Is this "reverse the limit" process right?

Question

Suppose you have two (nice enough) functions $f$ and $g$ and a constant $\lambda$ such that $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lambda$$ Is it true that $$\lim_{x\to\infty}\frac{f^{-1}(x)}{g^{-1}(x/\lambda)}=1$$

The "reasoning" goes like this: $$\frac{f(x)}{g(x)}\approx\lambda$$ $$f(x)\approx\lambda g(x)$$ $$x\approx f^{-1}(\lambda g(x))$$ $$g^{-1}(x/\lambda)\approx f^{-1}(x)$$ $$\frac{f^{-1}(x)}{g^{-1}(x/\lambda)}\approx 1$$

all of this supposing there is no problem in $x\mapsto f^{-1}(x)$ and $x\mapsto g^{-1}(x/\lambda)$

I think assuming "nice enough" (continuity, inverse, ...) and being a bit more precise like $$f(x)=\lambda g(x)+o(g(x))$$ will prove the statement.

What I'm more interested in is under what conditions does it remain true in discrete variable (and non-existing inverse function for $f$ or $g$ or both)

Thanks!

Answer 1

No. If you take $f(x) = \log(x)$ and $g(x) = \log(x)-1$ and $\lambda = 1$ then :

$$ \underset{x\rightarrow\infty}{\lim}\frac{f(x)}{g(x)} = 1 $$

But :

$$ \frac{f^{-1}(x)}{g^{-1}(x)} = \frac{e^x}{e^{x+1}} = e^{-1} \neq 1$$