Answer
We take $\delta=1/18$ here.
Work Step by Step
$$a=4/9 \hspace{1cm} b=4/7\hspace{1cm}c=1/2$$
Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-1/2|\lt\delta$, then $4/9\lt x\lt 4/7$.
The sketch is shown below.
- From $1/2$ to $4/9$: $1/2-4/9=1/18$
- From $1/2$ to $4/7$: $4/7-1/2=1/14$
So the nearer endpoint of $(4/9,4/7)$ to $c=1/2$ is $4/9$, and the distance is $1/18$.
So if we take $\delta=1/18$ or any smaller positive value, then $0\lt |x-1/2|\lt1/18$, and $x$ will only range in $(4/9,5/9)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $4/9$ and $4/7$ as requested.
In other words, $$0\lt |x-\frac{1}{2}|\lt\frac{1}{18}\hspace{2cm}\frac{4}{9}\lt x\lt \frac{4}{7}$$
