asymptotics of inverse function

Question

Suppose $f:[0,\infty)\to [0,\infty)$ is strictly increasing with $f(0)=0$ and it's given explicitly as a combination of elementary functions. How do you find the asymptotics of $f^{-1}(x)$ as $x\to 0$ if you can't invert $f$ explicitly? By asymptotics I mean an elementary function $g$ defined on an interval $(0,\epsilon)$ such that $\lim_{x\downarrow 0}\frac{f^{-1}(x)}{g(x)}=1$.

Answer 1

The method purposed by Greg Martin can be easily justified for some typical cases, for example, if $f(x)\sim g(x)$ and $g(x)=Cx^p$ for some $p>0$ and $C>0$.

In fact, suppose that $f(x)=g(x)q(x)$ with $g(x)=Cx^p$ and $q(x)\to1$. Then $$ x=(f(x)/C)^{1/p} q(x)^{-1/p} \sim (f(x)/C)^{1/p}. $$ Therefore $f^{-1}(y)\sim (y/C)^{1/p}=g^{-1}(y)$ as $y\to0$.

The situation changes if $f$ and $g$ converge to zero very slowly. Here is an example of $f$ and $g$ with $f/g\to1$ and $f^{-1}/g^{-1}\not\to1$. Put $$ f(x)=\frac{1}{\log(1/x)},\qquad g(x)=\frac{1}{\log(7/x)}. $$ Then $$ f^{-1}(y)=e^{-1/y},\qquad g^{-1}(y)=7e^{-1/y}. $$ In this example $$ \lim_{x\to0}\frac{f(x)}{g(x)}=1,\qquad \lim_{y\to0}\frac{g^{-1}(y)}{f^{-1}(y)}=7. $$