Answer
See explanations.
Work Step by Step
(a) The left side of the equation be be rewritten as $\frac{x^2+x-6}{x-2}=\frac{(x+3)(x-2)}{x-2}$, we can see that there is a "hole" at $x=2$ and the domain should be $(-\infty,2)\cup(2,\infty)$. On the right side, the domain would be $(-\infty,\infty)$. The left and right are not the same, and we can not simply cancel the common factor on the left side.
(b) The limit equation is correct because it evaluate the limit when $x$ approaches $2$, but not equal to $2$, We have: $\lim_{x\to 2}\frac{x^2+x-6}{x-2}=\lim_{x\to 2}\frac{(x+3)(x-2)}{x-2}=\lim_{x\to 2}(x+3)=2+3=5$
