## Calculus: Early Transcendentals 8th Edition

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