Answer
A. $\lim\limits_{x \to 3}$ f(x) = 0.5
B. The domain of f(x) is (-$\infty$,1) U (1,3) U (3,$\infty$).
If you analyze the graph, you may not see the hole (undefined y coordinate) at x = 3, you only see the limit, so you might think that the domain is (-$\infty$,1) U (1,$\infty$). Therefore, you must analyze the function to know that f(3) does not exist, so 3 cannot be included in the domain. Therefore, it is important to examine functions graphically as well as analytically.
Work Step by Step
A. To find $\lim\limits_{x \to 3}$ of f(x), you must cancel out a factor that is making this limit turn out as an error.
1. Write out the function: $\frac{x-3}{x^2-4x+3}$
2. Factor the function completely: $\frac{x-3}{(x-3)(x-1)}$
3. Cancel out the factors that can be cancelled: $\frac{1}{x-1}$
4. Now plug in 3 into the function: $\frac{1}{(3)-1}$ = $\frac{1}{2}$ or 0.5.
B. To find the domain of the function, graph it and exclude the critical points: the vertical asymptote at x = 1, and the undefined value at x = 3 where the limit equals 0.5. The answer will be (-$\infty$,1) U (1,3) U (3,$\infty$).
