The limit related to inverse functions How to formally prove, by Cauchy definition of limit, that $\lim\limits_{x\rightarrow 0^{+}}\frac{f(x)}{g(x)}=1$ implies that also $\lim\limits_{x\rightarrow 0^{+}}\frac{f^{-1}(x)}{g^{-1}(x)}=1$, where $f^{-1}$ (or $g^{-1}$) denotes the inverse function of $f$ (or $g$), functions $f$ and $g$ are both continuous and strictly increasing and $f(0)=g(0)=0$?
 A: Unless I am mistaken, the result does not hold. Here is a counterexample.
Define $f$ and $g$ on $I=[0,\mathrm e^{-1})$ by $f(0)=g(0)=0$ and, for every positive $x$ in $I$,
$$
\frac1{f(x)}=\log\left(\frac1x\right)-\log\log\left(\frac1x\right),\qquad\frac1{g(x)}=\log\left(\frac1x\right).
$$
Then $f$ and $g$ are continuous and increasing on $I$. Furthermore, $\log(u)\ll u$ when $u\to+\infty$, hence $\log\log\left(\frac1x\right)\ll\log\left(\frac1x\right)$ when $x\to0^+$, which implies that
$$
\lim\limits_{x\to0^+}\frac{f(x)}{g(x)}=1.
$$ 
The inverse functions $f^{-1}$ and $g^{-1}$ are defined on $J=[0,1)$ and $g^{-1}(y)=\mathrm e^{-1/y}$ for every $y$ in $J$. Writing $f^{-1}(y)$ as 
$$
f^{-1}(y)=\mathrm e^{-1/y}\cdot y\cdot (1+z(y)),
$$
one gets after some easy algebra that $z(y)$ solves the equation 
$$
z=y\cdot(\log(1+z)-\log(y)).
$$
From there, one sees that $z(y)\to0$  when $y\to0^+$ hence
$$
\lim\limits_{y\to0^+}\ \frac{f^{-1}(y)}{g^{-1}(y)}=\lim\limits_{y\to0^+}\ y\cdot(1+z(y))=0.
$$
