Answer
$(x\rightarrow \pm\infty)\Rightarrow \left(f^{-1}(x)\rightarrow \pm\dfrac{\pi}{2}\right)$
Work Step by Step
We are given the function:
$f(x)=\tan x$
For the inverse to exist, we must restrict the domain of $f$ so that the function is one-to-one:
$D_f=\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
$R_f=(-\infty,\infty)$
The domain and range of the inverse $f^{-1}(x)=\tan^{-1}(x)$ are:
$D_{f^{-1}}=(-\infty,\infty)$
$R_{f^{-1}}=\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
The graph of $f^{-1}$ is symmetrical with the graph of $f$ over the line $y=x$.
When $x$ gets close to $\pm\dfrac{\pi}{2}$, $f$ is very large or very small.
When $x$ is very large or very small, $f^{-1}$ gets close to $\pm\dfrac{\pi}{2}$.
