I am required to prove that $\lim_{z \to -a}$ Arg$(z)$ does not converge for $a > 0$. I've broken the question down into showing that $\lim_{z \to -a}$ Arg$(z) = \pi$ for $\Im (z) \geq 0$ and $\lim_{z \to -a}$ Arg$(z)=-\pi$ for $\Im(z) < 0$ (which is pretty obvious).
In order to prove these limits, I will need to use the epsilon-delta definition of a limit. Since I am new to these proofs in complex analysis, I'm unsure where to go after setting it out. For the first limit, $0 < |\text{Arg}(z) - \pi| < \varepsilon$. I am unsure where to go from here. Any guidance is greatly appreciated!
Let $z_1 = (-a + i[0]).$
It is assumed that Arg$(z_1) = \pi.$
In order to try to show that the Arg function is continuous at $z_1$
you would have to show that for any $\epsilon > 0$ there exists a $\delta > 0$
such that $0 < |z - z_1| < \delta \implies |Arg (z) - Arg (z_1)| < \epsilon.$
In order to refute the assertion of continuity of the Arg function at $z_1$, you have to show the contradiction.
This means that you have to exhibit a specific value for $\epsilon > 0$ such that
no matter how small you make $\delta > 0$ you will always be able to find
a $z$ such that $0 < |z - z_1| < \delta$ and $|Arg (z) - Arg (z_1)| \geq \epsilon.$
Choose $\epsilon = 1.$
For any $\delta > 0,$ choose $z = -a - i[\delta/2]$.
Then $0 < |z - z_1| < \delta$ and $-(\pi/2) > Arg(z) > -\pi.$
This means that $|Arg (z) - Arg (z_1)| \geq \epsilon.$
Therefore, you have just exhibited a specific value for $\epsilon$ such that
no satisfying value for $\delta$ can be found, no matter how small you set $\delta.$
You are done.
Note: showing that the Arg function is not continuous at $z_1$ is equivalent to showing that it is not the case that
the $\lim_{z \to z_1} ~Arg(z) = Arg (z_1).$