Determine the type of singularity of $z_0$ of $f(z)=\frac{\sin(z)}{e^{-z}+z-1},\:z_0=0$.
I tried to compute $\lim_{z\to 0}\frac{\sin(z)}{e^{-z}+z-1}$ directly but I get the indeterminate form $\frac{0}{0}$. However in the solution it is stated $f(z)\sim\frac{2}{z}$ and $\lim_{n\to\infty}\frac{2}{z}=0$
I do not understand the sort of solution that was presented to me via approximation.
Questions:
1) How is the approximation done? What is the procedure?
2) If I wanted to compute the limit, since I cannot use L'Hopital once $z$ is complex. How should I compute it then?
Thanks in advance!
Hint: $$\sin z = z - \frac{z^2}{3!} + \cdots$$ $$e^{-z} = 1 - z + \frac{z^2}2 + \cdots$$ $$e^{-z} + z - 1 = \cdots$$ $$ \frac{\sin z}{e^{-z} + z - 1} = \frac{z(1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots)}{z(\frac{z}{2} - \frac{z^2}{3!} + \frac{z^3}{4!} - \frac{z^4}{5!} + \cdots)} = \frac{2}z\,\frac{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots}{1 - \frac{2z}{3!} + \frac{2z^2}{4!} - \frac{2z^3}{5!} + \cdots} $$ And $$ 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots,\qquad 1 - \frac{2z}{3!} + \frac{2z^2}{4!} - \frac{2z^3}{5!} + \cdots $$ are analytic functions (convergent power series) with $$ \lim_{z\to 0}\, 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots = \cdots $$ $$ \lim_{z\to 0}\, 1 - \frac{2z}{3!} + \frac{2z^2}{4!} - \frac{2z^3}{5!} + \cdots = \cdots $$