#### Answer

$\delta=\dfrac{7}{16}$

#### Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=\dfrac{1}{4}$ of $L=1$, we must have
$$\dfrac{9}{16} \lt x \lt \dfrac{25}{16}.$$
Subtracting $c=1$ from all three sides gives
$$-\dfrac{7}{16} \lt x-1 \lt \dfrac{9}{16}.$$
Note that $$-\dfrac{7}{16} \lt x-1 \lt \dfrac{7}{16} \implies -\dfrac{7}{16} \lt x-1 \lt \dfrac{9}{16}.$$
Hence for $\delta = \dfrac{7}{16}$,
$$0 \lt |x-1| \lt \delta \implies 0 \lt |f(x)-1| \lt \epsilon.$$