A doubt in the rigorous definition of limits. I studied the definition of limits today, and I think I mostly understood it, but I have a little doubt. In the definition:
$f(x)$ is defined on some open interval containing $a$, except at possibly $a$. So, $\lim_{x\to a} f(x) = L $ if and only if for every number $\varepsilon>0$, there exists a corresponding number $\delta>0$ such that: 
$$\text{If   } 0<|x-a|<\delta \text{ then } |f(x)-L|<\varepsilon$$
First of all, if you see any mistake in my definition or understanding, please tell. So my question is, why can't we write $0<|x-a|\leq\delta$ and $|f(x)-L|\leq\varepsilon$ in the last line? What kinds of problems can arise from this? If the definitions are equivalent, can you explain/prove me why?

First assume:
 $\varepsilon>0$, there exists a corresponding number $\delta>0$ such that: 
$$\text{If   } 0<|x-a|<\delta \text{ then } |f(x)-L|<\varepsilon$$
For that definition take $\varepsilon=\frac{\epsilon}{2}$, then exist $\delta$ such:
$$\text{If   } 0<|x-a|<\delta \text{ then } |f(x)-L|<\frac{\epsilon}{2}$$
Then you can see:
$$0<|x-a|\leq\frac{\delta}{2}\Rightarrow 0<|x-a|<\delta\Rightarrow |f(x)-L|<\frac{\epsilon}{2}\leq{\epsilon}$$
Then take $\delta'=\frac{\delta}{2}$ and you have your definition with less or equal.
The other side is analogous
