Answer
$\left(\dfrac{10}{3}, 5\right)$
$\delta=\dfrac{2}{3}$
Work Step by Step
We are given $f(x)=\dfrac{1}{x}$, $L=\dfrac{1}{4}$, $c=4$, and $\epsilon=0.05=\dfrac{1}{20}$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|\dfrac{1}{x}-\dfrac{1}{4} \right| \lt \dfrac{1}{20}\\
-\dfrac{1}{20}\lt \dfrac{1}{x}-\dfrac{1}{4} \lt \dfrac{1}{20}\\
\dfrac{4}{20}\lt \dfrac{1}{x} \lt \dfrac{6}{20}\\
\dfrac{1}{5}\lt \dfrac{1}{x} \lt \dfrac{3}{10}\\
5 \gt x \gt \dfrac{10}{3}\\
\dfrac{10}{3} \lt x \lt 5.$$
Thus our interval is $\left(\dfrac{10}{3}, 5\right)$.
Now, to find our $\delta$, we note that
$$\dfrac{10}{3} \lt x \lt 5 \implies -\dfrac{2}{3} \lt x-4 \lt 1 .$$
and
$$-\dfrac{2}{3} \lt x-4 \lt \dfrac{2}{3} \implies -\dfrac{2}{3} \lt x-4 \lt 1.$$
Hence for $\delta=\dfrac{2}{3}$,
$$0 \lt |x-4| \lt \delta \implies \left|f(x)-\dfrac{1}{4}\right| \lt \epsilon.$$
