I'm assuming you're talking about this in the context of limits.
Let's suppose you know for a fact that $$\lim_{x \to a}f(x) = L\text{.} \tag{*}$$
By definition, this means that for every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x-a| < \delta$, then $|f(x) - L| < \epsilon$.
The for every part is key.
If we know that (*) holds (this assumption is important), we could say that for all $\epsilon > 0$, there is some $\delta > 0$ (let's assume we don't care about the actual value of $\delta$) such that if $|x - a| < \delta$, $|f(x) - L| < \dfrac{\epsilon}{2}$.
Why? Because since we're assuming $\epsilon$ is an arbitrary positive number, $\dfrac{\epsilon}{2}$ is an arbitrary positive number as well, no matter what $\epsilon > 0$ is!
The long story short is that you can stick anything in that inequality in place of $\epsilon$ as long as the limit exists (i.e,. (*) holds) and as long as what you have there is an arbitrary positive number!
To answer your question directly (and this has been addressed in the comments), the following is FALSE:
$$|x-a| < \delta \implies |f(x) - L| < \epsilon \nRightarrow |f(x) - L| < \dfrac{\epsilon}{2}\text{.}$$
What you can say is the following: suppose $\lim_{x \to a}f(x) = L$. (Notice how I have this assumption available to me.) Then, for every $\epsilon > 0$, there is a $\delta > 0$ such that if $|x - a| < \delta$, then $|f(x) - L| < \dfrac{\epsilon}{2}$.