Answer
$\left(\dfrac{4}{25}, \dfrac{9}{25}\right)$
$\delta=\dfrac{9}{100}$
Work Step by Step
We are given $f(x)=\sqrt{x}$, $L=\dfrac{1}{2}$, $c=\dfrac{1}{4}$, and $\epsilon=0.1=\dfrac{1}{10}$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|\sqrt{x}-\dfrac{1}{2} \right| \lt \dfrac{1}{10}\\
-\dfrac{1}{10}\lt \sqrt{x}-\dfrac{1}{2} \lt \dfrac{1}{10}\\
\dfrac{4}{10}\lt \sqrt{x} \lt \dfrac{6}{10}\\
\dfrac{2}{5}\lt \sqrt{x} \lt \dfrac{3}{5}\\
\dfrac{4}{25}\lt x \lt \dfrac{9}{25}.$$
Thus our interval is $\left(\dfrac{4}{25}, \dfrac{9}{25}\right)$.
Now, to find our $\delta$, we note that
$$\dfrac{4}{25} \lt x \lt \dfrac{9}{25} \implies \dfrac{4}{25}-\dfrac{1}{4} \lt x-\dfrac{1}{4} \lt \dfrac{9}{25}-\dfrac{1}{4}\\
\implies -\dfrac{9}{100} \lt x-\dfrac{1}{4} \lt \dfrac{11}{100}.$$
Hence for $\delta=\dfrac{9}{100}$,
$$0 \lt \left|x-\dfrac{1}{4}\right| \lt \delta \implies \left|f(x)-\dfrac{1}{2}\right| \lt \epsilon.$$
