If $f(s) =\frac{s}{\sqrt{1-s^2}}$, how to evaluate $\lim_{s\to \pm\infty}f^{-1}(s)$? Let $f:[0,1]\to\mathbb{R}$ be such that
$$f(s) =\frac{s}{\sqrt{1-s^2}}.$$
I wrote on my notes during the calculus class that
$$\lim_{s\to +\infty}f^{-1}(s)=1\quad\mbox{ and }\quad \lim_{s\to -\infty}f^{-1}(s)=-1.$$
Looking back, how it is possible to evaluate these limits? Actually, I’m not even able to explicitly compute $f^{-1}(s)$.
Could someone please help me to understand it?
Thank you in advance!
 A: $$f(x) = \frac x{\sqrt{1-x^2}}$$
Now, from above it is really clear that the domain for function $\in (-1, 1)$
So, yes I can substitute $x = \sin(t)$ now, here I do not mean that I'm converting the function $f(x)$ to some sort of polar function it's just I'll find the value of $t$ that corresponds to the input of function$(x)$
Say, what is the value of $f(x = a)$ where $ a \in (-1, 1)$
$$f(x = a) = \frac a{\sqrt{1-a^2}}$$
But we can do the same as $t = \arcsin(a)$ and $f(t = \arcsin(a)) = \frac {\sin(t)}{|\cos(t)|}$
$$A = (a, f(a)) = \left(t, \frac {\sin(t)}{|\cos(t)|}\right)$$

*

*Solution
As, the domain of a function is restricted to $(-1, 1)$ you can use the idea of asymptotes;
It means there must exist the constant tend of $y$
$$y = f(s) = \frac s{\sqrt{1-s^2}}$$
$$\implies y' = f'(s) = (1-s^2)^{-\frac 32}$$
$$\implies y'' = f''(s) = \frac {3s}{(1-s^2)^\frac 52}$$
Solve for the value of $s$ for which $f(s)$, $f'(s)$, and $f''(s)$ will tend to $\infty$
Thus, $s = 1$
A: Apart from the conventional solution suggested by Sarvesh, by observations
WLOG $s=\sin t $
$$f(\sin t)=\cdots=\tan t$$
$\implies $ one of the values of $f^{-1}(\tan t)$. will be $$=\sin t$$
Now let $t=\arctan u\implies \tan t=u,\sec t=+\sqrt{1+u^2}$ considering the principal value of $ u$
$$\implies f^{-1}(u)=\sin(\arctan u)=\sin t=?$$
Notice that for real $u,\sqrt{u^2}=|u|$ and for $u<0,|u|=-u$
Also we actually don't need the explicit value of $f^{-1}(u)$ as we know
$$\lim_{u\to\pm\infty}\arctan u=?$$
A: First confirm that $f:(-1,1)\to\mathbb{R}$ is bijective (one to one and onto) and continuous. Then $f^{-1}$ is also continuous and since $(f^{-1}\circ f)(t)=t$ for all $t\in(-1,1)$,
$$\lim_{t\to 1}(f^{-1}\circ f)(t)=\lim_{t\to1}t=1$$
and
$$\lim_{t\to-1}(f^{-1}\circ f)(t)=\lim_{t\to-1}t=-1$$
Now since $\lim_{t\to1}f(t)=+\infty$ and $\lim_{t\to1}f(t)=-\infty$ and $f^{-1}$ is continuous,
$$\lim_{s\to \infty}f^{-1}(s)=1\text{ and }\lim_{s\to -\infty}f^{-1}(s)=-1$$
