Solution
= Solution
The <Laurent series> is
$$
e^{-1/z}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}z^{-n}=1-\frac1z+\frac1{2z^2}-\frac1{6z^3}+\cdots,
\qquad z\ne0.
$$
Its <principal part> contains infinitely many nonzero negative powers, so \b[zero is an essential singularity]. The first three displayed terms are not an asymptotic truncation as $z\to0$; they are the beginning of the convergent <Laurent series> in powers of $1/z$.