The function $$ f(z)=\frac{\cos(z)}{z^2-\pi^2/4} $$ has two removable singularities at $z=\pm\pi/2$. I'm wondering is there a formal approach to determine the singularities are removable? Intuitively, I determined they're removable because at those points we have a $0/0$ form.
Thanks!
It's not just that they are of the form $0/0$, it's that both $\cos(z)$ and $z^2 - \pi^2/4$ have simple zeros at $\pm \pi/2$. That is, there are nonzero constants $a$ and $b$ such that $\cos(z) \sim a (z-p)$ and $z^2 - \pi^2/4 \sim b (z-p)$ as $z \to p$ (where $p = \pi/2$ or $-\pi/2$), so that $\cos(z)/(z^2-\pi/4) \to a/b$ as $z \to p$.