Consider the following functions and determine which kind of singularities they have in $z_0$. If it is a removeable singularity, then calculate the limit; if it is a pole, then give the order of the pole and the main part. $$ f(z)=\frac{1}{1-e^{z}}, z_0=0~~~~~~~~~~~~~~~~~~~~g(z)=\frac{1}{z-\sin z}, z_0=0 $$
Concerning $f(z)$, I wrote $$ f(z)=\frac{1}{\sum\limits_{i=1}^{\infty}\frac{z^i}{i!}} $$ and then considered $$ \lim\limits_{z\to 0}\frac{z^k}{\sum\limits_{i=1}^{\infty}\frac{z^i}{i!}}=0~\forall~k\in\mathbb{N}_0 $$ So the smallest $k\in\mathbb{N}_0$ for which the limit exists is $k=0$ and so here $z_0$ is a removable singularity. Is that right? And what is meant with calculating the limit now?
I'm cringing at your use of $i$ as an index while dealing with complex numbers, but I digress.
The limit $$\lim_{z\to 0}f(z)=\lim_{z\to 0}\cfrac{-1}{\sum\limits_{n=1}^\infty\frac{z^n}{n!}}$$ fails to exist, since the denominator vanishes. (Note also the negative sign that you omitted.) However, for $z\ne 0,$ we can write $$zf(z)=\cfrac{-1}{\sum\limits_{n=1}^\infty\frac{z^{n-1}}{n!}}=\cfrac{-1}{1+\sum\limits_{m=1}^\infty\frac{z^m}{(m+1)!}},$$ whence $\lim\limits_{z\to 0}zf(z)=-1.$ Hence, $f$ has a pole of order $1$ at $z=0$. Can you take it from there?