Big O notation for e^(-1/x²)

The following statement confuses me:

In order to try to verify this, I used the Taylor expansion for about yielding

which can be simplified to

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

I am guessing that terms who have negative exponents approach zero faster than just as approach zero faster than . Is this analogy mathematically valid? I am not sure at all since when whereas when .