Let us consider two differentiable functions \(f\) and \(g\) defined in an open interval \((a,b)\), where \(b\) might be \(\infty\). If

\[\lim\limits_{x \to b^-} f(x) = \lim\limits_{x \to b^-} g(x) = \infty\] and if \(g^\prime(x) \neq 0\) in some interval \((c,b)\), then a version of l’Hôpital’s rule states that \(\lim\limits_{x \to b^-} \frac{f^\prime(x)}{g^\prime(x)} = L\) implies \(\lim\limits_{x \to b^-} \frac{f(x)}{g(x)} = L\).

We provide a counterexample when \(g^\prime\) vanishes in all neighborhood of \(b\). The counterexample is due to the Austrian mathematician Otto Stolz.

We take \((0,\infty)\) for the interval \((a,b)\) and \[

\begin{cases}

f(x) &= x + \cos x \sin x\\

g(x) &= e^{\sin x}(x + \cos x \sin x)

\end{cases}\] which derivatives are \[

\begin{cases}

f^\prime(x) &= 2 \cos^2 x\\

g^\prime(x) &= e^{\sin x} \cos x (x + \cos x \sin x + 2 \cos x)

\end{cases}\] We have \[

\lim\limits_{x \to \infty} \frac{f^\prime(x)}{g^\prime(x)} = \lim\limits_{x \to \infty} \frac{2 \cos x}{e^{\sin x} (x + \cos x \sin x + 2 \cos x)} = 0,\] however \[

\frac{f(x)}{g(x)} = \frac{1}{e^{\sin x}}\] doesn’t have any limit at \(\infty\) as it oscillates between \(\frac{1}{e}\) and \(e\).

## Mathematical exceptions to the rules or intuition