#### Answer

$\delta=1.44$, (or any smaller positive number).

#### Work Step by Step

Since
$1.6^{2}=2.56,\quad (\sqrt{2.56}=1.6\quad f(2.56)=1.6)$ and
$2.4^{2}=6.76,\quad (\sqrt{6.76}=2.4\quad f(6.76)=2.4)$
the unknown interval bounds around x=4 are 2.56 and 6.76.
Now, from the graph we see that for $x\in(2.56, 6.76)$ or
$2.56< x < 6.76$
$-1.44< x-4 < 2.76$
it follows that
$1.6 < \sqrt{x} < 2.4$
$-0.4< \sqrt{x} -2 < 0.4$
$|\sqrt{x}-2|<0.4.$
We want a positive number $\delta$ such that
$-1.44\leq\ \ -\delta < x-4 < \delta\ \ < 2.76, $
So, if we observe the stricter vicinity of $\pm 1.44 $
around x=4, it follows
$-1.44\leq\ \ -1.44 < x-4 < 1.44\ \ < 2.76, $
$|x-4|<1.44$
ensures that x is still in $(2.56, 6.76)$
which also ensures that $|\sqrt{x}-2|<0.4.$
We take $\delta=1.44$, (or any smaller positive number).