If $\lim_{x\to x_0} \frac{f(x)}{g(x)} =L $ , then what we can say about $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}$?

Question

$f$ and $g$ are differentiable functions so that they have the first derivative in a neighborhood of a point $x_0$, and so that $g(x)\neq0$ and $g'(x)\neq0$ in a neighborhood of a point $x_0$.

It also holds: $\lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = 0$ .

If $\lim_{x\to x_0} \frac{f(x)}{g(x)} =L $ , then what we can say about $\lim_{x\to x_0} \frac{f'(x)}{g'(x)}$ ?

The same question I have if $\lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = \infty $ .

I am aware that if I would have the opposite, that $\lim_{x\to x_0} \frac{f'(x)}{g'(x)} =L $ , then I would have that $\lim_{x\to x_0} \frac{f(x)}{g(x} =L $ due to L'Hospitals rule, but I'm not sure what can I say about the opposite. I tried to find some examples when this is not true but I failed finding some...

Thank you in advance!

Answer 1

If the limit $\lim_{x\to x_0}\frac{f(x)}{g(x)}=L$, then, assuming that $\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0$, the limit$$\lim_{x\to x_0}\frac{f'(x)}{g'(x)},\tag1$$if it exists it must be equal to $L$. This follows from L'Hopital's Rule. But the limit $(1)$ doesn't have to exist. For instance, take $x_0=0$, $f(x)=x^2\left(2+\sin\left(\frac1{x^2}\right)\right)$ and $g(x)=x$. Then $\lim_{x\to0}\frac{f(x)}{g(x)}=0$, but the limit $\lim_{x\to0}\frac{f'(x)}{g'(x)}$ doesn't exist.