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How to formally prove, by Cauchy definition of limit, that $\lim\limits_{x\rightarrow 0^{+}}\frac{f(x)}{g(x)}=1$ implies that also $\lim\limits_{x\rightarrow 0^{+}}\frac{f^{-1}(x)}{g^{-1}(x)}=1$, where $f^{-1}$ (or $g^{-1}$) denotes the inverse function of $f$ (or $g$), functions $f$ and $g$ are both continuous and strictly increasing and $f(0)=g(0)=0$?

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1 Answer 1

Unless I am mistaken, the result does not hold. Here is a counterexample.

Define $f$ and $g$ on $I=[0,\mathrm e^{-1})$ by $f(0)=g(0)=0$ and, for every positive $x$ in $I$, $$ \frac1{f(x)}=\log\left(\frac1x\right)-\log\log\left(\frac1x\right),\qquad\frac1{g(x)}=\log\left(\frac1x\right). $$ Then $f$ and $g$ are continuous and increasing on $I$. Furthermore, $\log(u)\ll u$ when $u\to+\infty$, hence $\log\log\left(\frac1x\right)\ll\log\left(\frac1x\right)$ when $x\to0^+$, which implies that $$ \lim\limits_{x\to0^+}\frac{f(x)}{g(x)}=1. $$ The inverse functions $f^{-1}$ and $g^{-1}$ are defined on $J=[0,1)$ and $g^{-1}(y)=\mathrm e^{-1/y}$ for every $y$ in $J$. Writing $f^{-1}(y)$ as $$ f^{-1}(y)=\mathrm e^{-1/y}\cdot y\cdot (1+z(y)), $$ one gets after some easy algebra that $z(y)$ solves the equation $$ z=y\cdot(\log(1+z)-\log(y)). $$ From there, one sees that $z(y)\to0$ when $y\to0^+$ hence $$ \lim\limits_{y\to0^+}\ \frac{f^{-1}(y)}{g^{-1}(y)}=\lim\limits_{y\to0^+}\ y\cdot(1+z(y))=0. $$

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Ok, I followed by your suggestions and I'm able to prove that $f^{-1}(vy)\leq g^{-1}(y)\leq f^{-1}(uy)$ for y from some neighborhood of 0 but I really can't see in what exactly way I should use the continuity assumption to get desired result. Could you explain me it in more detailed way? Thank you. –  Mark Nov 7 '11 at 13:40
    
My so-called hint led to a dead end, sorry about that. See the revised version. –  Did Nov 7 '11 at 17:36
    
Good example (+1). For me, a slightly easier way to see what is going on here is to notice that $$\exp(-1/f(x)) = -x\log(x)$$ and $$\exp(-1/g(y)) = y$$ So if $f(x)=g(y)$, then $y=-x\log(x)$. Thus, $y/x=-\log(x)\to\infty$. –  robjohn Nov 7 '11 at 21:25
    
In fact, it seems that $$\frac{g^{-1}(z)}{f^{-1}(z)}=-\log(f^{-1}(z))$$ –  robjohn Nov 7 '11 at 21:35

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