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.