Answer
a) $\lim\limits_{x \to 0} h(x)=4$, Removable discontinuity
b) $\lim\limits_{x \to -1} h(x) =5$
Work Step by Step
Looking at the graph given in the problem, there is a hole at x=0 because there is an x in the denominator. It becomes a removable discontinuity when the function is simplified.
When x=-1 it evaluates to 5 with no problems.
