## Calculus: Early Transcendentals 8th Edition

(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.
(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.