Answer
(a) $1$, $2$
(b)Does not exist.
(c) See graph.
Work Step by Step
(a) Based on the piece-wise function, we have:
$$\lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}(x-1)=2-1=1$$
$$\lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}(x^2-4x+6)=2^2-4\times2+6=2$$
(b) Because $$\lim_{x\to 2^-}f(x)\ne\lim_{x\to 2^+}f(x)$$ $$\lim_{x\to 2}f(x)$$ does not exist.
(c) See graph.
