## Calculus: Early Transcendentals 8th Edition

$\delta=0.4$ (or any smaller positive number)
When $x\in(2.6,3.8)$, then $2.6 < x< 3.8$ $-0.4 < x-3 < 0.8$ and, we see from the graph that for these x, $f(x)\in(1.5,2.5)$, that is, $1.5 < f(x) < 2.5$ $-0.5 < f(x)-2 < 0.5$ $|f(x)-2| < 0.5\qquad(*)$ understanding that if x is within $0.4$ units from the left of $3$, or within $0.8$ units to the right of $3$, the condition (*) is satisfied. We will choose a $\delta$ within the interval $[0.4,0.8]$ such that $-0.4\leq\ \ -\delta < x-1 < \delta \ \ \leq 0.8$ Since $\delta$ will describe a vicinity of equal distance from $3$ from either side, we choose the stricter (closer to $3$) vicinity. We take $\delta=0.4$ (or any smaller positive number). It then follows $-0.4 < x-3 < 0.4\ \Rightarrow\ |x-3| <0.4$ which ensures that $x\in(2.6,3.8)$, which ensures $|f(x)-2| < 0.5$