Finding the $n$-th derivative at $0$ rigorously (Analysis) Let $\displaystyle f(x) =
\begin{cases}
e^{-1/x} , & x > 0 \\
0, &x \leq 0
\end{cases}$
Find (with justification) $f^{(n)} (0)$ for all $n \in \mathbb{N}$
(I am currently in the Differentiation chapter of Baby Rudin).
By L'Hôpital's Rule, it follows that $\displaystyle \lim_{x\to 0^+} \frac{e^{-1/x}}{x^n} = 0$ for all $n \in \mathbb{N}$. However, I am unsure on how to conclude that $f^{(n)} (0)=0$ for all $n \in \mathbb{N}$.

Show by induction $H(n): \forall x ≤ 0, f^{(n)}(x) = 0$ and $\forall x > 0, f^{(n)}(x) = P_n(\frac{1}{x})e^{-1/x}$ with $P_n$ a polynomial.
In the induction step, once you have proved $H(n)$ for $x>0$ and $x < 0$ 
You will have to show $f^{n+1}(x) = 0$: Write the definition of the slope at $0$ and look at the limits in $0^+$ and in $0^-$ applying these two results and limit comparison between $\exp$ and polynomial for $x >0$
