Intuitively, I know that the singularity at $z=1$ of $f(z)=\sin(\frac{z}{z-1})$ can’t be removed because $\lim_{z \rightarrow 1} f(z)$ isn‘t even defined. But in complex analysis we defined $a$ to be a removable singularity of any holomorphic function g:
$g: \Omega \setminus \{a\} \rightarrow \mathbb{C}$, $ \ \ \Omega \subset \mathbb{C}$ open
if $\lim_{z\rightarrow a} (z-a)g(z) = 0$.
And this is true for my function f because sin is bounded as z gets to 1 and $(z-a)$ goes to 0, isn‘t it?