## Thomas' Calculus 13th Edition

$\delta=\sqrt{5}-2$
We see from the graph that in order for $f(x)$ to be within $\epsilon=1$ of $L=4$, we must have $$\sqrt{3} \lt x \lt \sqrt{5}.$$ Subtracting $c=2$ from all three sides gives $$\sqrt{3}-2 \lt x-2 \lt \sqrt{5}-2.$$ Now, $\sqrt{3}-2 \approx -0.268$ and $\sqrt{5}-2 \approx 0.236$. This means $|\sqrt{5}-2| \lt |\sqrt{3}-2|$. Thus, $$-(\sqrt{5}-2) \lt x-2 \lt \sqrt{5}-2 \implies \sqrt{3}-2 \lt x-2 \lt \sqrt{5}-2.$$ Hence for $\delta = \sqrt{5}-2$, $$0 \lt |x-2| \lt \delta \implies 0 \lt |f(x)-4| \lt \epsilon.$$