limit of derivatives of a function I wanna show that for $f:(0,\infty)\rightarrow\mathbb R, x\mapsto\exp(-\frac1{x^2})$ the sum of the derivates of $f$, so $\sum\limits_{n=0}^\infty f^{(n)}(x)$, converges to $0$, so $$\lim\limits_{x\rightarrow0}\sum\limits_{n=0}^\infty f^{(n)}(x)=0$$
It's intuitively clear but I have some issues on writting it down. Can anybody help? Thanks!
 A: You can prove the following:
$$\eqalign{
  & y = \exp \left( { - \frac{1}{{{x^2}}}} \right)  \cr 
  & \log y =  - \frac{1}{{{x^2}}} \cr} $$
So that
$$y' = \frac{2}{{{x^3}}}\exp \left( { - \frac{1}{{{x^2}}}} \right)$$
Similarily:
$$y'' = \left( {\frac{2}{{{x^3}}} - \frac{3}{x}} \right)\left( {\frac{2}{{{x^3}}}\exp \left( { - \frac{1}{{{x^2}}}} \right)} \right)$$
Then, you can prove that, for any $k>0$
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^k}}}\exp \left( { - \frac{1}{{{x^2}}}} \right) = 0$$
Then proving (maybe by induction) that the derivatives will be linear combinations of that expression will do the job - basically, let 
$$F(x) = \exp \left(-\frac 1 {x^2}\right) \sum_{k=1}^r\frac{a_k}{x^k}$$
and prove that any $F^{(n)}$ will be of the same type (product rule and induction). 
ADD: I didn't spot the $\infty$ in the series. The fact that
$$\lim_{x \to 0}\sum_{k=0}^n f^{(k)}(x)=0 $$
for finite $n$ does not mean that 
$$\lim_{x \to 0}\sum_{k=0}^\infty f^{(k)}(x)=0$$
This is related to the notion of uniform convergence of series, which explains why the limit can't be evaluated termwise. As Robert pointed out, the limit might be interpreted as $$\int\limits_0^\infty  {\exp \left( { - x - \frac{1}{{{x^2}}}} \right)dx} $$
which evaluates to $\approx 0.293$
One obtains the above by some Taylor series manipulation,
Expand the function as a Taylor series 
$$f\left( {x + t} \right) = \sum\limits_{n = 0}^\infty  {{f^{\left( n \right)}}\left( x \right)} \frac{{{t^n}}}{{n!}}$$
Multiply by $e^{-t}$ and integrate over $(0,\infty)$:
$$\eqalign{
  & \int\limits_0^\infty  {{e^{ - t}}f\left( {x + t} \right)dt}  = \sum\limits_{n = 0}^\infty  {{f^{\left( n \right)}}\left( x \right)\frac{1}{{n!}}\int\limits_0^\infty  {{t^n}{e^{ - t}}dt} }   \cr 
  & \int\limits_0^\infty  {{e^{ - t}}f\left( {x + t} \right)dt}  = \sum\limits_{n = 0}^\infty  {{f^{\left( n \right)}}\left( x \right)\frac{1}{{n!}}n!}  = \sum\limits_{n = 0}^\infty  {{f^{\left( n \right)}}\left( x \right)}  \cr} $$
The integral function $$\int\limits_0^\infty  {\exp \left( { - t - \frac{1}{{{{\left( {x + t} \right)}^2}}}} \right)dt} $$
converges for any $x$ and is continuous so it might be reasonable to expect that 
$$\mathop {\lim }\limits_{x \to 0} \sum\limits_{n = 0}^\infty  {{f^{\left( n \right)}}\left( x \right)}  = \mathop {\lim }\limits_{x \to 0} \int\limits_0^\infty  {\exp \left( { - t - \frac{1}{{{{\left( {x + t} \right)}^2}}}} \right)dt}  = \int\limits_0^\infty  {\exp \left( { - t - \frac{1}{{{t^2}}}} \right)dt}  \approx 0.293$$
A: The series 
$$
\sum_{n=0}^\infty f^{(n)}(x)
$$
does not converge, at least for $x$ close to zero. Indeed, one shows by induction that
$$
f^{(n)}(x)=P_n(1/x)f(x) \quad \forall \ x>0, \ n \in \mathbb{N},
$$
where $\{P_n\}_{n \ge 0}$ is a sequence of polynomials such that $\deg P_n=3n$ and 
$$
P_0(t)=1, \quad P_n(t)=2t^3 P_{n-1}(t)+P_{n-1}'(t) \quad \forall \ t \in \mathbb{R}, \ n \in \mathbb{N}.
$$
For every $x>0$ and $n \in \mathbb{N}$ we have
$$
q_n(x)=\frac{P_{n+1}(1/x)}{P_n(1/x)}=\frac{2}{x^3}+\frac{P_n'(1/x)}{P_n(1/x)}.
$$
Thus
$$
\lim_{x \to 0+}q_n(x)=\infty.
$$
