Ytukyg

I have a singularity in $z=-2$, now I wolud like to find the kind of singularity, so I have to compute the limit:

$$lim_{zto-2}frac{z+2}{e^{frac{1}{(z+2)^2}}}$$

for me this limit is $0$, because I expand in Taylor the funcion $e^{frac{1}{(z+2)^2}}$ and then I obtain the null result.

But I'm not sure I can expand in Taylor the function because the Taylor expantion is everywhere valid except in $z+2=0$ and I also know that if $f(x)$ has a non removable singularity in $z_0$ I can tell that $e^{f(x)}$ has an essential singularity in $z_0$, now I have, in the denominator of the fraction, esactly this situation, maybe this can affect my result for the entire function, like "if I have somewhere an essential singularity in a part of the function, the function has an essential singularity"? Or maybe the presence of $z+2$ in the numerator kills the essential singularity? Or maybe knowing this can help me, the function $frac{1}{e^{frac{1}{(z+2)^2}}}$ has an essential singularity in $z=-2$?

Thank you all and sorry for bad english

If $z$ is real, then $exp(-1/(z+2)^2)to 0$ faster than any polynomial, so there is no Taylor polynomial.

If $z$ is complex, then it has no limit as you might approach along $z=-2+iy$.

Let $a_n$ be a complex number such that $a_n^{2}=frac 1 {2npi i}$. Through the sequence ${a_n-2}$ the function approaches $0$ but through the sequence $-2+frac i n$ it approaches $infty$ (in absolute value). This leaves with only one possibility: it has an essential singularity at $-1$.