In my book, the poles of $f(z)$ have been found out as $z=(2n+1)πi$. So, the singularity is non-isolated essential as $z=\infty$ is a limit point of these poles. I understand this.
However, if I compute the limit of $f(z)$ as $z\rightarrow∞$ I get, $$\lim_{x \to \infty} \frac{e^{-z} - 1}{e^{-z} + 1} = -1$$
This must mean that there is a removable singularity at $z=\infty$. How am I wrong here?