Answer: $1$

Identify the indeterminate form: As $x \to 0^+$, $x^x$ has the form $0^0$, which is indeterminate. Take the natural logarithm: Let $y = x^x$, so $\ln y = x \ln x$. We evaluate $\lim_{x \to 0^+} x \ln x$ by rewriting as $\frac{\ln x}{1/x}$, which is $\frac{-\infty}{\infty}$. Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} (-x) = 0$. Therefore $\ln y \to 0$, so $y \to e^0 = 1$. Why distractors fail: Option A ($0$) confuses the base approaching $0$ with the limit of the entire expression. Option C ($e$) would require $\ln y \to 1$, not $0$. Option D is wrong; the limit exists and equals $1$.

Answer: L'Hôpital's Rule requires the limit to produce an indeterminate form such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$; it does not apply to determinate forms.

Identify the misconception: The student's error is over-generalizing L'Hôpital's Rule to all fractional limits. The rule has a specific hypothesis that must be met. Why the correct answer works: Option B correctly identifies that L'Hôpital's Rule requires an indeterminate form. For example, $\lim_{x \to 1} \frac{x+1}{x+2} = \frac{2}{3}$ by direct substitution — applying L'Hôpital's here would give $\frac{1}{1} = 1$, which is wrong. Why distractors fail: Option A is false; the rule works for transcendental functions like $e^x$ and $\sin x$. Option C is false; sign does not matter. Option D is false; the rule can be applied repeatedly as long as the indeterminate form persists.

Answer: Simplify the original expression algebraically: cancel $x$ to get $\lim_{x \to 0} x\sin(1/x) = 0$

Recognize L'Hôpital's limitation: L'Hôpital's Rule is inconclusive when $\lim \frac{f'}{g'}$ does not exist. This does not imply the original limit fails to exist. Use algebraic simplification: $\frac{x^2 \sin(1/x)}{x} = x \sin(1/x)$. Since $|\sin(1/x)| \leq 1$, we have $|x \sin(1/x)| \leq |x| \to 0$. By the squeeze theorem (or direct argument), the limit is $0$. Why distractors fail: Option A incorrectly transfers L'Hôpital's failure to the original limit. Option B cannot help because further differentiation will not eliminate the oscillating term. Option D (the quotient rule) computes derivatives of the fraction itself, which is not the goal.

Answer: $\frac{\ln x}{1/x}$

Identify the form: The expression $x \ln x$ as $x \to 0^+$ gives $0 \cdot (-\infty)$, which is not directly eligible for L'Hôpital's Rule. We need to rewrite it as a fraction. Rewrite as a quotient: Write $x \ln x = \frac{\ln x}{1/x}$. As $x \to 0^+$, $\ln x \to -\infty$ and $1/x \to \infty$, giving $\frac{-\infty}{\infty}$, which qualifies for L'Hôpital's Rule. Why distractors fail: Option A ($\frac{x}{1/\ln x}$) gives $\frac{0}{0}$ superficially but $1/\ln x$ is harder to differentiate usefully and the computation is more cumbersome. Option C is algebraically valid but the squeeze theorem is not the indicated technique. Option D introduces unnecessary complexity and changes the problem.

Answer: $1$

Check the indeterminate form: As $x \to 0$, $\sin(0) = 0$ and $x \to 0$, so the form is $\frac{0}{0}$. L'Hôpital's Rule applies. Apply L'Hôpital's Rule: Differentiating numerator and denominator separately: $\lim_{x \to 0} \frac{\cos x}{1} = \cos(0) = 1$. Why distractors fail: Option A ($0$) may come from incorrectly substituting before recognizing the indeterminate form. Option C ($-1$) has no basis in the calculation. Option D is incorrect because the limit exists and equals $1$.

Answer: The numerator $f(x)$ and the denominator $g(x)$ separately

State L'Hôpital's Rule precisely: The rule says: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$. The numerator and denominator are differentiated independently. Why the correct answer works: Option C correctly describes taking $f'(x)$ and $g'(x)$ separately, then forming the new quotient $\frac{f'(x)}{g'(x)}$. Why distractors fail: Option A applies the quotient rule, which gives the derivative of the entire fraction — this is NOT what L'Hôpital's Rule does. Option B differentiates only the numerator, ignoring the denominator. Option D applies the product rule, which is irrelevant to the procedure.

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

Verify the initial form: At $x = 0$: $\tan 0 - 0 = 0$ and $0^3 = 0$, so the form is $\frac{0}{0}$. First application: $\lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2} = \lim_{x \to 0} \frac{\tan^2 x}{3x^2}$, which is $\frac{0}{0}$. Second application: $\lim_{x \to 0} \frac{2\sec^2 x \tan x}{6x}$, which is $\frac{0}{0}$. Third application: Differentiating again: the numerator derivative at $x=0$ evaluates to $2\sec^4(0) + 4\sec^2(0)\tan^2(0) = 2$. The denominator derivative is $6$. So the limit is $\frac{2}{6} = \frac{1}{3}$. Why distractors fail: Option A ($\frac{1}{6}$) is a common error from miscounting derivatives. Option B ($\frac{1}{2}$) and Option D ($1$) arise from differentiation mistakes in the intermediate steps.

Answer: $\frac{3}{5}$

Check the form: As $x \to 0$, $\sin(3x) \to 0$ and $\sin(5x) \to 0$, so the form is $\frac{0}{0}$. Apply L'Hôpital's Rule: $\lim_{x \to 0} \frac{3\cos(3x)}{5\cos(5x)} = \frac{3 \cdot 1}{5 \cdot 1} = \frac{3}{5}$. Why distractors fail: Option A ($\frac{5}{3}$) swaps the numerator and denominator coefficients. Option C ($0$) has no basis. Option D ($1$) ignores the different coefficients inside the sine functions.

Answer: $n$

Verify the form: At $x = 1$: $1^n - 1 = 0$ and $1 - 1 = 0$, so the form is $\frac{0}{0}$. Apply L'Hôpital's Rule: Differentiate: $\frac{d}{dx}(x^n - 1) = nx^{n-1}$ and $\frac{d}{dx}(x - 1) = 1$. The limit becomes $\lim_{x \to 1} \frac{nx^{n-1}}{1} = n \cdot 1^{n-1} = n$. Why distractors fail: Option A ($1$) only holds when $n = 1$. Option B ($n-1$) confuses the exponent after differentiation with the answer. Option D ($n^2$) has no mathematical basis.

Answer: The expression $\frac{0}{0}$ is undefined and the limit could be any finite value, $\infty$, $-\infty$, or may not exist, depending on the particular functions involved.

Define indeterminate form: An indeterminate form is one where the limiting value cannot be determined solely from the form itself. For $\frac{0}{0}$, different functions leading to this form can produce completely different limit values. Why the correct answer works: Option C correctly captures that the outcome depends on the specific functions $f(x)$ and $g(x)$, not on the symbolic form alone. For example, $\lim_{x \to 0} \frac{x}{x} = 1$ but $\lim_{x \to 0} \frac{x^2}{x} = 0$. Why distractors fail: Option A incorrectly assumes the limit must be $1$. Option B incorrectly assumes the limit must be $0$. Option D confuses $\frac{0}{0}$ with the form $\frac{c}{0}$ (where $c \neq 0$), which can indicate a vertical asymptote.

Answer: $0$

Identify the indeterminate form: As $x \to 0^+$, both $\frac{1}{\sin x}$ and $\frac{1}{x}$ tend to $\infty$, so the expression has the form $\infty - \infty$. Combine into a single fraction: $\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$. As $x \to 0^+$, numerator $\to 0$ and denominator $\to 0$, giving $\frac{0}{0}$. Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{1 - \cos x}{\sin x + x\cos x}$. This is $\frac{0}{0}$ again. Apply once more: $\lim_{x \to 0^+} \frac{\sin x}{\cos x + \cos x - x\sin x} = \frac{0}{2} = 0$. Why distractors fail: Option A ($1$) and Option D ($-1$) result from algebraic errors. Option B ($\infty$) incorrectly assumes the infinities don't cancel.

Answer: $\frac{0}{0}$

Recall the eligible indeterminate forms: L'Hôpital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ yields the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit equals $\lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists. Why $\frac{0}{0}$ is correct: The form $\frac{0}{0}$ is one of the two forms to which L'Hôpital's Rule applies directly, with no rewriting needed. Why the other options fail: Option A ($0 \cdot \infty$), Option C ($1^\infty$), and Option D ($\infty - \infty$) are all indeterminate forms, but they must first be rewritten as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before L'Hôpital's Rule can be applied.

Answer: $0$

Check the indeterminate form: As $x \to \infty$, $\ln x \to \infty$ and $x \to \infty$, giving the form $\frac{\infty}{\infty}$. Apply L'Hôpital's Rule: Differentiate: $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and $\frac{d}{dx}(x) = 1$. So $\lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$. Why distractors fail: Option A ($1$) might arise from incorrectly canceling. Option B ($\infty$) ignores that $x$ grows much faster than $\ln x$. Option D ($-\infty$) has no basis since both functions are positive for large $x$.

Answer: Verify that the limit produces an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

The prerequisite check: L'Hôpital's Rule has a strict prerequisite: the limit must produce one of the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Without confirming this, applying the rule is invalid. Why the correct answer works: Verifying the indeterminate form ensures the hypotheses of L'Hôpital's Rule are satisfied, which is the necessary first step. Why distractors fail: Option A is wrong because L'Hôpital's Rule differentiates the numerator and denominator separately, not as a quotient. Option C describes an algebraic technique that may help but is not a required first step for L'Hôpital's. Option D describes numerical estimation, which is unrelated to the rule's prerequisites.

Answer: $\frac{1}{2}$

Check the form: At $x = 0$: numerator $= e^0 - 1 - 0 = 0$, denominator $= 0$. Form is $\frac{0}{0}$. First application of L'Hôpital's Rule: $\lim_{x \to 0} \frac{e^x - 1}{2x}$. At $x = 0$ this is $\frac{0}{0}$ again. Second application: $\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$. Why distractors fail: Option A ($0$) results from an arithmetic error or stopping too early. Option C ($1$) comes from forgetting the coefficient $2$ in the denominator. Option D ($\frac{1}{3}$) has no basis in this calculation.

Answer: $1$

Check the indeterminate form: At $x = 0$: numerator $= e^0 - 1 = 0$ and denominator $= 0$. The form is $\frac{0}{0}$. Apply L'Hôpital's Rule: Differentiate: $\frac{d}{dx}(e^x - 1) = e^x$ and $\frac{d}{dx}(x) = 1$. So $\lim_{x \to 0} \frac{e^x}{1} = e^0 = 1$. Why distractors fail: Option A ($0$) confuses the numerator's value at $0$ with the limit. Option B ($e$) likely comes from evaluating $e^x$ at $x = 1$ instead of $x = 0$. Option D ($\infty$) has no basis in this computation.

Answer: L'Hôpital's Rule is inconclusive here; using algebraic simplification, the original limit equals $1$.

Recognize the failure mode: L'Hôpital's Rule states that the original limit equals $\lim \frac{f'}{g'}$ if that limit exists. When the derivative-limit does not exist, L'Hôpital's Rule gives no information and it does not mean the original limit fails to exist. Solve algebraically: $\frac{x + \sin x}{x} = 1 + \frac{\sin x}{x}$. Since $\frac{\sin x}{x} \to 0$ as $x \to \infty$, the limit is $1$. Why distractors fail: Option A incorrectly concludes the original limit does not exist — L'Hôpital's failure only means the method is inconclusive. Option C invokes an incorrect averaging argument. Option D is futile because further differentiation does not resolve the oscillation of $\cos x$.

Answer: $0$

Check the indeterminate form: As $x \to \infty$, both $x^2 \to \infty$ and $e^x \to \infty$, giving $\frac{\infty}{\infty}$. First application of L'Hôpital's Rule: $\lim_{x \to \infty} \frac{2x}{e^x}$, which is still $\frac{\infty}{\infty}$. Second application: $\lim_{x \to \infty} \frac{2}{e^x} = 0$, since $e^x \to \infty$. Why distractors fail: Option A ($\infty$) ignores that exponential growth dominates polynomial growth. Option B ($1$) and Option C ($2$) have no mathematical basis in this calculation.

Answer: $\frac{1}{2}$

Check the indeterminate form: At $x = 0$: $1 - \cos 0 = 0$ and $0^2 = 0$, giving $\frac{0}{0}$. First application of L'Hôpital's Rule: Differentiate: $\frac{\sin x}{2x}$. At $x = 0$ this is again $\frac{0}{0}$, so we apply L'Hôpital's Rule a second time. Second application: Differentiate again: $\frac{\cos x}{2}$. As $x \to 0$: $\frac{\cos 0}{2} = \frac{1}{2}$. Why distractors fail: Option A ($0$) results from stopping too early or a substitution error. Option B ($1$) comes from forgetting the factor of $2$ in the denominator derivative. Option D ($2$) inverts the correct answer.