A student evaluates lim(x→0) x/sin(2x) using L'Hôpital's Rule. They apply the quotient rule to compute d/dx[x/sin(2x)] = (sin(2x) − 2x·cos(2x))/sin²(2x) and then take the limit as x→0. What mistake did the student make?
AL'Hôpital's Rule does not apply here because the form is not indeterminate
BThe student used the quotient rule instead of differentiating numerator and denominator separately; the rule requires lim f′(x)/g′(x), not lim (f/g)′(x)
CThe student should have applied the rule twice before evaluating the limit
DThe form is ∞/∞, not 0/0, so a different conversion is needed first
L'Hôpital's Rule says replace f/g with f′/g′ — differentiate top and bottom separately. Using the quotient rule gives (f/g)′, which is a different expression entirely and will produce the wrong answer. Here, the correct application gives 1/(2cos(2x)) → 1/2 as x→0.
Question 2 Multiple Choice
Which of the following limits requires rewriting into 0/0 or ∞/∞ form before L'Hôpital's Rule can be applied?
Alim(x→0) sin(x)/x
Blim(x→∞) eˣ/x²
Clim(x→0⁺) x·ln(x)
Dlim(x→1) (x² − 1)/(x − 1)
lim x·ln(x) as x→0⁺ is 0·(−∞) — a product, not a ratio. L'Hôpital's Rule requires a ratio, so you must rewrite it as ln(x)/(1/x), which gives −∞/∞ form, then apply the rule. Options A, C (actually C is the answer), D are already ratios in 0/0 or ∞/∞ form.
Question 3 True / False
Applying L'Hôpital's Rule to lim(x→0) (x+1)/x is valid because substituting x = 0 produces a fraction with 0 in the denominator.
TTrue
FFalse
Answer: False
The form is 1/0, which is NOT indeterminate. A nonzero numerator over a shrinking denominator tells you the limit diverges to ±∞ — no ambiguity about the limit's value. L'Hôpital's Rule only applies to 0/0 and ∞/∞. Applying it to 1/0 gives the wrong answer.
Question 4 True / False
When applying L'Hôpital's Rule, you differentiate the numerator and denominator as separate functions, not as a quotient.
TTrue
FFalse
Answer: True
This is the central procedural point: lim f(x)/g(x) = lim f′(x)/g′(x). You compute f′ and g′ independently, then form their ratio. Using the quotient rule — computing (f/g)′ = (f′g − fg′)/g² — is the most common algebraic error and produces a different, incorrect expression.
Question 5 Short Answer
Explain why L'Hôpital's Rule cannot be applied to a limit of the form 3/0, even though substitution produces an undefined expression.
Think about your answer, then reveal below.
Model answer: 3/0 is not indeterminate — it means the limit is ±∞. The numerator stays near 3 while the denominator vanishes, so the ratio grows without bound. Indeterminate forms like 0/0 or ∞/∞ give no information about the limit because two competing tendencies are in tension (both numerator and denominator approach 0 or ∞ simultaneously). L'Hôpital's Rule resolves that tension; there is no tension to resolve in 3/0.
The word 'indeterminate' is precise: the form alone cannot determine the limit's value, so more information (derivatives) is needed. A form like 3/0 tells you exactly what happens: the expression blows up. Applying L'Hôpital there would give 0/1 = 0, which is simply wrong.