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