Showing that the $n$th derivative of this piecewise function vanishes at $0$

Question

Let $$f(x)= \begin{cases} e^{-1/x} & x>0 \\ 0 & x\le 0 \end{cases} $$ I want to show that $f^{(n)}(0)=0$ for every $n\in\mathbb{N}$.

My idea is to do this by induction on $n$. I can compute the first derivative by taking the limit of $$\frac{f(x)-f(0)}{x-0}=\frac{e^{-1/x}}{x}$$ as $x\rightarrow 0^+$. I'm not sure how to show this is $0$, which is why the induction step is unclear to me. If the statement is true for some $n$, then I can compute $f^{(n+1)}(0)$ by taking the limit $$\frac{f^{(n)}(x)-f^{(n)}(0)}{x-0}=\frac{f^{(n)}(x)}{x}$$ as $x\rightarrow 0^+$ and show that this is also $0$. Can anyone help?

Answer 1

$\lim_{x \to 0+} \frac{e^{-1/x}}{x} = \lim_{x \to 0+} \frac{1/x}{e^{1/x}}$ so we just apply L'Hopitals rule to get $0$ as the answer.

Now use induction to prove that $f^{(n)}(x)/x = \frac{e^{-1/x}}{P_n(x)}$ where $P_n(x)$ is a polynomial of degree $n$.

Answer 2

$$A=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}\frac{e^{-1/x}}{x}$$ for this let $\frac{1}{x}=u$ then $$A=\lim_{x\to 0^+}\frac{e^{-1/x}}{x}=ue^{-u}=\lim_{u\to +\infty}\frac{u}{e^u}=0$$ Note that $e^x=\sum \frac{x^n}{n!}$ thus $e^{-1/x}=\sum \frac{{(\frac{-1}{x})}^n}{n!}$ so $$\frac{e^{-1/x}}{x}=\sum \frac{(-1)^n}{x^{n+1}n!}$$