Let $f(0),g(0)\to0$, then $\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{g^{-1}(x)}{f^{-1}(x)}$
I'm looking for a proof to this
I'm not sure if this is correct or has limitations on what $f$ and $g$ can be but I'm trying to prove something larger relating arbitrary functions to hyperellipses and this has me stuck. I haven't taken analysis yet so this may be really simple but I'm not sure what to.
(Assuming $\lim_{x\to0}\frac{f'(x)}{g'(x)}$ exist)Here $f, g$ are differentiable around $0$. We can apply L'hospital rule and we get,
$\lim_{x\to0}\frac{f(x)}{g(x)} = \lim_{x\to0}\frac{f'(x)}{g'(x)} $. By Inverse Function theorem we have $f'(x) = \frac{1}{(f^{-1})'(f(x))}$ and $g'(x) = \frac{1}{(g^{-1})'(g(x))}$. Now since $f(x),g(x) \rightarrow 0$ as $x \rightarrow 0$. Therefore we get, $$\lim_{x\to0}\frac{f(x)}{g(x)} = \lim_{x\to0}\frac{(g^{-1})'(x)}{(f^{-1})'(x)}$$
Again since $f^{-1}(0) = g^{-1}(0) = 0$, we have $\lim_{x\to0}\frac{g^{-1}(x)}{f^{-1}(x)} = \lim_{x\to0}\frac{(g^{-1})'(x)}{(f^{-1})'(x)}$. Thus we get $$\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{g^{-1}(x)}{f^{-1}(x)}$$