## Big O notation for e^(-1/x²)

The following statement confuses me:

$e^{-\frac 1{x^2}} \text { is in } \text{O}(x^n) \text{ for all }n \, .$

In order to try to verify this, I used the Taylor expansion for $e^x$ about $x = 0$ yielding

$e^{-\frac 1{x^2}} = 1 + \left( -\frac 1{x^2} \right) + \frac {\left( -\frac 1{x^2} \right)^2}{2!} + \frac {\left( -\frac 1{x^2} \right)^3}{3!} + \, \ldots \, ,$

which can be simplified to

$e^{-\frac 1{x^2}} = 1 - x^{-2} + \frac 1{2!}x^{-4} - \frac 1{3!}x^{-6} + \, \ldots$

How can we now claim that this function is in $\text{O}(x^n)$ for all $n$? Correct me if I am wrong but being in $\text{O}(x^n)$ when $x \rightarrow 0$ means that the function approaches $0$ at least as fast as $x^n$ (faster sure, but no less than $x^n$!). Let's suppose $n=1$; the function then apparently is in $\text{O}(x)$ i.e. the function approaches $0$ at least as fast as x. Does this imply that the terms with negative exponents (who have a greater absolute value than $1$) reach $0$ faster than $x$? How is that possible when we know that negative exponents mean division by $x$ and if now $x \rightarrow 0$ then the quotient must approach infinity?

I am guessing that terms who have negative exponents approach zero faster than $x$ just as $x^2$ approach zero faster than $x$. Is this analogy mathematically valid? I am not sure at all since $x^{-2} \rightarrow \pm \infty$ when $x \rightarrow 0$ whereas $x^2 \rightarrow 0$ when $x \rightarrow 0$.