Let be given in $\mathbb{C}\setminus\{0\}$ the functions : $\displaystyle{\frac{1}{\sin(z)^{2}}; \frac{\tan(z)}{z^{2}}, \frac{1}{e^{\frac{1}{z^{2}}}}, \frac{z^{3}}{1-\cos(z)}} $ and the book says to evaluate the type of singularity at $z_{0}=0$ without the usage of Laurent series.
My solutions are as follows:
$1.$ since $\lim \limits_{z \rightarrow z_{0}} |f(z)| \rightarrow \infty$ this must be a pole. I guess it is order 2 , but how to proove that?
$2.$ since $\lim\limits_{z\rightarrow z_{0}} |\frac{\tan(z)}{z^{2}}| \rightarrow \infty$ so this must be a pole, again i dont know what the order is.. but I guess it is 2 .
$3.$ since $\lim\limits_{z\rightarrow z_{0}} \left|\frac{1}{e^{1/z^2}}\right| \rightarrow 0$ this exists in $\mathbb{R}$ so this must be a removable singularity.
$4.$ using de lhopital 4 times gives : $\lim |\frac{6}{-\sin(z)}| \rightarrow \infty$ so this is also a polish singularity.
Again I don't know how to determine the order of the polish ones.
Please tell me.