Answer: The value $f(x)$ approaches as $x$ approaches $4$ from values less than $4$

Interpret the notation: The superscript $-$ in $\lim_{x \to 4^-}$ indicates a left-hand limit, meaning $x$ approaches $4$ from values strictly less than $4$. Why the correct answer works: Option B correctly states that the notation describes the value $f(x)$ tends toward when $x$ comes from the left side (values below $4$). Why distractors fail: Option A confuses the limit with the function value at the point. Option C describes $\lim_{x \to 4^+}$, the right-hand limit. Option D misreads the superscript as a negative sign on the number itself.

Answer: The reviewer is incorrect; this situation describes a removable discontinuity, which is perfectly valid.

Recall the relationship between limits and function values: A limit $\lim_{x \to a} f(x) = L$ only requires that $f(x) \to L$ as $x \to a$. The actual value $f(a)$ can be anything — or even undefined. Why the correct answer works: When $\lim_{x \to 0} f(x) = 3$ but $f(0) = 7$, the function has a removable discontinuity at $x = 0$. This is a standard and valid scenario. Why distractors fail: Option A states a false claim — limits need not equal function values. Option C is wrong because removable discontinuities can occur in non-piecewise functions too (e.g., $\frac{x^2-1}{x-1}$ at $x=1$ if redefined). Option D is nonsensical — the limit can certainly exist when $f(a)$ is defined.

Answer: $5$

Recall the definition of a limit: The limit $\lim_{x \to a} f(x)$ depends only on the values $f(x)$ approaches as $x$ gets close to $a$, not the actual value $f(a)$. Apply to the graph: From the graph, as $x \to 3^-$ and $x \to 3^+$, the function approaches $y = 5$. Both one-sided limits agree, so $\lim_{x \to 3} f(x) = 5$. Why distractors fail: Option A ($2$) confuses the limit with the function value $f(3)=2$. Option C ($3$) confuses the limit value with the $x$-value. Option D is wrong because both one-sided limits agree at $5$, so the limit exists.

Answer: $\lim_{x \to 2} f(x) = 7$, because both one-sided limits exist and are equal

Key principle for two-sided limits: $\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$. Why the correct answer works: Both one-sided limits equal $7$, so by the definition of the two-sided limit, $\lim_{x \to 2} f(x) = 7$. No information about $f(2)$ is needed. Why distractors fail: Option B gives the right value but the wrong reasoning — the limit's existence does not depend on $f(2)$. Option C incorrectly requires checking $f(2)$. Option D absurdly adds one-sided limits.

Answer: Student B, because Student A's chosen $x$-values happened to give special results that may not represent the general trend

Understand the function's behavior: $\lfloor 1/x \rfloor \cdot x$ behaves irregularly. For $x = 1/n$, $\lfloor 1/x \rfloor = n$ so $f(x)=1$. But for other $x$-values, $\lfloor 1/x \rfloor \cdot x$ takes different values, generally approaching $1$ but not always equaling $1$. Why the correct answer works: Student A happened to choose values of the form $x = 1/n$ (or near it), which give the constant output $1$. Student B's varied choices reveal more nuanced behavior, making their analysis more robust. This highlights the danger of selecting convenient table values. Why distractors fail: Option A is wrong because three conveniently chosen points don't prove a limit. Option C is wrong because tables can give misleading results depending on which $x$-values are chosen. Option D is wrong — limits involving the floor function can often still be estimated or proven.

Answer: The claim is true only when $f$ is continuous at $x = a$.

Recall the definition of continuity: A function $f$ is continuous at $x = a$ if and only if $\lim_{x \to a} f(x) = f(a)$. This requires the limit to exist, $f(a)$ to be defined, and the two to be equal. Why the correct answer works: The student's statement matches the continuity condition exactly. It holds when $f$ is continuous at $a$, but not in general — a function can be defined at $a$ with a removable discontinuity where the limit differs from $f(a)$. Why distractors fail: Option A is wrong because piecewise or pathological functions can have $f(a)$ defined but the limit different. Option C is too restrictive — differentiability implies continuity, but continuity does not require differentiability. Option D is wrong because many functions (all continuous ones) satisfy the claim.

Answer: Only at $x = 0$

Check $x = 0$: $\lim_{x \to 0^-}(2x+1) = 1$ and $\lim_{x \to 0^+}(x^2+1) = 1$. These agree, so $\lim_{x \to 0} g(x) = 1$. Check $x = 3$: $\lim_{x \to 3^-}(x^2+1) = 10$ and $\lim_{x \to 3^+}(12-x) = 9$. These disagree ($10 \neq 9$), so $\lim_{x \to 3} g(x)$ does not exist. Why distractors fail: Option B is wrong because the limit at $x=3$ does not exist. Option C is wrong because the limit fails at $x=3$. Option D is wrong because the limit does exist at $x=0$.

Answer: A table only samples finitely many points and cannot rule out unexpected behavior between or beyond those samples

Understand the limitation of numerical evidence: A table of values, no matter how many entries, represents a finite sample. The function could behave erratically at values not included in the table. Why the correct answer works: A rigorous proof of a limit requires reasoning about all $x$ near $a$ (as in the $\epsilon$-$\delta$ definition), not just selected samples. A table provides evidence, not proof. Why distractors fail: Option A is wrong — tables work as estimates for any function type. Option C is wrong — tables can estimate both one-sided and two-sided limits. Option D is wrong — the limit does not depend on the value at $x = 5$.

Answer: $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$

Identify the key principle: A two-sided limit $\lim_{x \to a} f(x)$ exists if and only if the left-hand limit and the right-hand limit both exist and are equal. Why the correct answer works: When $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$, the one-sided limits disagree, which directly prevents the two-sided limit from existing. Why distractors fail: Option A: $f(a)$ being undefined does not prevent the limit from existing (e.g., $f(x) = \frac{x^2-1}{x-1}$ at $x=1$). Option C describes a removable discontinuity — the limit still exists, it just doesn't equal $f(a)$. Option D: a removable discontinuity means the limit exists but differs from $f(a)$.

Answer: Because $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely often as $x \to 0$, never settling on a single value

Understand the behavior near $x = 0$: As $x \to 0$, $1/x \to \pm\infty$, and $\sin$ continues to cycle between $-1$ and $1$ without converging. Why the correct answer works: The infinite oscillation means no single number $L$ can be the limit. For any proposed $L$, there are values of $x$ arbitrarily close to $0$ where $\sin(1/x)$ is far from $L$. Why distractors fail: Option A: being undefined at a point doesn't prevent a limit from existing. Option C: while $1/x \to \infty$, the sine function bounds the output — the issue is oscillation, not unboundedness. Option D: neither one-sided limit exists either, because the oscillation occurs from both sides.

Answer: $4$

Examine the table from the left: As $x$ approaches $2$ from the left: $f(1.9)=3.61$, $f(1.99)=3.96$, $f(1.999)=3.996$. The values are approaching $4$. Examine the table from the right: As $x$ approaches $2$ from the right: $f(2.001)=4.004$, $f(2.01)=4.04$, $f(2.1)=4.41$. The values are also approaching $4$. Why distractors fail: Options A and C pick specific table entries rather than the value being approached. Option D uses the value farthest from $x=2$ and ignores the trend.

Answer: $5$

Compute the left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-}(x^2+1) = 4 + 1 = 5$. Compute the right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+}(3x-1) = 6 - 1 = 5$. Conclude: Both one-sided limits equal $5$, so $\lim_{x \to 2} f(x) = 5$. Why distractors fail: Option A ($4$) comes from $2^2$ without adding $1$. Option C ($3$) might result from a computation error like $3(2)-3$. Option D is wrong because the one-sided limits agree.

Answer: Valid; even though the sign changes, the magnitudes approach $0$ from both sides

Observe the magnitudes: Despite sign changes, $|f(x)|$ decreases: $0.0544 \to 0.00507 \to 0.000827$. From both sides, the outputs shrink toward $0$. Why the correct answer works: For a limit to equal $0$, we need $|f(x)|$ to approach $0$, regardless of sign oscillation. The table confirms this trend. Why distractors fail: Option A confuses oscillation in sign with a non-existent limit — oscillation only prevents a limit when the values don't converge. Option C is irrelevant; being undefined at a point doesn't prevent a limit. Option D is factually wrong — the values are not all positive.

Answer: $3$

Read the numerical trend: From both sides, $f(x)$ values approach $3.0000$ as $x \to 0$. Why the correct answer works: The table entries converge to $3$ from both sides, so $\lim_{x \to 0} \frac{\sin(3x)}{x} = 3$. This is consistent with the known result $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$. Why distractors fail: Option A ($0$) confuses substituting $x=0$ directly. Option B ($1$) confuses this with $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Option D picks a table value far from the target rather than the limiting value.

Answer: $\lim_{x \to -1} h(x) = 2$ and $\lim_{x \to 3} h(x)$ does not exist

Analyze the limit at $x = -1$: From both sides, the graph approaches $y = 2$. The one-sided limits agree, so $\lim_{x \to -1} h(x) = 2$. Analyze the limit at $x = 3$: From the left, the graph approaches $y = 4$. From the right, it approaches $y = 1$. Since $4 \neq 1$, the two-sided limit at $x = 3$ does not exist. Why distractors fail: Option A incorrectly claims the limit at $x = 3$ is $4$ (ignoring the right-hand side). Option C incorrectly says the limit at $x = -1$ does not exist. Option D incorrectly claims neither limit exists.

Answer: $\lim_{x \to 1^-} f(x) = 3$, $\lim_{x \to 1^+} f(x) = 6$, $\lim_{x \to 1} f(x)$ does not exist

Read the one-sided limits from the graph: Tracing from the left, the curve approaches $y = 3$ as $x \to 1^-$. Tracing from the right, the curve approaches $y = 6$ as $x \to 1^+$. Determine the two-sided limit: Since $3 \neq 6$, the one-sided limits disagree, so $\lim_{x \to 1} f(x)$ does not exist. Why distractors fail: Options A and B both claim the two-sided limit exists (as $3$ or $6$), which is impossible when the one-sided limits differ. Option D swaps the left and right limits.

Answer: $f(x) = \begin{cases} 3x & \text{if } x \neq 1 \\ 5 & \text{if } x = 1 \end{cases}$

Check Option A: For $x \neq 1$: $f(x) = 3x$. So $\lim_{x \to 1^-} 3x = 3$ and $\lim_{x \to 1^+} 3x = 3$. At $x = 1$: $f(1) = 5$. All three conditions are satisfied. Check Option B: $\lim_{x \to 1^-} 3 = 3$ ✓, but $\lim_{x \to 1^+} 5 = 5 \neq 3$ ✗. Check Options C and D: Option C: $\lim_{x \to 1^-} 5x = 5 \neq 3$ ✗. Option D: $\lim_{x \to 1^-}(x+2) = 3$ ✓, but $\lim_{x \to 1^+}(x+4) = 5 \neq 3$ ✗, and $f(1) = 5$ ✓ but the right-hand limit fails.

Answer: The claim is incorrect; having a limit equal to $f(a)$ guarantees continuity, not differentiability.

Distinguish continuity from differentiability: From the given information, $\lim_{x \to 2} f(x) = 4 = f(2)$, which means $f$ is continuous at $x = 2$. Continuity is necessary but not sufficient for differentiability. Why the correct answer works: A function can be continuous at a point but fail to be differentiable (e.g., $f(x) = |x|$ at $x = 0$ has a sharp corner). The graphical information about limits alone tells us about continuity, not the existence of the derivative. Why distractors fail: Option A overstates what a graph's appearance can guarantee. Option C is false — $|x|$ is continuous but not differentiable at $0$. Option D misreads the problem — the limit does exist here.

Answer: $x$: $2.9, 2.99, 2.999, 3.001, 3.01, 3.1$ with $f(x)$: $1.9, 1.99, 1.999, 6.001, 6.01, 6.1$

Identify the requirements: The table must show $f(x) \to 2$ as $x \to 3^-$ and $f(x) \to 6$ as $x \to 3^+$, with $x$-values sufficiently close to $3$ on each side. Why the correct answer works: Option A has $x$-values progressively approaching $3$ from both sides. From the left, $f(x)$ approaches $2$. From the right, $f(x)$ approaches $6$. This clearly shows the jump. Why distractors fail: Option B shows $f(x) \to 2$ from both sides, which would suggest the limit exists (no jump). Option C only has two values on each side without showing the convergence pattern. Option D uses $x$-values too far from $3$ to meaningfully estimate the limit at $x = 3$.

Answer: Yes, because the values from both sides are converging toward $5$

Analyze the left-side trend: $g(0.9) = 4.8$, $g(0.99) = 4.98$, $g(0.999) = 4.998$. These values approach $5$ from below. Analyze the right-side trend: $g(1.001) = 5.002$, $g(1.01) = 5.02$, $g(1.1) = 5.2$. These values approach $5$ from above. Why the correct answer works: Both sides converge toward $5$, which provides strong numerical evidence that $\lim_{x \to 1} g(x) = 5$. Why distractors fail: Option A is wrong because the limit does not depend on $g(1)$. Option B is wrong — the left and right values need not be equal; they must approach the same limiting value. Option D gives an incomplete reason: closeness of two particular values isn't the same as a convergence trend.