Answer
a. See explanations.
b. Yes, $1$
Work Step by Step
a. By examining the graph, we see a discontinuity at $x=-1$. For a given $\delta\gt0$, $|x+1|\lt\delta$ we get $-1-\delta\lt x\lt\delta-1$. In the interval of $-1-\delta\lt x\lt -1$, the function $g(x)\lt1$ which gives $|g(x)-2|\geq1$, thus we can set $\epsilon=1$ or less so that $|g(x)-2|\geq\epsilon$ for any $\delta\gt0$ which means that $\lim_{x\to3}g(x)\ne2$
b. As $x\to -1, x\ne -1$, the actually function value at $x=-1$ will not affect the limit at this point. We can see from the graph that the limit exists and $\lim_{x\to -1}g(x)=1$
