Not to be nitpicky: it is not the Taylor polynomial (it does not approach any continuous function). It is a result by weierstrass on polynomial approximation on

Not to be nitpicky: it is not the Taylor polynomial (it does not approach any continuous function). It is a result by weierstrass on polynomial approximation on a closed interval. f(x)=exp(-x^2) has the same Taylor expansion as g(x)=0 at x=0.