As $x \to 0$ this approaches zero since the zero of $e^{\frac{-1}{x^2}}$ beats out the pole of $\frac{1}{Q_k(x)}$? In the Robert Strichartz's book "A Guide to Distribution Theory and Fourier Transforms" at the page $4$, we have an interesting exercise : 

$$ \psi(x) =
  \begin{cases}
    e^{\frac{-1}{x^2}}      &amp; x&gt;0\\
    0 &amp; x \leq 0\\
  \end{cases}
$$

has continuous derivatives of all orders $$\frac{d^k (e^{\frac{-1}{x^2}})}{dx} = \frac{P_k(x)}{Q_k(x)} e^{\frac{-1}{x^2}},$$ whre $P_k(x)$ and $Q_k(x)$ are two polynomials such that deg($P_k(x)$) $&lt;$ deg ($Q_k(x)$). Is there anyone could explain (at least a hint) to me why as $x \to 0$ this approaches zero since the zero of $e^{\frac{-1}{x^2}}$ beats out the pole of $\frac{1}{Q_k(x)}$ and what is its utility in the problem??
Let us first prove (by induction) a more explicit formula for the derivatives of $f = \textrm e ^{- \frac 1 {x^2}}$ ($x&gt;0$): if $k \ge 1$, then $f^{(k)} = \dfrac {P_k (x)} {x^{3k}} \ \textrm e ^{- \frac 1 {x^2}}$ with $\deg P_k \le 2k-2$ and $P_k (0) = 2^k$.

Checking it for $k=1$ is easy. Next, assume the formula valid for $k$ and obtain it for $k+1$ (use Leibniz's formula):

$$f^{(k+1)} = \left( \frac {P_k (x)} {x^{3k}} \ \textrm e ^{- \frac 1 {x^2}} \right)' = \frac {P_k ' (x) x^3 - 3k x^2 P_k (x) + 2 P_k (x)} {x^{3(k+1)}} \ \textrm e ^{- \frac 1 {x^2}} .$$

Notice that the three terms in the numerator have degrees $2k$, again $2k$ and finally $2k-2$, therefore the degree of the numerator $P_{k+1}$ is at most the largest of them, i.e. $\deg P_{k+1} \le 2k$. Also, a quick look at the formula of the numerator shows that $P_{k+1} (0) = 2 P_k (0) = 2^{k+1}$.

It is clear then that when computing $\lim \limits _{x \to 0^+} f^{(k)} (x)$ we may safely ignore the numerator (which will only contribute a $2^k$) and focus on $\lim \limits _{x \to 0^+} \dfrac {\textrm e ^{- \frac 1 {x^2}}} {x^{3k}}$, which upon making the change of variable $t = \frac 1 {x^2}$ becomes $\lim \limits _{t \to \infty} \dfrac {t^{\frac 3 2 k}} {\textrm e ^t} = 0$ (the exponential is stronger than any power). This shows that $\lim \limits _{x \to 0^+} f^{(k)} (x) = 0$, from which it follows easily that all the derivatives of $\psi$ in $0$ exist and are continuous. Since $\psi$ obviously has continuous derivatives of any order on $\Bbb R \setminus \{0\}$, the above shows that $\psi$ has continuous derivatives of any order on the whole $\Bbb R$ (i.e. $\psi \in C^\infty (\Bbb R)$) and $\psi ^{(k)} (0) = 0$.

Now, it $\psi$ analytic? It is clearly so on $(-\infty, 0)$ (being constant) and on $(0, \infty)$ (being a composition of the analytic functions $\exp$ and $-\frac 1 {x^2}$). Nevertheless, it cannot be analytic on any neighbourhood of $0$ because if it were, then it would be equal to its own Taylor series in $0$, which is $0$ (because $\psi ^{(k)} (0) = 0 \ \forall k \ge 0$), so we would have $\psi = 0$ around $0$ - clearly not true.

To conclude, what this (classic) example constructs is a smooth function that is not analytic - clearly not a trivial thing.
As
$x \to 0$,
$\dfrac1{x^2} \to +\infty$
so
$e^{\frac1{x^2}} \to \infty$
so
$e^{-\frac1{x^2}}
=\dfrac1{e^{\frac1{x^2}}} 
\to 0$
and does so very strongly.