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!

  • Why do you have $n$ in the limit? – Thomas Andrews Aug 18 at 23:28
  • 6
    Where does $g^{-1}(x/\lambda)\approx f^{-1}(x)$ come from? – Jack M Aug 19 at 0:36
  • @ThomasAndrews thanks! i wanted to write $x$ – Pedro Aug 19 at 1:05
  • @JackM from setting $x\mapsto g^{-1}(x/\lambda )$ – Pedro Aug 19 at 1:06
  • @Pedro I don't understand what that means. We have $g(x)\approx \frac{f(x)}{\lambda}$, but not $g(x)\approx\frac x\lambda$ – Jack M Aug 19 at 10:35
up vote 17 down vote accepted

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 :

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

  • Nice counterexample! Thanks – Pedro Aug 19 at 1:10

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