Singularity of $\frac{z+2}{e^{\frac{1}{(z+2)^2}}}$

Question

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

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

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

Answer 1

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 $z=-2+iy$.

Answer 2

Let $a_n$ be a complex number such that $a_n^{2}=\frac 1 {2n\pi i}$. 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 $-1$.