## Thomas' Calculus 13th Edition

$(\sqrt{2.9}, \sqrt{3.1})$ $\delta=\sqrt{3.1}-\sqrt{3}$
We are given $f(x)=x^2$, $L=3$, $c=\sqrt{3}$, and $\epsilon=0.1$. To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$: $$\left|x^2-3 \right| \lt 0.1\\ -0.1 \lt x^2-3 \lt 0.1\\ 2.9\lt x^2 \lt 3.1\\ \sqrt{2.9}\lt x \lt \sqrt{3.1} .$$ Thus our interval is $(\sqrt{2.9}, \sqrt{3.1})$. Now, to find our $\delta$, we note that $$\sqrt{2.9} \lt x \lt \sqrt{3.1} \implies \sqrt{2.9}-\sqrt{3} \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}.$$ Notice $\sqrt{2.9-\sqrt{3}} \approx -0.0291$ and $\sqrt{3.1}-\sqrt{3} \approx 0.0286$. This means $|\sqrt{3.1}-\sqrt{3}| \lt |\sqrt{2.9}-\sqrt{3}|$. Thus, $$-(\sqrt{3.1}-\sqrt{3}) \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}\\ \implies \sqrt{2.9}-\sqrt{3} \lt x-\sqrt{3} \lt \sqrt{3.1}-\sqrt{3}.$$ Hence for $\delta=\sqrt{3.1}-\sqrt{3}$, $$0 \lt |x-\sqrt{3}| \lt \delta \implies |f(x)-3| \lt \epsilon.$$