Answer
$ \delta=0.1$ (or any smaller positive number)
Work Step by Step
When $x\in(0.7,1.1)$, then
$0.7 < x< 1.1$
$-0.4 < x-1 < 0.1$
and, we see from the graph that for these x,
$f(x)\in(0.8,1.2)$, that is,
$0.8 < f(x) < 1.2$
$-0.2 < f(x)-1 < 0.2$
$|f(x)-1| < 0.2\qquad(*)$
understanding that if x is within 0.4 units from the left of 1, or
within 0.1 units to the right of 1, the condition (*) is satisfied.
We will choose a $\delta$ within the interval $[-0.4,0.1]$ such that
$-0.4\leq-\delta < x-1 < \delta \leq 0.1$
Since $\delta$ will describe a vicinity of equal distance from 1 from either side, we choose the stricter (closer to 1) vicinity.
We take $\delta=0.1$ (or any smaller positive number).
It then follows
$-0.4< -0.1 < x-1 < 0.1\ \Rightarrow\ |x-1| <0.1$
which ensures that $x\in(0.7,1.1)$, which ensures $|f(x)-1| < 0.2$
$ \delta=0.1$ (or any smaller positive number)
