Answer
$$\lim_{x\to3}h(x)=2$$
Work Step by Step
$$h(x)=\frac{x^2-2x-3}{x^2-4x+3}$$
(a) The table is shown below.
As we can see from 2 tables below, as $x$ gets more decimals and approaches $3$, the value of $h(x)$ also gets more decimals and apparently approaches $2$. So I would estimate here that $\lim_{x\to3}h(x)=2$.
(b) The graph is shown below.
Again, looking at the graph, the closer $x$ approaches $3$, the closer $h(x)$ approaches $2$.
(c) $$\lim_{x\to3}h(x)=\lim_{x\to3}\frac{x^2-2x-3}{x^2-4x+3}=\lim_{x\to3}\frac{(x+1)(x-3)}{(x-1)(x-3)}$$ $$\lim_{x\to3}h(x)=\lim_{x\to3}\frac{x+1}{x-1}=\frac{3+1}{3-1}=\frac{4}{2}=2$$
Therefore, $$\lim_{x\to3}h(x)=2$$
