Q:

$x$ is smaller than all positive numbers

We have $0\le x \lt h$ for all $h \gt 0$. Prove that $x=0$. I know it is obvious but I can't prove it.
My try : I think it has some connections with concept of limit but I can't prove it.

A:

Approach 1: 
Suppose that $x\ne 0$ then take $h=\frac{x}{2}$ and then
$$h<x$$
what is a contradiction.
Approach 2:
Another approach is use the sequence $h_n=\frac{1}{n}$, $0_n=0$ and use Squeeze Theorem:
$$0_n\le x<h_n$$
because $h_n\to 0$ and $0_n\to0$.

