Let $f(x)$ be a continuous, real-valued, one-to-one function on some domain $(a,b)$ which diverges at one end of the domain (say $b$). As $f$ is one-to-one I expect to be able to define a inverse function $f^{-1}$ such that $f^{-1}(f(x)) = x$ for $x$ in the domain of $f$.
Is the limit $$ \lim_{x\to b} f^{-1}(f(x)) \,,$$ well defined (and presumably equal to $b$), and if so is there a general procedure for evaluating $$ \lim_{x\to b} f^{-1}(c \, f(x)) \,,$$ for some constant $c$?
I have in mind $$\lim_{x\to \pi/2} \arctan( c \, \tan(x) )$$ in particular if there is some further condition that $f$ must meet to proceed.
If $f:(a,b)\longrightarrow \mathbb{R}$ is continuous and injective then it must be monotone. If $f$ 'diverges' near $b$ it must be unbounded near $b$; $+\infty$ if $f$ is increasing and $-\infty$ if $f$ is decreasing. We assume the first case for simplicity, but the other case can be treated in much the same manner.
Since $f$ is injective, it admits an inverse $f^{-1}:\text{Im}(f)\longrightarrow (a,b)$. It will also be monotone: because $f$ is continuous, it maps connected sets to connected sets, so $\text{Im}(f)\subset \mathbb{R}$ will also be an interval. We have then that $f^{-1}(x)$ approaches $b$ as $x\to +\infty$.
The limit $\lim_{x\to b}f^{-1}(f(x))$ clearly always exists because it is simply $\lim_{x\to b} x$. When $c>0$, $\lim_{x \to b}f(x)=\lim_{x \to b}c\cdot f(x)=+\infty$, so $\lim_{x\to b}f^{-1}(c\cdot f(x))=b$.
However, for $c\leq 0$ the limit may not be defined.