Add 1 in each and simplify

\(\displaystyle{\left|{f{{\left({x}\right)}}}-{1}\right|}{ < }{0.2}\)

\(\displaystyle-{0.2}{ < }{f{{\left({x}\right)}}}-{1}{ < }{0.2}\)

\(\displaystyle{0.8}{ < }{f{{\left({x}\right)}}}{ < }{1.2}\)

The graph shows that if \(\displaystyle{0.8}{ < }{f{{\left({x}\right)}}}{ < }{1.2}\) x must verify \(\displaystyle{0.7}{ < }{x}{ < }{1.1}\)

On the other hand we have x that must verify this unequality:

\(\displaystyle{\left|{x}-{1}\right|}{ < }\delta\)

\(\displaystyle-\delta{ < }{x}-{1}{ < }\delta\)

If we can find \(\displaystyle\delta\) that verify \(\displaystyle-{0.3}\leq-\delta{ < }{x}-{1}{ < }\delta\leq{0.1}\), we can make sure that \(\displaystyle{0.7}{ < }{x}{ < }{1.1}\)

so \(\displaystyle\delta\) must verify two conditions: \(\displaystyle\delta\leq{0.1}\) and \(\displaystyle-{0.3}\leq-\delta\)

which means that \(\displaystyle\delta\leq{0.1}\) and \(\displaystyle{0.3}\geq\delta\)

\(\displaystyle{0.7}{ < }{x}{ < }{1.1}\)

\(\displaystyle{0.7}-{1}{ < }{x}-{1}{ < }{1.1}-{1}\)

\(\displaystyle-{0.3}{ < }{x}-{1}{ < }{0.1}\)

We can choose \(\displaystyle\delta={0.1}\), so to be sure two conditions are satisfied, but if you want, you can choose any value.