Answer
0
Work Step by Step
1. f is a closed function.
We know by Th.10.1 that it is continuous, that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. evaluating: $f($2$)$, (plugging $x=$2) , we see that $x=$2 is NOT in the domain of f.
The form is indeterminate, 0/0 so we try simplifying.
Since f is defined for $x>2$, (x-2) is positive on that domain.
Writing (x-2)=($\sqrt{x-2})^{2}$ we can reduce...
$\displaystyle \lim_{x\rightarrow 2+} \displaystyle \frac{x-2}{\sqrt{x-2}}=\lim_{x\rightarrow 2+}\sqrt{x-2}=$... plug 2...= 0
