$\displaystyle f(x)=\frac{e^{\frac{1}{z-1}}}{e^z-1}$

if we go $\displaystyle z=0^+$ f(z)=+infinity

if we go $\displaystyle z=0^-$ f(z)=-infinity

we dont have a limit here

so why its a pole

??

by my definition a pole is when $\displaystyle f(z)=\frac{g(z)}{(a-z)^n}$

if g is analitical at g(a) then 'a' in th n'th order pole

but in my case there is no (a-z)^n representation

and i proved that there is no limit

in my exam i have much complicated functions

and i cant develop it into a series every time

i tried to solve this one strictly by the limit tests

where am i wrong in my limit test

where is my conclusion wrong

??