Answer: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

Test each by direct substitution: Substitute the target value into each expression. Option A gives $\frac{1+1}{1+3} = \frac{2}{4}$, which is defined. Option C gives $\frac{3}{1} = 3$. Option D gives $(-1)^3 + 2(-1) = -3$. Identify the indeterminate form: For Option B, substituting $x = 2$ gives $\frac{4-4}{2-2} = \frac{0}{0}$, which is the indeterminate form. Why distractors fail: Options A, C, and D all yield finite, well-defined values upon direct substitution, so they are not indeterminate.

Answer: Rewrite as $\frac{\sin(2x)}{2x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{2}{3}$ and apply the standard trig limit

Simplify the expression: Note $\frac{x}{x} = 1$ for $x \neq 0$, so the limit is $\lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)}$, which is $\frac{0}{0}$. Why the correct strategy works: Rewriting: $\frac{\sin(2x)}{\sin(3x)} = \frac{\sin(2x)}{2x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{2}{3}$. Each ratio approaches $1$, so the result is $\frac{2}{3}$. Why distractors fail: Option A: The double-angle formula applies to $\sin(2x)$ but doesn't simplify $\sin(3x)$ directly. Option C: Removing $x/x$ still leaves $\sin(2x)/\sin(3x)$, which is $0/0$. Option D: The Squeeze Theorem is unnecessarily complex here.

Answer: $6$

Check for indeterminate form: Substituting $x=9$: $\frac{9-9}{\sqrt{9}-3} = \frac{0}{0}$. Rationalize the denominator: Multiply by $\frac{\sqrt{x}+3}{\sqrt{x}+3}$: $\frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} = \frac{(x-9)(\sqrt{x}+3)}{x-9} = \sqrt{x}+3$. Evaluate: $\lim_{x \to 9}(\sqrt{x}+3) = 3 + 3 = 6$. Why distractors fail: Option A ($3$) only computes $\sqrt{9}$ without the added $3$. Option B ($0$) mistakenly uses the numerator's value before simplification. Option D ($9$) confuses the target value with the limit.

Answer: The expression is an indeterminate form and requires further algebraic work

Define indeterminate form: The form $\frac{0}{0}$ means both the numerator and denominator approach zero. This alone does not determine the limit's value. Why the correct answer works: An indeterminate form signals that additional techniques — such as factoring, rationalizing, or simplifying — are needed to resolve the limit. Why distractors fail: Option A is wrong because $0/0$ does not mean the limit fails to exist; it may well exist after simplification. Option B is wrong because the limit could be any value. Option D is wrong because $0/0$ typically indicates a discontinuity (like a hole), not continuity.

Answer: $1$

Rewrite using standard forms: $\frac{\sin(x^2)}{x \sin x} = \frac{\sin(x^2)}{x^2} \cdot \frac{x}{\sin x}$. Evaluate each factor: As $x \to 0$: $\frac{\sin(x^2)}{x^2} \to 1$ (since $x^2 \to 0$) and $\frac{x}{\sin x} \to 1$. Product $= 1 \cdot 1 = 1$. Why distractors fail: Option A ($0$) incorrectly treats the product as vanishing. Option C ($2$) may arise from miscounting factors. Option D is wrong; the limit exists and is finite.

Answer: $\frac{1}{3}$

Divide numerator and denominator by $x$: $\frac{\sin x}{2x + \sin x} = \frac{\frac{\sin x}{x}}{2 + \frac{\sin x}{x}}$. Apply the standard trig limit: As $x \to 0$: $\frac{\sin x}{x} \to 1$. So the expression becomes $\frac{1}{2 + 1} = \frac{1}{3}$. Why distractors fail: Option A ($1/2$) may arise from ignoring the $\sin x$ term in the denominator. Option C ($1$) ignores the $2x$ in the denominator. Option D ($0$) is incorrect since both numerator and denominator approach $0$ at the same rate.

Answer: $-\frac{1}{16}$

Check for indeterminate form: Substituting $x = 4$: $\frac{1/4 - 1/4}{4 - 4} = \frac{0}{0}$. Simplify the complex fraction: Combine the numerator: $\frac{1}{x} - \frac{1}{4} = \frac{4 - x}{4x}$. So the expression becomes $\frac{4-x}{4x(x-4)} = \frac{-(x-4)}{4x(x-4)} = \frac{-1}{4x}$ for $x \neq 4$. Evaluate the limit: $\lim_{x \to 4} \frac{-1}{4x} = \frac{-1}{16}$. Why distractors fail: Option B ($1/16$) drops the negative sign from the $4 - x = -(x - 4)$ step. Option C ($-1/4$) forgets the $x$ factor in the denominator. Option D ($1/4$) makes both errors.

Answer: The limit equals $1$

Recall the standard trigonometric limit: One of the most fundamental results in calculus is $\lim_{x \to 0} \frac{\sin x}{x} = 1$, where $x$ is measured in radians. Why the correct answer works: This result can be proven geometrically using the Squeeze Theorem and is a foundational building block for evaluating trigonometric limits. Why distractors fail: Option A ($0$) confuses the limit with the value of $\sin(0) = 0$. Option C is incorrect because the limit does exist. Option D ($\pi$) has no basis in the derivation.

Answer: $6$

Check for indeterminate form: Substituting $x = 3$ gives $\frac{9 - 9}{3 - 3} = \frac{0}{0}$, so we need to simplify. Factor and cancel: $\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$. Then $\lim_{x \to 3}(x+3) = 6$. Why distractors fail: Option A ($0$) likely comes from only evaluating the numerator. Option B ($3$) may result from substituting into the wrong simplified expression. Option D is wrong because the limit does exist after cancellation.

Answer: No, because $\frac{1-\cos x}{x^2}$ must be analyzed as its own expression; the correct limit is $\frac{1}{2}$

Analyze the student's error: The student's logic treats $\frac{1-\cos x}{x^2}$ as $\frac{1}{x} \cdot \frac{1-\cos x}{x}$, concluding the first factor diverges and the second goes to zero. But limits of products require both factors, and this form is indeterminate. Correct evaluation: Using the identity $1-\cos x = 2\sin^2(x/2)$: $\frac{1-\cos x}{x^2} = \frac{2\sin^2(x/2)}{x^2} = \frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \to \frac{1}{2}(1)^2 = \frac{1}{2}$. Why distractors fail: Option A incorrectly applies arithmetic of finite numbers to limits. Option C is factually wrong: $1-\cos x \approx x^2/2$, so they vanish at the same rate. Option D is wrong because the limit is finite.

Answer: $\frac{3}{7}$

Rewrite using the standard limit: $\frac{\sin(3x)}{\sin(7x)} = \frac{\sin(3x)}{3x} \cdot \frac{7x}{\sin(7x)} \cdot \frac{3x}{7x} = \frac{\sin(3x)}{3x} \cdot \frac{7x}{\sin(7x)} \cdot \frac{3}{7}$. Evaluate each factor: As $x \to 0$: $\frac{\sin(3x)}{3x} \to 1$ and $\frac{7x}{\sin(7x)} \to 1$. So the limit is $1 \cdot 1 \cdot \frac{3}{7} = \frac{3}{7}$. Why distractors fail: Option A ($7/3$) inverts the ratio. Option C ($1$) ignores the different coefficients. Option D ($0$) incorrectly assumes the expression vanishes.

Answer: The student incorrectly concluded that an indeterminate form means the limit does not exist, when in fact $\frac{0}{0}$ signals the need for further simplification

Identify the misconception: The form $\frac{0}{0}$ is indeterminate — it does not tell us whether the limit exists or what its value is. It is a signal, not a conclusion. Correct approach: Factor: $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. Then $\lim_{x \to 1}(x+1) = 2$. The limit exists and equals $2$. Why distractors fail: Option A is wrong because the student didn't attempt to factor, so no formula was misapplied. Option C is misleading; L'Hôpital's Rule would give the same answer but is not the issue — the error is in interpreting $0/0$ as a final answer. Option D is wrong because the substitution was done correctly.

Answer: $12$

Identify the indeterminate form: Substituting $x = -2$: $\frac{(-2)^3 + 8}{-2+2} = \frac{0}{0}$. Factor using the sum of cubes: $x^3 + 8 = x^3 + 2^3 = (x+2)(x^2 - 2x + 4)$. Canceling $(x+2)$: $\lim_{x \to -2}(x^2 - 2x + 4)$. Evaluate: $(-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12$. Why distractors fail: Option A ($0$) confuses the numerator's value before factoring. Option C ($-12$) is a sign error in the sum-of-cubes formula. Option D ($8$) comes from $2^3$ without completing the factoring.

Answer: $15$

Rewrite using the given limit: $\frac{f(3x)}{x} = 3 \cdot \frac{f(3x)}{3x}$. Apply substitution: Let $u = 3x$. As $x \to 0$, $u \to 0$. So $\frac{f(3x)}{3x} = \frac{f(u)}{u} \to 5$. Compute: The limit is $3 \cdot 5 = 15$. Why distractors fail: Option A ($5$) forgets the factor of $3$. Option B ($5/3$) divides by $3$ instead of multiplying. Option D ($0$) incorrectly assumes the limit vanishes.

Answer: $\frac{1}{4}$

Check for indeterminate form: Substituting $x = 0$: $\frac{\sqrt{4}-2}{0} = \frac{0}{0}$. Rationalize the numerator: Multiply by $\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}$: $\frac{(x+4)-4}{x(\sqrt{x+4}+2)} = \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2}$. Evaluate the limit: $\lim_{x \to 0} \frac{1}{\sqrt{x+4}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}$. Why distractors fail: Option B ($1/2$) may come from forgetting the $+2$ in the denominator. Option C ($0$) incorrectly treats the limit as zero. Option D ($4$) inverts the correct answer.

Answer: $4$

Expand the numerator: $(2+h)^2 - 4 = 4 + 4h + h^2 - 4 = 4h + h^2 = h(4+h)$. Cancel and evaluate: $\frac{h(4+h)}{h} = 4 + h$ for $h \neq 0$. Then $\lim_{h \to 0}(4+h) = 4$. Why distractors fail: Option A ($0$) substitutes $h=0$ before simplifying. Option B ($2$) likely confuses the base value with the limit. Option D ($8$) may come from incorrectly expanding $(2+h)^2$ as $4+8h+h^2$.

Answer: $\frac{3}{2}$

Identify the indeterminate form: At $x = 1$: $\frac{1-1}{1-1} = \frac{0}{0}$. Factor both numerator and denominator: $x^3 - 1 = (x-1)(x^2+x+1)$ and $x^2 - 1 = (x-1)(x+1)$. After canceling $(x-1)$: $\frac{x^2+x+1}{x+1}$. Evaluate: $\frac{1+1+1}{1+1} = \frac{3}{2}$. Why distractors fail: Option B ($1$) may come from incorrectly canceling. Option C ($2/3$) inverts the fraction. Option D ($0$) confuses the indeterminate form with the limit value.

Answer: The limit equals $4$ because, after canceling the common factor, the simplified function $g(x) = x + 2$ agrees with $f(x)$ for all $x \neq 2$, and limits only depend on values near, not at, the point

Key principle about limits: The limit of $f(x)$ as $x \to a$ depends only on the values of $f$ near $a$, not at $a$ itself. Why the correct answer works: $\frac{x^2-4}{x-2} = x+2$ for all $x \neq 2$. Since the limit only concerns behavior near $x=2$, we evaluate $\lim_{x \to 2}(x+2) = 4$. Why distractors fail: Option A is wrong because evaluating the limit does not redefine the function. Option C is wrong because there is no convention that assigns a value to $f(2)$. Option D reflects a fundamental misconception: a function being undefined at a point does not prevent the limit from existing.

Answer: (I) Factoring, (II) Rationalizing, (III) Simplifying complex fractions

Classify each limit: (I) has $x^2-16 = (x-4)(x+4)$, a polynomial that can be factored. (II) has $\sqrt{x+1}-1$, a radical expression requiring the conjugate. (III) has a difference of rational expressions in the numerator, requiring complex fraction simplification. Why the correct answer works: Option B correctly pairs each limit with its most appropriate technique: factoring for (I), rationalizing for (II), and simplifying complex fractions for (III). Why distractors fail: Option A swaps the techniques for (I) and (II) and incorrectly suggests direct substitution for (III), which gives $0/0$. Option C suggests direct substitution for (II), which also gives $0/0$. Option D assigns complex fractions to (I), which has no fractions in the numerator.

Answer: Factoring fails because the numerator involves a radical, not a polynomial; multiply by the conjugate $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ instead

Why factoring does not apply: Factoring is used for polynomials. The expression $\sqrt{1+x} - 1$ is irrational and cannot be factored in the traditional sense. Correct technique: Multiply by $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$: $\frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1} \to \frac{1}{2}$. Why distractors fail: Option A incorrectly suggests direct substitution, which still gives $0/0$. Option C writes an expression that equals $(\sqrt{1+x})^2 - 1 = x$, which is the conjugate method — not factoring — and incorrectly labels it as factoring. Option D is wrong because the limit exists and equals $1/2$.

Answer: Multiply by the conjugate $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$ to get $\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}$, then cancel and substitute

Check for indeterminate form: At $x = 0$: $\frac{\sqrt{1}-\sqrt{1}}{0} = \frac{0}{0}$. Rationalize: Multiply by $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$. Numerator: $(1+x)-(1-x) = 2x$. So $\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})} = \frac{2}{\sqrt{1+x}+\sqrt{1-x}}$. Evaluate: $\lim_{x \to 0} \frac{2}{\sqrt{1}+\sqrt{1}} = \frac{2}{2} = 1$. Why distractors fail: Option A: You cannot factor $x$ from a radical expression directly. Option C: There is no trigonometric component. Option D: Direct substitution yields $0/0$, so it fails.

Answer: $0$

Recall the standard result: The standard trigonometric limit states that $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$. Why the correct answer works: Near $x = 0$, $1 - \cos x \approx \frac{x^2}{2}$, so $\frac{1-\cos x}{x} \approx \frac{x}{2} \to 0$. Why distractors fail: Option A ($1$) confuses this with $\lim \frac{\sin x}{x} = 1$. Option B ($-1$) has no justification. Option D ($1/2$) confuses this with $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$, a different limit.

Answer: $-\frac{1}{9}$

Check for indeterminate form: At $x=0$: $\frac{1/3 - 1/3}{0} = \frac{0}{0}$. Simplify the complex fraction: Combine: $\frac{1}{x+3} - \frac{1}{3} = \frac{3-(x+3)}{3(x+3)} = \frac{-x}{3(x+3)}$. Dividing by $x$: $\frac{-x}{3x(x+3)} = \frac{-1}{3(x+3)}$. Evaluate the limit: $\lim_{x \to 0} \frac{-1}{3(x+3)} = \frac{-1}{3 \cdot 3} = -\frac{1}{9}$. Why distractors fail: Option A ($1/9$) drops the negative sign. Option C ($-1/3$) forgets the extra factor of $3$ in the denominator. Option D ($1/3$) makes both errors.

Answer: $3$

Check for indeterminate form: Substituting $x=5$: $\frac{25-35+10}{5-5}=\frac{0}{0}$. Factor and simplify: $x^2 - 7x + 10 = (x-5)(x-2)$, so $\frac{(x-5)(x-2)}{x-5} = x - 2$ for $x \neq 5$. Then $\lim_{x \to 5}(x-2) = 3$. Why distractors fail: Option A ($5$) confuses the target value with the limit. Option C ($-3$) is a sign error. Option D ($2$) comes from the factored term $(x-2)$ evaluated at $x=0$ or confused with the other factor.

Answer: Substitute $x = 3$ directly into the expression

Identify the type of function: $2x^2 - 5x + 1$ is a polynomial, and polynomials are continuous everywhere. Why direct substitution works: Since the function is continuous at $x = 3$, we can evaluate the limit by substituting directly: $2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4$. Why distractors fail: Option A (factoring) is unnecessary since there is no indeterminate form. Option B (conjugate) is used for radicals, not polynomials. Option D (L'Hôpital's Rule) requires an indeterminate form like $0/0$, which does not arise here.

Answer: $\frac{1}{6}$

Check for indeterminate form: At $x=0$: numerator $= \sqrt{9}-3 = 0$, denominator $= 0+0 = 0$. So $\frac{0}{0}$. Rationalize and factor: Multiply by $\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$: numerator becomes $(x+9)-9 = x$. Denominator: $x(x+1)(\sqrt{x+9}+3)$. Cancel $x$: $\frac{1}{(x+1)(\sqrt{x+9}+3)}$. Evaluate: $\lim_{x \to 0} \frac{1}{(x+1)(\sqrt{x+9}+3)} = \frac{1}{(1)(6)} = \frac{1}{6}$. Why distractors fail: Option B ($1/3$) forgets the $(x+1)$ factor. Option C ($0$) incorrectly simplifies. Option D ($1/54$) incorrectly computes the denominator as $9 \cdot 6$.

Answer: $\frac{4}{3}$

Separate the factors: $\frac{\sin(4x)(x+1)}{3x} = \frac{\sin(4x)}{4x} \cdot \frac{4}{3} \cdot (x+1)$. Evaluate each limit: As $x \to 0$: $\frac{\sin(4x)}{4x} \to 1$, $\frac{4}{3}$ is constant, and $(x+1) \to 1$. Product: $1 \cdot \frac{4}{3} \cdot 1 = \frac{4}{3}$. Why distractors fail: Option B ($3/4$) inverts the coefficient ratio. Option C ($0$) mistakenly treats $\sin(4x) \to 0$ as making the whole limit zero. Option D ($4$) forgets the $\frac{1}{3}$ from the denominator.

Answer: $1$

Rewrite using sine and cosine: $\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}$. Apply known limits: As $x \to 0$: $\frac{\sin x}{x} \to 1$ and $\frac{1}{\cos x} \to \frac{1}{1} = 1$. So the limit is $1 \cdot 1 = 1$. Why distractors fail: Option A ($0$) confuses $\tan(0)=0$ with the limit of the ratio. Option B ($1/2$) has no valid derivation. Option D is wrong because the limit clearly exists.

Answer: It converts the radical expression into a difference of squares, removing the radical and revealing a common factor that can be canceled

The conjugate identity: For an expression like $\sqrt{a} - \sqrt{b}$, multiplying by $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}$ uses the identity $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = a - b$. Why the correct answer works: This eliminates the radical, producing a polynomial or rational expression. A common factor with the denominator often emerges, which can then be canceled to resolve the $0/0$ form. Why distractors fail: Option A is wrong because the denominator is not eliminated; a factor is canceled. Option C is wrong because multiplying by $1$ (the conjugate over itself) does not change the expression's value. Option D is wrong because no trigonometric conversion is involved.

Answer: Multiplying the expression by a conjugate to eliminate radicals

Define rationalizing: Rationalizing involves multiplying the numerator and denominator by the conjugate of an expression containing a radical, converting it into a difference of squares. Why the correct answer works: For example, to evaluate a limit involving $\sqrt{x+1} - 1$, we multiply by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ to eliminate the radical. Why distractors fail: Option A describes factoring, a different technique. Option C describes a small-angle approximation, not rationalizing. Option D describes partial fractions, which is an integration technique.

Answer: $5$

Rewrite to match the standard form: $\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}$. Apply the standard trig limit: As $x \to 0$, $5x \to 0$, so $\frac{\sin(5x)}{5x} \to 1$. Therefore the limit is $5 \cdot 1 = 5$. Why distractors fail: Option A ($0$) confuses the limit with the value of $\sin(0)$. Option B ($1$) forgets the factor of $5$. Option D ($1/5$) inverts the coefficient.