Singularity of $\frac{1 - e^{-z}}{z}$

Question

I think the function $f\colon\mathbb{C}\to\mathbb C,\ z\mapsto \frac{1 - e^{-z}}{z}$ has removable singularity at 0. Since lim of the function is 1 (By L-Hospital Rule) Since the limit is finite, so we can redefine the function, so that it has no singularity.. Am I Correct? Or please Help..