Answer
The interval $I=(5,5+\delta)$ with $\delta=\epsilon^2$ would get $\sqrt{x-5}\lt\epsilon$
The limit being referred to here is $$\lim_{x\to5^+}\sqrt{x-5}=0$$
Work Step by Step
Given $\epsilon\gt0$, first, we look at the inequality: $$\sqrt{x-5}\lt\epsilon$$
- Square both sides: $$x-5\lt\epsilon^2$$ $$x\lt5+\epsilon^2$$
Therefore, if $x\lt5+\epsilon^2$, then $\sqrt{x-5}\lt\epsilon$
Take $\delta=\epsilon^2$, we would have $$x\lt5+\delta\Rightarrow\sqrt{x-5}\lt\epsilon$$
Combining with the interval $I=(5,5+\delta)$ with $\delta=\epsilon^2$: $$5\lt x\lt5+\delta\Rightarrow\sqrt{x-5}\lt\epsilon$$
According to the defintion of one-side limit, the limit being referred to here is $$\lim_{x\to5^+}\sqrt{x-5}=0$$
