Counterexample

Is this true?

If lim f/g exists (0/0 form) then lim f′/g′ exists

No — it is false.

The counterexample

$f(x) = x^{2} \sin(1/x), g(x) = x$ as x → 0

Why it fails

f/g = x sin(1/x) → 0, but f′/g′ = 2x sin(1/x) − cos(1/x) has no limit. L'Hôpital goes one way only.

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Keep practicing

Converse assumed

The theorem runs one way. You used it in the other.

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differentiation

Go deeper· Explained in full in Differentiation — MVT, Darboux, Taylor and the classic spoilers· Asked in CSIR NET June 2025· From Continuity and Differentiation › Differentiability, mean value theorems, Taylor, L'Hôpital

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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