Using your graphing calculator and your knowledge of functions and algebra, find all values x = a where the function is discontinuous. For each such x value, evaluate the limit as x approaches a. (5 pts each) a.6 ( ) 6 x f x x + = + b.2 3 ( ) 9 x f x x −

Let's find the points of discontinuity and evaluate the limits for each given function.

a. $f(x) = \frac{x+6}{x+6}$

1. Step 1: Simplify the function
   Notice that the numerator and denominator are the same, so we can simplify the expression, provided that $x + 6 \neq 0$.
   $f(x) = \frac{x+6}{x+6} = 1 \quad \text{for} \quad x \neq -6$

2. Step 2: Identify the discontinuity
   The function is undefined when the denominator is zero, i.e., when $x + 6 = 0$. This gives: $x = -6$ So, the function is discontinuous at $x = -6$.

3. Step 3: Evaluate the limit as $x \to -6$
   Since the simplified form of the function is $f(x) = 1$ for all $x \neq -6$, the limit exists as $x \to -6$ and is: $\lim_{x \to -6} f(x) = 1$

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b. $f(x) = \frac{x-2}{x^2 - 9}$

1. Step 1: Factor the denominator
   The denominator is a difference of squares, so we can factor it: $f(x) = \frac{x-2}{(x-3)(x+3)}$

2. Step 2: Identify the discontinuities
   The function is undefined when the denominator is zero, i.e., when $(x-3)(x+3) = 0$. This gives: $x = 3 \quad \text{or} \quad x = -3$ So, the function is discontinuous at $x = 3$ and $x = -3$.

3. Step 3: Evaluate the limit as $x \to 3$
   As $x \to 3$, the denominator approaches zero, and the function behaves as: $\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{x-2}{(x-3)(x+3)}$ Here, the numerator is $x-2$, which equals $1$ when $x = 3$, while the denominator tends to zero. Therefore, the function has a vertical asymptote at $x = 3$, and the limit does not exist (it goes to $\pm \infty$).

4. Step 4: Evaluate the limit as $x \to -3$
   As $x \to -3$, the denominator again approaches zero, but the numerator approaches $-5$ (since $x-2$ becomes $-3-2 = -5$). Thus, we have: $\lim_{x \to -3} f(x) = \lim_{x \to -3} \frac{-5}{(-3-3)(-3+3)} = \frac{-5}{0} = \infty$ So, the limit does not exist as $x \to -3$ due to a vertical asymptote.

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Summary:

1. For part (a): The function is discontinuous at $x = -6$, and $\lim_{x \to -6} f(x) = 1$.
2. For part (b): The function is discontinuous at $x = 3$ and $x = -3$, and the limits at both points do not exist (due to vertical asymptotes).

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Would you like more details or have any questions?

Here are 5 relative questions to expand on this topic:

1. What types of discontinuities exist (removable, jump, infinite)?
2. How do vertical asymptotes affect the behavior of limits?
3. How do you check for removable discontinuities?
4. How can factoring help in identifying points of discontinuity?
5. How would you graph these functions to visualize the discontinuities?

Tip: Always simplify functions first before determining discontinuities—it helps to clearly see undefined points and asymptotic behavior.