Real analysis question $e^{-1/x^2}$ Let $f$ be defined on $\mathbb{R}$ by $f(x) = e^{-1/x^2}$ for $x$ not equal to $0$. and $f(0)= 0$. Prove that $f^{(n)}(0)=0$ for all $n = 1, 2,3$ ...
Do I need to use Taylor expansion from calculus class?
Any hint would be appreciated. 
 A: First note that for $x \neq 0$, we have for $n>0$, $$f^{(n)}(x) = \dfrac{P_{2n-2}(x)\exp(-1/x^2) }{x^{3n}}$$ where $P_k(x)$ is a polynomial in $x$ with degree $k$. The proof for this follows immediately from induction.
The main ingredient to prove that $f^{(n)}(0) = 0$ is the following limit
$$\lim_{x \to \infty} x^m \exp(-x^2) = 0$$
First let us prove that $f^{(1)}(x) = 0$. We have that
$$f^{(1)}(x) = \lim_{h \to 0} \dfrac{\exp(-1/h^2) - 0}h = \lim_{x \to \infty} x \exp(-x^2) = 0$$
Assume $f^{(n)}(0) = 0$ for $n=k$. Then we have that $$f^{(k+1)}(0) = \lim_{h \to 0} \dfrac{f^{k}(h) - f^{(k)}(0)}{h} = \lim_{h \to 0} \dfrac{P_{2n-2}(h)\exp(-1/h^2)}{h^{3k+1}}$$
Now note that
$$\lim_{h \to 0} \dfrac{P_{2n-2}(h)\exp(-1/h^2)}{h^{3k+1}} = P_{2n-2}(0) \times \left(\lim_{x \to \infty} x^{3k+1} \exp(-x^2) \right) = 0$$
Hence, by induction we have that $f^{(n)}(0) = 0$.
A: You can’t use a Taylor expansion: this function isn’t represented by one centred at $0$, precisely because $f^{(n)}(0)=0$ for $n\in\Bbb Z^+$. Use the definition of the derivative:
$$f\,'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}h=\lim_{h\to 0}\frac{e^{-1/h^2}}h\;.$$
For $f^{(n)}(0)$ with $n>1$ you’ll need to know something the $n$-th derivative of $e^{-1/x^2}$. Don’t try to find a general formula for it; instead, just show by induction that the $n$-th derivative of $e^{-1/x^2}$ is of the form $p_n(x^{-1})e^{-1/x^2}$, where $p_n$ is a polynomial.
A: Use induction.  Check that it is true for $n=1$ (this is easy).  Assume it holds for $n$, and then compute limits for $f^{(n+1)}$ using the definition of the derivative.  
