Lagrange Remainder the taylor expansion of $e^{-1/x^2}$

Question

I know $e^{-1/x^2}$ and $f(x) = 0$ if $x = 0$, is an example of a smooth function that cannot be represented all the coefficients are $0$, but I am curious about the Lagrange remainder. Doesn't the remainder still approach $0$ because $n!$ grows faster than any derivative of $e^{-1/x^2}$. So in that case, since the remainder goes to $0$ as n approaches infinity, that would imply this function is represented by the Taylor expansion right? Am I misunderstanding the remainder theorem?

Answer 1

Since $(\forall n\in\Bbb Z_+):f^{(n)}(0)=0$, every Taylor polynomial is the null polynomial, then the remainder at any point $x$ is $f(x)$ itself (for every $n\in\Bbb Z_+$).