How is this asymptotic relation?

[pivoxa15]

Homework Statement

e^((ln(x))^2)=O(1)?
as x->0

Homework Equations

g(x)=O(f(x)) => |g(x)|<K|f(x)| as x->0 in this case but can approach any value as desired.
K<infinity

The Attempt at a Solution

We can try numbers for small x but non zero that satisfy the inequality however are we allowed to manipulate infinities? Somehow I think that the equality shouldn't stand.

I know that x^2 does not equal O(x) when x->infinity but does equal when x->0

We can look at it this way, g(x)=O(f(x)) => |g(x)|<K|f(x)| as x->0, K<infinity => lim x->0 |g(x)/f(x)| is finite.

 

[Dick]

The limit is infinity as x->0. How can this be O(1)?