A student evaluates lim_{x→1} (x² + 1)/(x + 1) by applying L'Hôpital's rule: differentiating numerator and denominator to get lim 2x/1 = 2. What is wrong?
AThe student differentiated the denominator incorrectly — it should be 1/(x+1)²
BL'Hôpital's rule requires differentiating the entire fraction, not numerator and denominator separately
CThe original limit is not an indeterminate form — direct substitution gives (1 + 1)/(1 + 1) = 1 — so the rule cannot be applied; the correct answer is 1, not 2
DThe rule only applies when x → 0, not x → 1
L'Hôpital's rule requires that the original limit be an indeterminate form (0/0 or ±∞/∞). Here, direct substitution gives 2/2 = 1 — no indeterminate form exists. Applying the rule anyway yields the wrong answer (2 ≠ 1). This is the most common misuse of the rule: students apply it mechanically without first checking whether the limit is genuinely indeterminate.
Question 2 Multiple Choice
In the proof of L'Hôpital's rule for the 0/0 case, the Cauchy Mean Value Theorem plays which role?
AIt proves directly that lim f'(x)/g'(x) = lim f(x)/g(x) at the limit point by evaluating both limits at a
BIt guarantees that g'(x) is nonzero on the entire interval, preventing division by zero in the ratio
CFor each x near a, it provides a point c between a and x where f(x)/g(x) = f'(c)/g'(c); as x → a, c is squeezed to a, linking the limit of f/g to the limit of f'/g'
DIt establishes that f and g must both be continuous on the interval, which is the key hypothesis for the rule
The proof rewrites f(x)/g(x) as (f(x) − f(a))/(g(x) − g(a)) (since both approach zero), then applies the Cauchy MVT to find c ∈ (a, x) where this equals f'(c)/g'(c). As x → a, c is squeezed between a and x and must also approach a. If lim_{x→a} f'(x)/g'(x) = L, then f'(c)/g'(c) → L too, completing the proof. The squeeze on c is the essential mechanism.
Question 3 True / False
If lim_{x→a} f'(x)/g'(x) does not exist (for example, because it oscillates), then L'Hôpital's rule cannot be applied — even if lim_{x→a} f(x)/g(x) itself does exist.
TTrue
FFalse
Answer: True
L'Hôpital's rule is a one-directional implication: if the limit of f'/g' exists (or is ±∞), then the limit of f/g equals it. The converse is false. The classic example is lim_{x→0} (x² sin(1/x))/x = lim_{x→0} x sin(1/x) = 0, but the ratio of derivatives oscillates without a limit. The original limit exists; the derivative limit does not. The rule's hypothesis is not satisfied, but the limit itself is still well-defined — you just can't use L'Hôpital to find it.
Question 4 True / False
L'Hôpital's rule directly handles indeterminate forms like 0 · ∞ and 1^∞ in the same way as 0/0, without any algebraic rearrangement.
TTrue
FFalse
Answer: False
The rule is stated for fractions in the form 0/0 or ±∞/∞. Forms like 0 · ∞, ∞ − ∞, 0⁰, 1^∞, and ∞⁰ must first be algebraically converted. For example, 0 · ∞ is rewritten as 0/(1/∞) = 0/0; exponential forms are handled by taking logarithms to reduce to 0 · ∞ and then rewriting again. Each conversion must be checked to confirm the result is genuinely 0/0 or ∞/∞ before applying the rule.
Question 5 Short Answer
Why does the rigorous statement of L'Hôpital's rule include the condition that lim f'(x)/g'(x) must exist, and what error does applying the rule without checking this condition risk?
Think about your answer, then reveal below.
Model answer: The rule's conclusion (that lim f/g equals lim f'/g') only holds when lim f'/g' converges to a finite limit or ±∞. If the derivative ratio oscillates, the rule gives no valid conclusion about f/g. The error is applying the rule in a situation where it says nothing — and potentially writing down a 'result' that is simply wrong, confusing an oscillating derivative expression for the limit of the original function.
A concrete failure: for f(x) = x + sin(x)cos(x) and g(x) = x, the ratio f'/g' = (1 + cos(2x))/1 oscillates between 0 and 2 near ∞, while f/g → 1. The rule's hypothesis fails, so its conclusion cannot be invoked. Checking hypotheses is not a formality — it is what separates valid analysis from pattern-matching.