Let $f(z)= \dfrac 1 {e^{1/z}+1}$. Does this have a removable singularity at $z=\infty$?
I found that all the singulairties (other than $\infty$) are bounded. So $z=\infty$ is a isolated singularity of $f(z)$. But does $f(1/z)=\dfrac 1 {e^{1/(1/z)}+1}=\dfrac 1 {e^{z}+1}$ have removable singularity at $z=0$ because just it has $z$ in the denominator?