Let $\omega_0 \in \mathbb{R}$ and $\omega_0 > 0$. Be $G(s)$ a transfer function defined as:
$$G(s)=\frac{1-\frac{s}{\omega_0}}{1+\frac{s}{\omega_0}}$$
We're interested into evaluating its phase response in a continuous LTI system. First, we split $G(j\omega)$ into its real and imaginary parts (assuming $\tau=\frac{\omega}{\omega_0}$):
$$G(j\omega) = \frac{1}{1+\tau^2}\left[1-\tau^2+j\left(-2\tau\right)\right]$$
Then using the definition of phase of a transfer function we have:
$$\angle G(j\omega)=\arctan\left(\frac{\mathbb{Im}\{G(j\omega)\}} {\mathbb{Re}\{G(j\omega)\}}\right)=\arctan\left(\frac{-2\tau}{1-\tau^2}\right)$$
So, for any $\tau \gg 1$ (which implies that $\omega \gg 1$) then the following should hold for Taylor's series:
$$\angle G(j\omega) \simeq \arctan\left(\frac{2}{\tau}\right) \simeq \frac{2}{\tau}$$ which approaches $0$ as $\omega$ gets bigger.
Here comes the discrepancy. Let $\omega_0 = 1$ (for plotting reasons), why does Wolfram Alpha give me the following Bode plot?
Also, I expected one discontinuity point at $\omega = \omega_0 = 1$ (and another one for $\omega = -\omega_0$ not shown in the plot), but I guess I messed up something since the beginning.
2018-12-12 (in response to Sam F.) Here's how I've separated the real from the imaginary part of $G(j\omega)$.
$$ \begin{align} G(j\omega) &=\frac{1-j\frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}} \\ &=\frac{1-j\tau}{1+j\tau} \\ &=\frac{1-j\tau}{1+j\tau}\cdot\frac{1-j\tau}{1-j\tau} \\ &=\frac{(1-j\tau)^2}{1^2-(j\tau)^2} \\ &=\frac{1^2+(j\tau)^2-2\cdot 1 \cdot j\tau }{1+\tau^2} \\ &=\frac{1- \tau^2+j(-2\tau)}{1+\tau^2} \\ \end{align} $$