Answer
1) The open interval is $(\sqrt{15},\sqrt{17})$
2) $\delta=\sqrt{17}-4$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-4|\lt\delta\Rightarrow|(x^2-5)-11|\lt1$$
1) Find the interval around $4$ on which $|(x^2-5)-11|\lt1$ holds.
Solve the inequality: $$|(x^2-5)-11|\lt1$$ $$-1\lt(x^2-5)-11\lt1$$ $$10\lt x^2-5\lt12$$ $$15\lt x^2\lt17$$
Here we examine the interval around $4$, so we automatically take $x$ to be greater than $0$.
Therefore, $$\sqrt{15}\lt x\lt\sqrt{17}$$
The open interval around $4$ is $(\sqrt{15},\sqrt{17})$.
2) Give a value for $\delta$
The nearer endpoint to $4$ is $\sqrt{17}$, and the distance between them is $\sqrt{17}-4$.
So if we take $\delta=\sqrt{17}-4$ or any smaller positive number, then $0\lt|x-4|\lt\sqrt{17}-4$, meaning all $x$ would be placed in the interval $(\sqrt{15},\sqrt{17})$ so that $|(x^2-5)-11|\lt1$.
In other words, $$0\lt |x-4|\lt\sqrt{17}-4\Rightarrow|(x^2-5)-11|\lt1$$
