Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values? Does the ratio of the $x-$values of two continuous functions $f(x), g(x) \in \mathbb{R^+}$ tell you anything about the ratio between their $y-$values?
In general, obviously not. But suppose the following is also true:
$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0$,
$\lim_{x \to \infty} \frac{f(x)}{g(x)}$ converges to $L_1$ and $\lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}$ converges to $L_2$.
Then what can we say about $L_2$ in terms of $L_1$? e.g. If $L_1 = \infty$ then does $L_2 = \infty$? If $L_1$ is finite then is $L_2$ finite?
 A: To begin with, let us assume that $f,g$ are injective so that we have no problems with defining the inverses. Then by continuity, they are monotonic. To vahe nno problems with $x\to 0^+$, we better assume they are positive (and hence decreasing).
Even then, we cannot conclude a lot about $L_2$ from $L_1$:

*

*By  swapping $f\leftrightarrow g$, we find that if $(L_1,L_2)$ is possible, then so is $(1/L_1,1/L_2)$.

*Of course, if $1<L_1\le \infty$, then the graph of $f$ is eventually above that of $g$, hence $f^{-1}$ must also be above $g^{-1}$, i.e., $1\le L_2\le\infty$.

*In other words, $L_1<1<L_2$ and  $L_1>1>L_2$ are impossible.

We can construct $f,g$ with limits to our liking as follows:
Let $\{a_n\}_{n\in\Bbb N}$ and $\{b_n\}_{n\in\Bbb N}$ be convergent (possibly to $\infty$) sequences of numbers $>1$. Let $x_n=\prod_{k=1}^n a_k$, $y_n=\prod_{k=1}^n b_k^{-1}$.
Assume $x_n\to \infty$ and $y_n\to 0$.
Let $f$ be the piecewise linear interpolation through the points $(x_n,y_n)$ and $g$ the piecewise linear interpolation through the points $(x_n,y_{n+1})$. Then one directly checks that $\frac{f(x)}{g(x)}=\frac{y_n}{y_{n+1}}=b_{n+1}$ at $x=x_n$.
One verifies that on the interval $[x_n,x_{n+1}]$, the quotient $\frac{f(x)}{g(x)}$ only varies between $b_{n+1}$ and $b_{n+2}$ (the quotient is essentially piecewise a hyperbola). We conclude that $$L_1=\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{n\to\infty} b_n.$$ By the same argument,
$$ L_2=\lim_{y\to0^+}\frac{f^{-1}(y)}{g^{-1}(y)}=\lim_{n\to\infty} a_n.$$
By picking $a_n=a^n$ with $1<a<\infty$ we can thus achieve $L_1=a$.
By picking $a_n=n!$, we achieve $L_1=\infty$, and by $a_n=1+\frac1n$ we achieve $L_1=1$. In summary, we can achieve any $L_1\in[1,\infty]$ by a suitable choice of  $\{a_n\}_{n\in\Bbb N}$.
Likewise, we can achieve any $L_2\in[1,\infty]$ by a suitable choice of  $\{b_n\}_{n\in\Bbb N}$.
As the choices are independent, any combination $(L_1,L_2)\in[1,\infty]\times [1,\infty]$ can be achieved (and likewise any $(L_1,L_2)\in [0,1]\times[0,1]$).
A: If $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = 0$ and
\begin{equation}
  \lim_{x \to \infty} \frac{f(x)}{g(x)}=L_1 \text{ and }
\lim_{y \to 0+} \frac{f^{-1}(y)}{g^{-1}(y)}=L_2\tag{1}
\end{equation}
then if $L_1>1$, $L_2\geq 1$ and if $L_1<1$, $L_2\leq 1$.  Since
$L_1>1$ implies that for sufficiently large $x$, $f(x)>g(x)$, for
sufficiently small $y$, $f^{-1}(y)>g^{-1}(y)$ ($x$ must be bigger for
$f$ to shrink to $y$). Thus, $L_2\geq 1$. The second conclusion is
similarly shown. Too many examples:

*

*Let $L_1$ and $L_2$ be arbitrary positive real numbers either
both greater than one or both less than one. Set
$\alpha=\log{L_2}/\log{L_1}$ and note that $\alpha>0$. Let
$f(x)=L_1x^{-1/\alpha}$ and $g(x)=x^{-1/\alpha}$ so that
$f^{-1}(y)=L_2y^{-\alpha}$ and $g^{-1}(y)=y^{-\alpha}$, and hence (1) holds.

*Let $L_1=1$ and $L_2$ be an arbitrary positive real number. Let
$a=-\log{L_2}$ and set $f(x)=(\log x+a)^{-1}$ and $g(x)=(\log x)^{-1}$.
Since $f^{-1}(y)=L_2e^{1/y}$ and $g^{-1}(y)=e^{1/y}$,  (1) holds.

*Let $L_1=1$ and $L_2=\infty$. Set
$f(x)=(\log\log x-\log 2)^{-1}$ and $g(x)=(\log\log x)^{-1}$.  Since
$f^{-1}(y)=e^{2e^{1/y}}$ and $g^{-1}(y)=e^{e^{1/y}}$, (1) holds.

*Let $L_1=1$ and $L_2=0$. Set
$f(x)=(\log\log x+\log 2)^{-1}$ and $g(x)=(\log\log x)^{-1}$.  Since
$f^{-1}(y)=e^{\frac12 e^{1/y}}$ and $g^{-1}(y)=e^{e^{1/y}}$, (1) holds.

*Let $L_1=\infty$ and $1<L_2<\infty$. Set
$f(x)=e^{-x/L_2}$ and $g(x)=e^{-x}$.  Since
$f^{-1}(y)=-L_2\log y$ and $g^{-1}(y)=-\log y$, (1) holds.

*Let $L_1=0$ and $0<L_2<1$. Set
$f(x)=e^{-x/L_2}$ and $g(x)=e^{-x}$.  Since
$f^{-1}(y)=-L_2\log y$ and $g^{-1}(y)=-\log y$, (1) holds.

*Let $L_1=0$ and $L_2=0$. Set
$f(x)=x^{-2}$ and $g(x)=x^{-1}$.  Since
$f^{-1}(y)=y^{-1/2}$ and $g^{-1}(y)=y^{-1}$, (1) holds.

*Let $L_1=\infty$ and $L_2=\infty$. Set
$f(x)=x^{-1}$ and $g(x)=x^{-2}$.  Since
$f^{-1}(y)=y^{-1}$ and $g^{-1}(y)=y^{-1/2}$, (1) holds.

