Answer
$$\lim_{x\to-2}F(x)=-1$$
Work Step by Step
$$F(x)=\frac{x^2+3x+2}{2-|x|}$$
(a) The table is shown below.
As we can see from 2 tables below, as $x$ gets more decimals and approaches $-2$, the value of $F(x)$ also gets more decimals and apparently approaches $-1$. So I would estimate here that $\lim_{x\to-2}F(x)=-1$.
(b) The graph is shown below.
Again, looking at the graph, the closer $x$ approaches $-2$, the closer $F(x)$ gets to $-1$.
(c) $$\lim_{x\to-2}F(x)=\lim_{x\to-2}\frac{x^2+3x+2}{2-|x|}$$
For $x\ge0$, $|x|=x$ and for $x\lt0$, $|x|=-x$
However, here we need to calculate the limit of $F(x)$ as $x\to-2$, and we care only the values of $x$ very close around $-2$, which is below $0$.
So we can say here that we choose $x\lt0$, and $|x|=-x$
$$\lim_{x\to-2}F(x)=\lim_{x\to-2}\frac{x^2+3x+2}{2-(-x)}=\lim_{x\to-2}\frac{(x+1)(x+2)}{2+x}$$
$$\lim_{x\to-2}F(x)=\lim_{x\to-2}(x+1)=-2+1=-1$$
Therefore, $$\lim_{x\to-2}F(x)=-1$$
