TITLE: Maclaurin series expansion for $e^{-1/x^2}$
QUESTION [13 upvotes]: I am currently extremely confused on how to proceed with the Maclaurin series expansion for my current function.
I got my derivatives and I got my formula, however, plugging them in gives me a non-possible answers since division by $0$ is not possible.

REPLY [4 votes]: you can depend on the common form of $e^x$ and put $x=-1/x^2$ as follow
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$$
$$e^{-1/x^2}=1-\frac{1}{x^2}+\frac{1}{2!x^4}-\frac{1}{3!x^6}+\frac{1}{4!x^8}+...$$