Answer
(a)
(i) $3$
(ii) $0$
(iii) does not exist
(iv) $0$
(v) $0$
(vi) $0$
(b) $f$ is discontinuous at $x=0$ and $x=3$
(c) We can see a sketch of $f(x)$ below.
Work Step by Step
(a)
(i) $\lim\limits_{x \to 0^+}f(x) = \lim\limits_{x \to 0^+} (3-x) = 3$
(ii) $\lim\limits_{x \to 0^-}f(x) = \lim\limits_{x \to 0^+} \sqrt{-x} = 0$
(iii) $\lim\limits_{x \to 0}f(x)$ does not exist since the left limit and the right limit are not equal at this point.
(iv) $\lim\limits_{x \to 3^-}f(x) = \lim\limits_{x \to 3^-} (3-x) = 0$
(v) $\lim\limits_{x \to 3^+}f(x) = \lim\limits_{x \to 3^+} (x-3)^2 = 0$
(vi) $\lim\limits_{x \to 3}f(x) = 0~~$ since both the left limit and the right limit equal $0$ at this point.
(b) $f$ is discontinuous at $x=0$ and $x=3$
Note that $f(3)$ does not exist.
(c) We can see a sketch of $f(x)$ below.
