Is this "reverse the limit" process right?
Solution 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 :
$$ \frac{f^{-1}(x)}{g^{-1}(x)} = \frac{e^x}{e^{x+1}} = e^{-1} \neq 1$$