I want to prove that
If $f(x)\sim g(x)$ as $x\to+\infty$ then $f^{-1}(x)\sim g^{-1}(x)$ as $x\to +\infty$.
or equivalently
$\lim_{x\to+\infty}\frac{f(x)}{g(x)}=1$ implies $\lim_{x\to+\infty}\frac{f^{-1}(x)}{g^{-1}(x)}=1$.
We may need to assume some conditions on $f$ and $g$. At the first glance, we note that $f$ and $g$ should be invertible for large $x$. Furthermore, $g$ and $g^{-1}$ are assumed to be nonzero for large $x$. We may need more! My question is
What are the minimum sufficient conditions for this to be true?
To give an example, we know that $\cosh x=\frac{1}{2}(e^x+e^{-x})$. As $x\to+\infty$ it is clear that $\cosh x\sim \frac{1}{2}e^x$. Knowing that $\cosh^{-1}x=\ln (x+\sqrt{x^2-1})$, we further observe that $x+\sqrt{x^2+1}\sim 2x$ and since both sides have the requirements of L'Hopital theorem we can conclude $\ln (x+\sqrt{x^2-1})\sim \ln 2x$ or equivalently $\cosh^{-1}x\sim \ln 2x$ as $x\to+\infty$. So the assertion seems to be true for this example.
My motivation for this question is that I read in section 1.4 of this book that we can make an asymptotic approximation for the inverse function without directly computing it. Indeed, this assertion let us to just compute $g^{-1}$ which may be much easier than computing $f^{-1}$ itself.
For a counterexample, consider $f(x) = \log(x)$ and $g(x) = \log(2x)$ as $x \to \infty$. Then $f(x) \sim g(x)$ but $f^{-1}(y) = \exp(y)$ and $g^{-1}(y) = \exp(y)/2$ so $f^{-1}(y)/g^{-1}(y) \to 2$, not $1$, as $y \to \infty$.
It's also easy to find examples where $f^{-1}/g^{-1}$ does not tend to a limit at all, or goes to $\infty$.