit $L$, so the squeezed function $f(x)$ must also approach $L$. Why distractors fail: Option B is wrong because the Squeeze Theorem guarantees existence of a limit, not continuity — $f(a)$ could differ from $L$ or be undefined. Option C is wrong for the same reason: bounding by continuous functions does not force $f$ to be continuous. Option D is wrong because the theorem requires both bounding limits to be equal; if $L_1 \neq L_2$, no conclusion about $\lim f(x)$ can be drawn.

Answer: All $n \geq 2$

Use the limit definition of the derivative: We need $\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^n \sin(1/h)}{h} = \lim_{h \to 0} h^{n-1} \sin(1/h)$. Apply the Squeeze Theorem: For $n \geq 2$, $|h^{n-1} \sin(1/h)| \leq |h|^{n-1} \to 0$ as $h \to 0$. By the Squeeze Theorem, the limit is $0$, so $f'(0) = 0$. Check $n = 1$: When $n = 1$, the difference quotient becomes $\sin(1/h)$, which oscillates between $-1$ and $1$ as $h \to 0$. This limit does not exist, so $f$ is not differentiable at $0$ when $n = 1$. Why distractors fail: Option A is wrong because $n = 1$ is precisely the case where differentiability fails. Option B is wrong for the same reason. Option D is wrong because $n = 3$ (odd, $\geq 2$) also works.

Answer: The claim is invalid: L'Hôpital's Rule requires a $0/0$ or $\infty/\infty$ form, and $x^2 \sin(1/x)$ is not naturally in such a form; even if rearranged, differentiating $\sin(1/x)$ produces another oscillating expression.

Check applicability of L'Hôpital's Rule: L'Hôpital's Rule requires an indeterminate form $0/0$ or $\infty/\infty$. The expression $x^2 \sin(1/x)$ is a product, not a quotient. Rewriting as $\frac{\sin(1/x)}{1/x^2}$ attempts to create an $\text{oscillating}/\infty$ form, which is not an indeterminate form L'Hôpital addresses. Why Option C is correct: Even if one attempts to rewrite, differentiating $\sin(1/x)$ yields $-\frac{\cos(1/x)}{x^2}$, which still oscillates. L'Hôpital's Rule does not resolve the problem, whereas the Squeeze Theorem handles it elegantly. Why distractors fail: Option A is wrong because $\sin(1/x)$ does not approach $0$ as $x \to 0$, so the form is not $0/0$. Option B is wrong as shown above. Option D is wrong because L'Hôpital's Rule applies to any differentiable functions, not just rational ones.

Answer: $-x^2 \leq x^2 \sin(1/x) \leq x^2$

Bound the oscillating factor: Since $-1 \leq \sin(1/x) \leq 1$ for all $x \neq 0$, multiplying throughout by $x^2 \geq 0$ gives $-x^2 \leq x^2 \sin(1/x) \leq x^2$. Apply the Squeeze Theorem: Both $\lim_{x \to 0}(-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, so the Squeeze Theorem yields $\lim_{x \to 0} x^2 \sin(1/x) = 0$. Why distractors fail: Option A uses bounds of $\pm x$, which are not correct since multiplying $\sin(1/x) \in [-1,1]$ by $x^2$ gives $\pm x^2$, not $\pm x$. Option C is a true inequality but useless for squeezing because the bounds $-1$ and $1$ do not approach the same limit (they are constants with different signs). Option D is incorrect because $x^2 \sin(1/x)$ can be negative, so the lower bound of $0$ is wrong.

Answer: $-|x|^3 \leq x^3 \cos(1/x)\sin(1/x) \leq |x|^3$, so the limit is $0$.

Bound the oscillating product: Since $|\cos(1/x) \sin(1/x)| \leq 1$ for all $x \neq 0$ (each factor is bounded by $1$ in absolute value, and their product is bounded by $1$), we have $|x^3 \cos(1/x)\sin(1/x)| \leq |x|^3$, giving $-|x|^3 \leq x^3\cos(1/x)\sin(1/x) \leq |x|^3$. Why Option A is correct: The bounds $-|x|^3$ and $|x|^3$ are valid for all $x \neq 0$, including negative $x$, and both approach $0$ as $x \to 0$. By the Squeeze Theorem, the limit is $0$. Why distractors fail: Option B uses bounds of $\pm x^6$, which are not correctly derived — bounding $|\cos\sin| \leq 1$ and multiplying by $|x^3|$ gives $|x|^3$, not $x^6$. Option C assumes the product is always non-negative, which is false since both $x^3$ and $\cos(1/x)\sin(1/x)$ can be negative. Option D uses constant bounds of $\pm 1$, which do not share the same limit ($-1 \neq 1$), so the Squeeze Theorem cannot conclude the limit is $0$ from these bounds.

Answer: $0$

Substitute $u = x - 3$: Let $u = x - 3$. Then as $x \to 3$, $u \to 0$, and $f = u^2 \cos(1/u)$. This is a standard Squeeze Theorem form. Apply bounding: Since $-1 \leq \cos(1/u) \leq 1$ and $u^2 \geq 0$, we get $-u^2 \leq u^2 \cos(1/u) \leq u^2$. Both bounds approach $0$ as $u \to 0$. Why distractors fail: Options A and C ($\pm 1$) correspond to extreme values of cosine but ignore the vanishing $(x-3)^2$ factor. Option B is incorrect; the Squeeze Theorem tames the oscillation.

Answer: Although $\sin(1/x)$ oscillates, it is multiplied by $x$, which approaches $0$ and forces the product to $0$ by the Squeeze Theorem.

Identify the error in the student's reasoning: The student ignores the effect of the factor $x$. While $\sin(1/x)$ alone oscillates without a limit, the product $x \sin(1/x)$ is bounded: $-|x| \leq x \sin(1/x) \leq |x|$. Why Option B is correct: Since $\lim_{x \to 0} |x| = 0$, the Squeeze Theorem forces $\lim_{x \to 0} x \sin(1/x) = 0$. The oscillation is "squeezed out" by the vanishing factor. Why distractors fail: Option A is wrong because the limit does exist and equals $0$. Option C is wrong because the limit is $0$, not $1$, and the reasoning about "dampened by continuity" is not a valid principle. Option D is wrong because the Squeeze Theorem is precisely the standard tool for handling oscillating functions multiplied by vanishing factors.

Answer: Yes, because the Squeeze Theorem shows $\lim_{x \to 0} f(x) = 0 = f(0)$, which is exactly the definition of continuity at $x = 0$.

Recall the definition of continuity: A function $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. This requires: (1) $f(a)$ is defined, (2) the limit exists, (3) they are equal. Verify all three conditions: Here $f(0) = 0$ is defined. The Squeeze Theorem gives $\lim_{x \to 0} f(x) = 0$. Since $0 = 0$, all three conditions are met. Why Option C is correct: The student's reasoning is complete: the Squeeze Theorem establishes the limit, and the piecewise definition supplies $f(0) = 0$, matching the limit. Why distractors fail: Option A overlooks that the student also verified $f(0) = 0$; together with the limit, this is sufficient for continuity. Option B is wrong because $f(0) = 0$ by the piecewise definition. Option D is wrong because $x^2 \sin(1/x)$ is not the same as $x^2$; the Squeeze Theorem is essential to handling the oscillating factor.

Answer: $0$

Establish bounds: For all $x > 0$, $-1 \leq \sin x \leq 1$. Dividing by $x > 0$: $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$. Apply the Squeeze Theorem: Both $\lim_{x \to \infty}(-1/x) = 0$ and $\lim_{x \to \infty}(1/x) = 0$. By the Squeeze Theorem, $\lim_{x \to \infty} \frac{\sin x}{x} = 0$. Why distractors fail: Option A ($1$) confuses this with the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which is a completely different limit. Option B ($-1$) has no basis. Option C is wrong because even though $\sin x$ oscillates, dividing by $x \to \infty$ forces the fraction to $0$.

Answer: $0$

Establish bounds: Since $-1 \leq \sin(1/x^2) \leq 1$ and $\sqrt{x} \geq 0$ for $x > 0$, we have $-\sqrt{x} \leq \sqrt{x}\,\sin(1/x^2) \leq \sqrt{x}$. Apply the Squeeze Theorem: Both $\lim_{x \to 0^+}(-\sqrt{x}) = 0$ and $\lim_{x \to 0^+} \sqrt{x} = 0$. By the Squeeze Theorem, the limit is $0$. Why distractors fail: Option A ($1/2$) has no basis in the computation. Option B is incorrect because the bounding argument shows the limit does exist. Option D ($1$) confuses the bound on sine with the overall limit.