Answer
The interval $I=(4-\delta,4)$ with $\delta=\epsilon^2$ would get $\sqrt{4-x}\lt\epsilon$
The limit being referred to here is $$\lim_{x\to4^-}\sqrt{4-x}=0$$
Work Step by Step
Given $\epsilon\gt0$, first, we look at the inequality: $$\sqrt{4-x}\lt\epsilon$$
- Square both sides: $$4-x\lt\epsilon^2$$ $$-x\lt\epsilon^2-4$$ $$x\gt4-\epsilon^2$$
Therefore, if $x\gt4-\epsilon^2$, then $\sqrt{4-x}\lt\epsilon$
Take $\delta=\epsilon^2$, we would have $$x\gt4-\delta\Rightarrow\sqrt{4-x}\lt\epsilon$$
Combining with the interval $I=(4-\delta,4)$ with $\delta=\epsilon^2$: $$4-\delta\lt x\lt4\Rightarrow\sqrt{4-x}\lt\epsilon$$
According to the defintion of one-side limit, the limit being referred to here is $$\lim_{x\to4^-}\sqrt{4-x}=0$$
