A Limit That Looks Innocent
A student is asked to evaluate L = lim(x→0) (1 − cos x)/x². They substitute x = 0 and get 0/0, then try x = 0.1 and x = 0.01, obtaining 0.499958 and 0.49999958 respectively, and conclude L = 0.5. A second student uses L'Hôpital once and gets lim(x→0) sin x/(2x), then stops, saying 'it is still 0/0, so the limit does not exist'. A third claims the limit is 0 because 'cos 0 = 1, so the numerator is zero'. The examiner wants a rigorous answer plus an error diagnosis for each student.
1. Evaluate L rigorously and identify which student's conclusion is correct.
Read the worked answer
Method (double angle): 1 − cos x = 2 sin²(x/2).
So (1 − cos x)/x² = 2 sin²(x/2)/x² = 2 · [sin(x/2)/(x/2)]² · (1/4) = ½ · [sin(x/2)/(x/2)]².
As x → 0, sin(u)/u → 1 with u = x/2, so L = ½ · 1² = ½.
Method (L'Hôpital twice): 0/0 form → differentiate: sin x/(2x), still 0/0 → differentiate again: cos x/2 → 1/2 as x → 0. Same answer.
Verdict: the first student's numerical answer 0.5 is correct, but the justification (a table of values) is not a proof.
2. Diagnose the error in each of the second and third students' reasoning.
Read the worked answer
Second student: correct first application of L'Hôpital, but wrong conclusion. Getting 0/0 again is not evidence that the limit fails to exist - it only means L'Hôpital can be applied again (the hypotheses still hold near 0). Applying it once more gives cos x/2 → ½. The error is treating 'indeterminate' as 'undefined'.
Third student: confuses the value of the numerator at a point with the behaviour of the quotient near that point. Both numerator and denominator vanish at x = 0; the limit asks how fast each vanishes. Here 1 − cos x ≈ x²/2, exactly the same order as x², so the ratio tends to ½, not 0. Substituting x = 0 into a 0/0 form is invalid.
Judgement: numerical evidence can suggest a value but cannot establish it; the algebraic or L'Hôpital argument is what settles the question.