Proving or disproving $\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{y\to 0}\frac{df(g^{-1}(y))}{dy}$

Question

I'm trying to prove or dispove that (in $\mathbb{R}$) $$\lim_{x\to 0}\frac{f(x)}{g(x)}\stackrel{?}{=}\lim_{y\to 0}\frac{dh(y)}{dy}$$ with $$h(y)=f(g^{-1}(y))$$ $$g(g^{-1}(x))=g^{-1}(g(x))=x$$ $$f(0)=\lim_{x\to 0}f(x)=0$$ $$g(0)=\lim_{x\to 0}g(x)=0.$$ I tried applying L'Hôpital's rule to the left side, followed by the inverse function rule and the chain rule; but to no avail. On the other hand, I'm failing to find a valid pair of functions that disproves the equation.

Any feedback is appreciated!

EDIT: Let's add the restrictions that both limits must exist and are finite real.

I would appreciate all suggestions for further reasonable restrictions that might be necessary to ensure the equality to be possibly true even if it limits the set of potential function pairs to those with more specific properties.

Maybe the more fitting question would be:

Under what restrictions can $\lim_{x\to 0}\frac{f(x)}{g(x)}\stackrel{?}{=}\lim_{y\to 0}\frac{dh(y)}{dy}$ be proven to be true?

Answer 1

As a counterexample, choose $$f(x)= \begin{cases} x^2\sin\left(\frac{1}{x}\right) & x\ne 0 \\ 0 & x=0 \end{cases}\quad\text{and}\quad g(x)=x\,\,\text{for all }x\in\mathbb{R} $$ Then $g^{-1}(x)=x$, so $h(y)=f(y)$ for all $y\in\mathbb{R}$ and $$h'(y)= \begin{cases} 2y\sin\left(\frac 1y\right)-\cos\left(\frac 1y\right) & y\ne 0 \\ 0 & y=0 \end{cases} $$ We have $\lim_{x\to 0}\frac{f(x)}{g(x)}=0$. However, $\lim_{y\to 0}h'(y)$ doesn't exist because $y\sin\left(\frac 1y\right)\to 0$ and $\cos\left(\frac 1y\right)$ has no limit as $y\to 0$.