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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

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.

The kind of mistake this is

Converse assumed

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

Drill statements like this

differentiation

Others that fail the same way

From Continuity and Differentiation › Differentiability, mean value theorems, Taylor, L'Hôpital

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