#### Answer

$\delta=0.1$
(or any smaller positive number)

#### Work Step by Step

To reword the problem,
we need to find a small interval around 1,
($1-\delta,1+\delta)$
so that the function values of all x's from that interval
fall within $(1-\epsilon,1+\epsilon).$
From the graph we read the following:
When $x\in(-0.7, 1.1)$, then $f(x)\in(0.8, 1.2)$.
Write can write this as
$x\in(1-0.3, 1+0.1)$, then $f(x)\in(1-0.2, 1+0.2)$
or given the $\epsilon=0.2,$
$x\in(1-0.3, 1+0.1)$, then $f(x)\in(1-\epsilon, 1+\epsilon)$, that is,
if $x\in(1-0.3, 1+0.1)$, then $|f(x)-1| < \epsilon.$
For $(1-0.1, 1+0.1)$, the subset of $(1-0.3, 1+0.1)$,
from $x\in(1-0.1, 1+0.1)$ it also follows that $|f(x)-1| < \epsilon, $
So, for the given $\epsilon=0.2$, we have found $\delta=0.1 > 0$ such that
if $ |x-1| < \delta$, then $|f(x)-1| < \epsilon$.
Note: we can take $\delta$ to be any number from $(0, 0.1)$