Answer
$$\lim_{x\to-6}G(x)=-0.125$$
Work Step by Step
$$G(x)=\frac{x+6}{x^2+4x-12}$$
(a) The table is shown below.
As we can see from 2 tables below, as $x$ gets more decimals and approaches $-6$, the value of $G(x)$ also gets more decimals and apparently approaches $-0.125$. So I would estimate here that $\lim_{x\to-6}G(x)=-0.125$.
(b) The graph is shown below.
Again, looking at the graph, the closer $x$ approaches $-6$, the closer $G(x)$ approaches $-0.125$. So my estimate again is $\lim_{x\to-6}G(x)=-0.125$.
(c) $$\lim_{x\to-6}G(x)=\lim_{x\to-6}\frac{x+6}{x^2+4x-12}=\lim_{x\to-6}\frac{x+6}{(x-2)(x+6)}$$ $$\lim_{x\to-6}G(x)=\lim_{x\to-6}\frac{1}{x-2}=\frac{1}{-6-2}=-\frac{1}{8}=-0.125$$
Therefore, $$\lim_{x\to-6}G(x)=-0.125$$
