If an invertible function approaches infinity at infinity, does its inverse approach infinity at infinity? Prove or disprove: if $f(x)$ is defined on $ℝ$ (not necessarily continuous) and has an inverse and $\lim_{x→∞} f(x)=∞ $ then $\lim_{x→∞}f^{-1}(x)=∞ $.
I think it's true.
I tried using Heine's theorem:
We know that $\lim_{x→∞}f(x)=∞ $, this means that for every sequence $x_n→∞$ we have $f(x_n)→∞$. So if by way of contradiction we assume that $\lim_{x→∞}f^{-1}(x)≠∞ $, this means that there exists a  sequence $y_n→∞$ such that $\lim_{n→∞}f^{-1}(y_n)≠∞ $ but now I'm stuck because it doesn't give me any information about $f(x)$.
 A: An idea towards a counterexample
"Because the function can be discontinuous, $f(x)$ might have a vertical asymptote at some point $a$. There might be a sequence $x_n$ that approaches $a$, such that $f(x_n)$ would approach infinity. However, $f^{−1}(f(x_n))$ approaches $a$." This is the idea written somewhere by the asker.
A counterexample where $f$ near $0$ can be arbitrarily large
Define $f:\Bbb R\to\Bbb R$ such that

*

*$f$ is the identity function except on $\{\frac12,\frac13, \frac14,\frac15,\frac16,\frac17,\cdots\}\cup\{2,3,4,5,6,7,\cdots\}$.

*$f$ maps ${2,3,4,\cdots}$ to $3,5,7,\cdots$ respectively.

*$f$ maps $\frac12,\frac14,\frac16,\cdots$ to $2,4,6,\cdots$ respectively.

*$f$ maps $\frac13,\frac15,\frac17,\cdots$ to $\frac12,\frac13,\frac14,\cdots$ respectively.

Verify that $f$ is a bijection from $\Bbb R$ to $\Bbb R$. Verify that $\lim_{x\to\infty}f(x)=\infty$. 
Let $i$ be a positive even integer. Since $f(\frac1{i})=i$, we have
$$\lim_{i\to\infty}f^{-1}(i)=\lim_{i\to\infty}\frac1{i}=0\not=\infty.$$
By the way, $f$ can be expressed as $$f(x)=\begin{cases}
\frac1{\frac{i+1}2}&\text{if }x=\frac1i,\text{ where }i\text{ is a positive odd integer},\\
i&\text{if } x=\frac1i,\text{ where }i\text{ is a positive even integer},\\
2x-1&\text{if } x\text{ is a positive integer},\\
x&\text{for other cases.}\end{cases}$$
An exercise, is the infinity special?
Suppose $f(x)$ is defined on $\Bbb R$ (not necessarily continuous) and has an inverse and $\lim_{x\to b} f(x)=a$ for $a,b\in\Bbb R$, is it true $\lim_{x\to a}f^{-1}(x)=b$? (If an invertible function approaches $a$ near $b$, does its inverse approach $b$ near $a$? )
