#### Answer

1) The open interval is $(-0.19,0.21)$
2) $\delta=0.19$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-0|\lt\delta\Rightarrow|\sqrt{x+1}-1|\lt0.1$$
1) Find the interval around $0$ on which $|\sqrt{x+1}-1|\lt0.1$ holds.
Solve the inequality: $$|\sqrt{x+1}-1|\lt0.1$$ $$-0.1\lt\sqrt{x+1}-1\lt0.1$$ $$0.9\lt\sqrt{x+1}\lt1.1$$
Square: $$0.81\lt x+1\lt1.21$$ $$-0.19\lt x\lt 0.21$$
The open interval around $0$ is $(-0.19,0.21)$.
2) Give a value for $\delta$
The nearer endpoint to $0$ is $-0.19$, and the distance between them is $0.19$.
So if we take $\delta=0.19$ or any smaller positive number, then $0\lt|x-0|\lt0.19$, meaning all $x$ would be placed in the interval $(-0.19,0.21)$ so that $|\sqrt{x+1}-1|\lt0.1$.
In other words, $$0\lt |x-0|\lt0.19\Rightarrow|\sqrt{x+1}-1|\lt0.1$$