Nature of singularity and value of residue

Question

I want to come to know the type of singularity of the complex function $f(z)= (e^{(1/z)})/\sin z$ at $z=0$. For $e^{(1/z)}$ the limit at $0$ does not exist. And also $sin z$ approaches $0$ when $z$ tends to $0$. Could this singularity be a pole despite non-existence of limit of function in the numerator? If it is not a pole then could we still talk about its residue? If I consider this singularity a pole of order $1$and evaluate the limit $zf(z)$ as $z$ tends to $0$, I end up with the limit of$e^{(1/z)}$ as $z$ tends to $0$ which does not exist. Now I am stuck. I tried using Laurent series expansions but that seemed daunting to me. How do I calculate residue for this function in a relatively quick manner?

Answer 1

Hint: if $f(z)$ has a pole of order $n$ at $z=0$, and $g(z)$ has a zero of order $m$ at $z=0$, then what does $f(z) g(z)$ have at $z=0$? Could it be an essential singularity?

As to the other question you mentioned: yes, it is quite possible to talk about a residue for an essential singularity.