#### Answer

$(-0.19,0.21)$
$\delta=0.19$

#### Work Step by Step

We are given $f(x)=\sqrt{x+1}$, $L=1$, $c=0$, and $\epsilon=0.1$.
To find the desired interval about $c$, we plug the above into $|f(x)-L| \lt \epsilon$ and solve for $x$:
$$\left|\sqrt{x+1}-1 \right| \lt 0.1\\
-0.1\lt \sqrt{x+1}-1 \lt 0.1 \\
0.9\lt \sqrt{x+1} \lt1.1 \\
0.81\lt x+1 \lt 1.21\\
-0.19 \lt x \lt 0.21.$$
Thus our interval is $(-0.19,0.21)$.
Now, to find our $\delta$, we note that
$$-0.19 \lt x-0 \lt 0.19 \implies -0.19 \lt x-0 \lt 0.21.$$
Hence for $\delta=0.19$,
$$0 \lt |x-0| \lt \delta \implies |f(x)-1| \lt \epsilon.$$