Find the Taylor series of $f(x) = e ^{- 1 / x^2}$ Find the Taylor series about 0, the function defined as:
$f(x) = e ^{- 1 / x^2}$ if $x \ne 0$ and $f(x) = 0$ if $x=0$ and What can i conclude of the resulting?
First i note that the function f is even then i calculate the derivatives: 
$$f'(x) = e ^{- 1 / x^2} (2 / x^3)^2$$
$$f'(x) = e ^{- 1 / x^2} (2 / x^3)^2 + e ^{- 1 / x^2} (- 6/ x^4) $$
but when when i analyzed the derivative in $x=0$ :
$\lim\limits_{x\rightarrow 0} \frac{f(x) - f(0)}{x-0} = \lim\limits_{x\rightarrow 0} \frac{e ^{- 1 / x^2}}{x} $ and this is and indeterminate  $\frac{0}{0}$ then for l'hopital:
$\lim\limits_{x\rightarrow 0} \frac{(e {^{- 1 /x^{2}}})'}{x'}  = \lim\limits_{x\rightarrow 0} \frac{(e {^{- 1 /x^{2}}})( \frac{2}{x^3})}{1} $
but i stuck here how can i get the taylor series if i can't find the derivative in this point, please help me.
 A: Notice that derivatives of $f(x)=e^{-1/x^2}$ will always have the form:
\begin{equation}
f^{(n)}(x)=e^{-1/x^2}\frac{P_n(x)}{Q_n(x)}
\end{equation}
where $P_n(x)$ and $Q_n(x)$ are polynomials, i.e. $n$-th derivative of $f(x)$ is $f(x)$ times rational function. This can be trivially proven by induction.
Let
\begin{equation}
g_n(x)=\frac{P_n(1/x)}{Q_n(1/x)}
\end{equation}
$g_n(x)$ is also rational function and now $f^{(n)}(x)=e^{-1/x^2}g(1/x)$.
$n$-th derivative of $f(x)$ in zero is:
\begin{equation}
f^{(n)}(0)=\lim_{x\rightarrow0}e^{-1/x^2}g(1/x)=
\lim_{y\rightarrow\pm\infty}e^{-y^2}g(y)=
\left[ \lim_{y\rightarrow\pm\infty}\frac{e^{y^2}}{g(y)} \right]^{-1}=\infty^{-1}=0
\end{equation}
Infinity appears because exponential function grows faster than any rational function.
Therefore, function $f(x)=e^{-1/x^2}$ has a strange property:
\begin{equation}
\lim_{x\rightarrow0}f^{(n)}(x)=0 ~~~~ \forall n \in \mathbb{N}_0
\end{equation}
This property is less strange once one notices that this function has an essential singularity at $x=0$. Consequently, any Taylor expansion of this function at $x=a$ will have radius of convergence equal to $|a|$.
A: Hint: You can rewrite the last limit by substituting $t=\tfrac1x$: 
$\qquad\displaystyle\lim_{x\to0^+} e^{-1/x^2} \cdot \tfrac2{x^3} = \lim_{t\to+\infty} e^{-t^2}2t^3$.
