## Thomas' Calculus 13th Edition

$\delta=\dfrac{7}{16}$
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 $\delta = \dfrac{7}{16}$, $$0 \lt |x-1| \lt \delta \implies 0 \lt |f(x)-1| \lt \epsilon.$$