## University Calculus: Early Transcendentals (3rd Edition)

We take $\delta=1$ here.
$$a=1 \hspace{1cm} b=7\hspace{1cm}c=2$$ Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-2|\lt\delta$ then $1\lt x\lt 7$ The sketch is shown below. The distance from $c=2$ to the nearer endpoint of $(1,7)$ is $1$. So if we take $\delta=1$ or any smaller positive value, then $0\lt |x-2|\lt1$, and $x$ will only range in $(1,3)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $1$ and $7$ as requested. In other words, $$0\lt |x-2|\lt1\hspace{2cm}1\lt x\lt 7$$