Letting $\epsilon = \frac{\epsilon}{2}$ I know this is minor, but how is it that you justify this formally?
$$
\begin{equation}
\begin{split}
| x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\
&\Rightarrow |f(x) - l| < \frac{\epsilon}{2}
\end{split}
\end{equation}
$$ 
To better illustrate what I mean, consider as a counter-example how easy it is to justify something like this:
$$
\begin{equation}
\begin{split}
| x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\
&\Rightarrow|f(x) - l| < 3\epsilon
\end{split}
\end{equation}
$$ 
 A: 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 $|f(x) - L| < \dfrac{\epsilon}{2}$.
A: Of course for a fixed $\epsilon, \delta$, the following is false:
$$
\begin{equation}
\begin{split}
| x - a | < \delta &\Rightarrow |f(x) - l| < \epsilon \\
&\Rightarrow |f(x) - l| < \frac{\epsilon}{2}
\end{split}
\end{equation}
$$
But these are equivalent:
$$
\forall \epsilon > 0 \exists \delta > 0\big[| x - a | < \delta \Rightarrow |f(x) - l| < \epsilon\big]
$$
and
$$
\forall \epsilon > 0\; \exists \delta > 0\big[| x - a | < \delta \Rightarrow |f(x) - l| < \frac{\epsilon}{2}\big]
$$
A: The statement as you've written it is false, but there is a way to work this.
I assume this is in the context of analysis, working with the ever-present definition of a limit: $\forall \epsilon > 0, \exists \delta > 0, |x-a|<\delta \implies |f(x) - L| < \epsilon$. The thing to note here is that the $\delta$ that exists depends on the originally chosen $\epsilon$, so we can imagine it as a function $\delta(\epsilon)$ picking a suitable $\delta$ for each input $\epsilon$.
For every $\epsilon$ we have $|x-a|<\delta(\epsilon) \implies |f(x) - L| < \epsilon$.
Therefore for every epsilon we have $|x-a|<\delta(\frac{\epsilon}{2}) \implies |f(x) - L| < \frac{\epsilon}{2}$
