Answer
$(\sqrt{2.9}, \sqrt{3.1})$
$\delta=\sqrt{3.1}-\sqrt{3}$
Work Step by Step
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 $\delta=\sqrt{3.1}-\sqrt{3}$,
$$0 \lt |x-\sqrt{3}| \lt \delta \implies |f(x)-3| \lt \epsilon.$$
