Following question and answer are from Thomas calculus book:
Find a value of $\delta >0$ such that for all $0< |x-x_0|< \delta \implies a<x<b$ . we have $a=1, b=7, x_0=5$ .
Step 1: $|x-5|<\delta \implies -\delta< x-5< \delta \implies -\delta+5<x<\delta+5$.
Step 2: $ \delta+5=7 \implies \delta=2,$ or $-\delta+5=1 \implies \delta=4$ The value of $\delta$ which assures $|x-5|< \delta \implies 1<x<7$ is the smaller value, $ \delta=2$
My question:when we consider $x=1, |x-5|=|1-5|=4$, and it is not less than $\delta=2$, then how come we take $\delta$ to be equal to $2$? where am I going wrong?