Answer
See explanations.
Work Step by Step
(a) Step 1. Identify the given piece-wise function: $h(x)=\begin{cases}x^2,\hspace0.8cm x\lt2 \\ 3,\hspace1cm x=2\\2,\hspace1cm x\gt2 \end{cases}$
Step 2. To show that $\lim_{x\to2}h(x)\ne4$, we need to find a value $\epsilon\gt0$ so that for any given value $\delta\gt0$, we can find a $x$ value in the interval of $|x-2|\lt\delta$ to give $|h(x)-4|\geq\epsilon$.
Step 3. In the interval of $2\lt x\lt2+\delta$, we have $h(x)=2$, and we can choose $\epsilon=2$ or less which will give $|h(x)-4|\geq\epsilon$ and this means that $\lim_{x\to2}h(x)\ne4$
(b) Repeat the above process: in the interval of $2\lt x\lt2+\delta$, we have $h(x)=2$, and we can choose $\epsilon=1$ or less which will give $|h(x)-3|\geq\epsilon$ and this means that $\lim_{x\to2}h(x)\ne3$
(c) For this case, we need to use the other interval of $2-\delta\lt x\lt2$, we have $h(x)=x^2$, and $|x^2-2|\geq|(2-\delta)^2-2|=|2-2\delta+\delta^2|$. As we are evaluating a limit with $x\to2$, the value of $\delta$ will be small (say $\delta\leq1$) and we have $|x^2-2|\geq|2-2+1|=1$. Thus by setting $\epsilon=1$ or less, we have $|x^2-2|\geq\epsilon$ and this means that $\lim_{x\to2}h(x)\ne2$
