Answer
(a) (i) $0$ (ii) $0$ (iii) $1$ (iv) $4$ (v) $6$ (vi) Does not exist.
(b) See graph
Work Step by Step
(a) Based on the piece-wise function, we have:
(i) For $x\to 0^+$, we should use the middle function, $\lim_{x\to 0^+}h(x)=\lim_{x\to 0^+}x^2=0^2=0$
(ii) For $x\to 0$, we need to test both left and right limits using the first and second equations respectively, $\lim_{x\to 0^-}h(x)=\lim_{x\to 0^-}x=0$, $\lim_{x\to 0^+}h(x)=\lim_{x\to 0^+}x^2=0$, thus $\lim_{x\to 0}h(x)=0$
(iii) For $x\to 1$, we should use the middle function, $\lim_{x\to 1}h(x)=\lim_{x\to 1}x^2=1$
(iv) For $x\to 2^-$, we should use the middle function, $\lim_{x\to 2^-}h(x)=\lim_{x\to 2^-}x^2=2^2=4$
(v) For $x\to 2^+$, we should use the third function, $\lim_{x\to 2^+}h(x)=\lim_{x\to 2^+}(8-x)=8-2=6$
(vi) For $x\to 2$, we need to test both left and right limits, since $\lim_{x\to 2^+}h(x)\ne \lim_{x\to 2^-}h(x)$ as shown above. $\lim_{x\to 2}h(x)$$ does not exist.
(b) See graph.
