#### Answer

Sample answer
$\delta=0.1$
(or any smaller positive number)

#### Work Step by Step

First, we address the question marks on the x-axis.
They stand for the numbers whose function values are $0.5$ and $1.5$
$x^{2}=0.5 \ \ \Rightarrow\ \ x=\sqrt{0.5}\approx$0.707106781187
$\sqrt{x}=2.4 \ \ \Rightarrow\ \ x=\sqrt{1.5}\approx$1.22474487139
So, the interval (to which x=$1$ belongs) is
$I=(\sqrt{0.5}, \sqrt{1.5})$
.
From the graph, $x\in I \displaystyle \ \ \Rightarrow\ \ |f(x)-1| < \frac{1}{5}$.
(all values of $x\in I$ have f(x) within $\displaystyle \pm\frac{1}{5} $ of $1$)
We take a subset from $I$,
one that we can write as$ (1-\delta,1+\delta)$,
such as
$I_{1}=(1-0.1, 1+0.1)$
(the problem asks to find a $\delta$,
not the greatest possible $\delta$,
which would be $\sqrt{1.5}-1$,
as $\sqrt{1.5}$ is closer to 1 than $\sqrt{0.5}$)
For $x\displaystyle \in I_{1}\ \ \Rightarrow\ \ x\in I\ \ \Rightarrow\ \ |f(x)-1| < \frac{1}{2}$.
So for $\displaystyle \epsilon=\frac{1}{2}, $we have found a number,
$\delta=0.1$, such that
$ 0 < |x-1| < \delta$, then $|x^{2}-1| < \epsilon$.