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!
Best Answer
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*}