The function $f(x) = \frac{(x-9)(x+5)(x-2)}{5(x-9)(x+2)}$ has potential points where the limit does not exist due to the terms in the denominator. We will evaluate the expression and simplify it to determine where the function is undefined or discontinuous.

The function simplifies to: $f(x) = \frac{(x+5)(x-2)}{5(x+2)}$ after canceling $(x-9)$ from the numerator and the denominator.

The limit $\lim_{x \to a} f(x)$ does not exist at points where:

1. The function is undefined.
2. There is a discontinuity.

The function is undefined at:

• $x = -2$ (from the denominator)

The function has a removable discontinuity at:

• $x = 9$ (originally in both the numerator and denominator but canceled out).

From the options provided:

• A. 2 - This value is defined and continuous, so the limit exists.
• B. 9 - The limit does not exist due to a removable discontinuity; the function is not defined here in the original function.
• C. 5 - This value is defined and continuous, so the limit exists.
• D. -5 - This value is defined and continuous, so the limit exists.
• E. -2 - The limit does not exist as the function is undefined here.

Therefore, the correct answers, where the limit does not exist, are B (9) and E (-2).

Would you like to explore more about the properties of this function or need further explanation on any part? Here are some related questions you might find interesting:

1. What is the behavior of $f(x)$ as $x$ approaches infinity?
2. How would you find the asymptotes of $f(x)$?
3. What are the critical points of $f(x)$ and how do they influence the graph of the function?
4. Can you explore the local maxima and minima of $f(x)$?
5. How does the graph of $f(x)$ look around $x = -2$ and $x = 9$?

Tip: Always check for points where functions are undefined or have discontinuities when working with rational functions, as they are critical for understanding the overall behavior of the function.