Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity (case (3)). If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.
$$\frac{e^z-1}{z}$$
Here is my answer:
- First, we get that z=0 is a singularity.
- Next, we have $$\lim_{z\to 0}\frac{e^z-1}{z}=\lim_{z\to 0}\frac{e^z}{1} = \lim_{z\to 0}{e^z} = 1$$
- Hence, by definition, it is a removable singularity.
