I want to show that the function
$$ g = \begin{cases} 0, \quad \text{if } x \le 0, \\ e^{-\frac{1}{x^2}}, \quad \text{if } x > 0, \end{cases} $$ is infinitely differentiable at $0$ and that all derivative vanishes: $g^{(n)}(0) = 0$.
I was having trouble expanding the function into its Taylor expansion. Any help is appeciated!
$$g'_+(0):=\lim_{x\to0^+}\frac{g(x)-g(0)}x=\lim_{x\to0^+}\frac{e^{-1/x^2}}x=0$$
since exponential beats polynomial (or you can do L'Hospital or whatever). Since clearly $\;g'(0)_-=0\;$ , the function is differentiable at zero. Now,
$$g'(x)=\begin{cases}0,&x\le0\\{}\\ \frac2{x^3}e^{-1/x^2},&x>0\end{cases}$$
Once again, and for the same reasons as in the case above, the second, third, etc. derivatives at zero exist, and they all are zero.
Deduce that a Maclaurin series is not going to work nice here...