Answer
$\delta=0.4$
(or any smaller positive number)
Work Step by Step
To reword the problem, given $\epsilon=0.5,$
we need to find a small interval around $3$,
($3-\delta,3+\delta)$
so that the function values of all x's from that interval
fall within $(2-\epsilon,2+\epsilon).$
From the graph we read the following:
When $x\in(2.6, 3.8)=(3-0.4,3+0.8)$,
then $f(x)\in(1.5, 2.5)$, that is
$|f(x)-2| < \epsilon=0.5$
The interval $(2.6, 3.8)$ around x=3, contains the interval
$(2.6, 3.4) = (3-0.4,3+0.4)$, for which
$x\in(3-0.4,3+0.4)\ \ \Rightarrow\ \ $
$0 < |x-3| < 0.4\ \ \Rightarrow\ \ |f(x)-2| < 0.5$.
So, for the given $\epsilon=0.5$, we have found $\delta=0.4 > 0$ such that
if $ 0 < |x-3| < \delta$, then $|f(x)-2| < \epsilon$.
Note: we can take $\delta$ to be any number from $(0, 0.4)$
