TITLE: Prove or disprove a converse to l'Hospital's rule
QUESTION [6 upvotes]: Assume that:
1)
$$ \lim_{x\to c}f(x) = \lim_{x\to c}g(x) = 0, $$
where $c$ is a finite real number contained in an interval $I=(a,b)$ (which means that we have excluded the situation that $c=\pm\infty$);

$f(x)$ and $g(x)$ are differentiable on the interval $I$ except possibly at the point $c$ above;

$$\lim_{x\to c}f^{'}(x) = \lim_{x\to c}g^{'}(x) = 0;$$

$$\lim_{x\to c}\frac{f(x)}{g(x)}$$ exits and is equal to $L$.


Then prove or disprove: $\lim_{x\to c}\frac{f^{'}(x)}{g^{'}(x)}$ exits and is equal to $L$.
It's seems that most counterexamples I have found are under one of the following two circumstances:
1)$c=\pm\infty$,  and
2)$c\neq\pm\infty$ but either $\lim_{x\to c}f^{'}(x)$ or $\lim_{x\to c}g^{'}(x)$ is allowed to be nonzero,
but it seems harder to find a counterexample given that both circumstances above are excluded. Can anyone be so kind to help me? Thanks in advance!

REPLY [8 votes]: Take
$$
f(x)=x^3\sin\frac 1x,\qquad  g(x)=x^2,\qquad c=0.
$$
Then
\begin{align*}
\lim_{x\to 0}\frac{f(x)}{g(x)}&=0,\\[12pt]
\lim_{x\to 0} f(x)=\lim_{x\to 0} f'(x)&=0\\[12pt]
\lim_{x\to 0} g(x)=\lim_{x\to 0} g'(x)&=0\\[12pt]
\lim_{x\to 0}\frac{f'(x)}{g'(x)}&\quad\text{does not exist}
\end{align*}