Answer
$a.\quad 0$
$b.\quad f$ is not defined when $x\lt 2$,
so the left-sided limit does not exist.
Work Step by Step
$a.\quad $
Approaching $x=2$ from the right,
$f(x)=\sqrt{x-2} $ is defined,
and, by the the fractional power rule (Th.2.3.7),
$\displaystyle \lim_{x\rightarrow 2^{+}}\sqrt{x-2}=\sqrt{\lim_{x\rightarrow 2^{+}}(x-2)}=\sqrt{2-2}=0$
$b.\quad $
Approaching $x=2$ from the left,
$f(x)=\sqrt{x-2} $is NOT defined,
because $x-2$ is negative for $x\lt 2.$
Thus, $\displaystyle \lim_{x\rightarrow 2^{-}}\sqrt{x-2}$ does not exist.
