Answer
a. See the table, $2.000$
b. See the graph, $2.0$
c. $2$
Work Step by Step
a. From the table, $\lim\limits_{x \to 3}\frac{x^2-2x-3}{x^2-4x+3}\approx 2.000$
b. From the graph, $\lim\limits_{x \to 3}\frac{x^2-2x-3}{x^2-4x+3}\approx 2.0$
c. $\lim\limits_{x \to 3}\frac{x^2-2x-3}{x^2-4x+3}=\lim\limits_{x \to 3}\frac{(x-3)(x+1)}{(x-3)(x-1)}=\lim\limits_{x \to 3}\frac{x+1}{x-1}=2$
