Answer
A. $\lim\limits_{x \to 3}$ f(x) = $\frac{1}{6}$
B. The domain of the function is (-$\infty$, -3) U (-3,3) 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 (-∞,-3) U (-3,∞). Therefore, you must analyze the function to know that f(3) does not exist, so 3 cannot be included in the domain as well. Therefore, it is important to examine functions graphically as well as analytically.
Work Step by Step
A. To find $\lim\limits_{x \to 3}$ f(x), you must cancel out a factor that is making this limit turn out as an error.
1. Write out the function: $\frac{x-3}{x^2-9}$
2. Factor the function completely. Use the difference of squares rule to factor the trinomial in the denominator: $\frac{x-3}{(x+3)(x-3)}$
3. Cancel out the factors that can be cancelled: $\frac{1}{(x+3)}$
4. Now plug in 3 into the function: $\frac{1}{((3)+3)}$ = $\frac{1}{6}$.
B. To find the domain of the function, graph it and exclude the critical points: the vertical asymptote at x = -3, and the undefined value at x = 3 where the limit equals $\frac{1}{6}$. The answer will be (-∞,-3) U (-3,3) U (3,∞).
