# Proofs using the Taylor expansion, zero series and limits

If we have the function $f: (0,\infty) \rightarrow\ R$, $f(x) = e^{\frac{-1}{x^2}}$

Show that for $n≥0$, there exists a polynomial function $p_{(n)}(t)$ such that $f^{(n)} (x) =p_{(n)}(\frac{1}{x})f(x)$ for all $x>0$

Given any polynomial function $p(t)$, show that $\lim_{x\to0} p(\frac{1}{x})f(x)=0$

Show that f is infinitely often differentiable and that the Taylor series of $f$ at $0$ is the zero series $0+0x+0x^2+...$

I think I need to use induction to do this, I already know that $\lim_{x\to\infty} \frac{x^n}{e^{x^2}}=0$ for $n≥0$ Any help is appreciated.

• Welcome to MSE. :) Take a look at this post. It will help you to handle the first parts. :) – H. R. Dec 23 '15 at 16:58

Hint: if $$f^{(n)}(x) = p_n(1/x)e^{-1/x^2}$$ then $$f^{(n+1)}(x) =p_n'(1/x)(-1/x^2)e^{-1/x^2} + p_n(1/x)e^{-1/x^2}(2/x^3)$$ an the derivative of a polynomial is...