$f,g$ convex, increasing real functions with $\frac{f(x)}{g(x)}\to 1.$ Does $\frac{f^{-1}(x)}{g^{-1}(x)}\to 1\ ?$ Let $f,g:\mathbb{R}\to\mathbb{R}$ be convex strictly increasing real functions (so we have both $f(x)\to\infty $ and $g(x)\to\infty$ as $x\to\infty),$ and suppose further that $\frac{f(x)}{g(x)}\to 1.$

Then is it true that $\frac{f^{-1}(x)}{g^{-1}(x)}\to 1\ ?$

I know that since $f,g$ are convex with domain $\mathbb{R}$, they are continuous, which is why I left this out of the question.
And if $f,g$ did not have to be convex, then $f(x) = \ln(x),\ g(x) = \ln(x)+1$ would be a counter-example, as $\frac{f^{-1}(x)}{g^{-1}(x)}\to e\neq 1.$
I'm not coming up with a counter-example, and am also not sure how to approach the question otherwise as there seems to be a lot of things to consider in order to prove it true.
 A: Fix $\epsilon \in ]0,1[$.
Since
$$\frac{f(g^{-1}(x))}{x} = \frac{f(g^{-1}(x))}{g(g^{-1}(x))} \to 1 \textrm { as } x \to +\infty,$$
one has for every large enough positive $x$
$$f((1-\epsilon)g^{-1}(x)) \le (1-\epsilon)f(g^{-1}(x)) + \epsilon f(0) < x, \text{ by convexity of } f,$$
so $$(1-\epsilon)g^{-1}(x) \le f^{-1}(x), \text{ since $f^{-1}$ is increasing.}$$
Switching the roles of $f$ and $g$ shows that $g^{-1}(x)/f^{-1}(x) \to 1$ as $x \to +\infty$.
A: Yes.
We have
$$f(f^{-1}(x)) = g(g^{-1}(x)).$$
Thus by convexity, for any $x$ such that $f^{-1}(x) \geq g^{-1}(x)$, we have
$$g(g^{-1}(x)) = f(f^{-1}(x)) \geq \frac{f^{-1}(x)}{g^{-1}(x)}f(g^{-1}(x)) - \frac{f^{-1}(x) - g^{-1}(x)}{g^{-1}(x)}f(0).$$
Or
$$\frac{f^{-1}(x)}{g^{-1}(x)} \leq \frac{g(g^{-1}(x)) - f(0)}{f(g^{-1}(x)) - f(0)}.$$
As $x \to \infty$, we have $g^{-1}(x) \to \infty$, so $\frac{g(g^{-1}(x)) - f(0)}{f(g^{-1}(x)) - f(0)} \to 1$. We conclude that
$$\limsup_{x \to \infty} \frac{f^{-1}(x)}{g^{-1}(x)} \leq 1.$$
Similarly,
$$\limsup_{x \to \infty} \frac{g^{-1}(x)}{f^{-1}(x)} \leq 1.$$
and we conclude the desired result.
