Answer
$\lim_{x\to5^+}\sqrt {x-5}=0$
Work Step by Step
Step 1. Let $f(x)=\sqrt {x-5}$, for $5\lt x\lt 5+\delta$ and $\sqrt {x-5}\lt\epsilon$, we have $x-5\lt\epsilon^2$ or $x\lt \epsilon^2+5$
Step 2. Comparing the requirements for the upper boundary of $x$, we can set $5+\delta=\epsilon^2+5$ which gives $\delta=\epsilon^2$
Step 3. By setting the $\delta$ value and working backwards, we see that $x$ lies in the interval $I$ and $\sqrt {x-5}\lt\epsilon$
Step 4. We can identify that this procedure is verifying the limt $\lim_{x\to5^+}f(x)=0$ or the right -hand limit, which is zero.
