Calculus 8th Edition

$\delta=0.1$ (or any smaller positive number)
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 $(0, 0.1)$