## University Calculus: Early Transcendentals (3rd Edition)

The interval $I=(5,5+\delta)$ with $\delta=\epsilon^2$ would get $\sqrt{x-5}\lt\epsilon$ The limit being referred to here is $$\lim_{x\to5^+}\sqrt{x-5}=0$$
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 $$\lim_{x\to5^+}\sqrt{x-5}=0$$