The tangent to calculate is: $$\phi=-\tan^{-1}\left[\frac{2\zeta\frac{\omega}{\omega_n}}{1-(\frac{\omega}{\omega_n})^2}\right]$$
Where $\omega_n$ is constant.
If
$\omega=0 \to\phi=0$
$\omega=\omega_n \to\phi=-\frac{\pi}{2}$
$\omega=\infty \to \phi=-\pi$
This results are correct. What I want to understand is why the limit:
$$\lim_{\omega \to 0} \left[-\tan^{-1}\left(\frac{2\zeta\frac{\omega}{\omega_n}}{1-\left(\frac{\omega}{\omega_n}\right)^2}\right)\right]=-\tan^{-1}\left(\frac{0}{1}\right)=0$$
and the limit
$$\lim_{\omega \to \infty} \left[-\tan^{-1} \left( \frac{2\zeta\frac{\omega}{\omega_n}}{1-\left(\frac{\omega}{\omega_n} \right)^2}\right)\right] =-\tan^{-1}\left(\frac{2\zeta}{\infty}\right)=-\pi$$
are consider differently in the calculus of the inverse tangent. Surely the two results must be opposite, but why one of the $tan^{-1}(0)$ is equal to $0$ and the other is $\pi$?
I am guessing you are dealing with the phase response of an under-damped second order linear system. The response in the frequency ($s$) domain would be $\hat{h}(s) = { \omega_n^2 \over s^2+2 \zeta\omega_n s + \omega_n^2 }$.
The phase of $\hat{h}(j\omega) = { \omega_n^2 \over \omega_n^2-\omega^2 +j 2\zeta\omega_n \omega} = { 1 \over 1- { \omega^2 \over \omega_n ^2 } + j 2 \zeta { \omega \over \omega_n } }$ can be seen to be the same as the phase of $ 1- { \omega^2 \over \omega_n ^2 } - j 2 \zeta { \omega \over \omega_n }$ (that is, the phase of ${1 \over z}$ is the same as the phase of $\overline{z}$).
Apart from the obvious discontinuities, an issue with using $(x+jy) \to \arctan { y \over x }$ to compute the phase is that it does not distinguish opposite quadrants ($(-(x+jy))$ will give the same phase as $x+jy$, even though they are $\pi$ apart). For $\zeta = {1 \over 2}$ we have
which shows the issue when passing through $\omega = \omega_n$.
If we plot $x \mapsto 1-x^2-j 2 \zeta x$ instead, we get
from which it is clear that the phase converges to $-\pi$ as $x \to \infty$. It is also clear that the phase for $x=0$ (which corresponds to $1-0j$) is zero.
In this particular example, since the locus of $\omega \to \hat{h}(j \omega)$ does not cross the negative real axis, we can use the principal value of $\arg$ to compute the phase. It is straightforward to show that $\lim_{x \to \infty} \arg (1-x^2-j 2 \zeta x) = -\pi$.