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We are given the function:

\[
f(x) = \frac{x^2 - 9}{3 - x}
\]

and are asked to find the limit as \( x \) approaches \( 3^{-} \) (from the left-hand side), and then select the appropriate value from the table.

### Step-by-Step Solution:

1. **Simplify the function**:
   We start by factoring the numerator:
   \[
   x^2 - 9 = (x - 3)(x + 3)
   \]
   So, the function becomes:
   \[
   f(x) = \frac{(x - 3)(x + 3)}{3 - x}
   \]

2. **Rewrite the denominator**:
   Notice that \( 3 - x \) can be written as:
   \[
   3 - x = -(x - 3)
   \]
   This gives us:
   \[
   f(x) = \frac{(x - 3)(x + 3)}{-(x - 3)}
   \]

3. **Cancel the common term**:
   We can cancel \( (x - 3) \) from the numerator and denominator (as long as \( x \neq 3 \)) to simplify the function:
   \[
   f(x) = -(x + 3)
   \]

4. **Evaluate the limit**:
   Now, we evaluate the limit of \( f(x) = -(x + 3) \) as \( x \to 3^{-} \):
   \[
   \lim_{x \to 3^{-}} -(x + 3) = -(3 + 3) = -6
   \]

### Conclusion:
The value of the function approaches \(-6\) as \(x \to 3^{-}\). Therefore, the correct option is:

\[
\boxed{c.} -6.01, -6.001, -6.0001, -6.00001, -6.000001
\]

Would you like more details or clarifications?

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Here are 5 related questions to deepen understanding:

1. What happens to the function as \( x \to 3^{+} \) (from the right)?
2. How do you find limits involving discontinuities in a function?
3. What are vertical asymptotes, and does this function have one?
4. How does factoring help simplify rational functions like this?
5. Can you think of another function with a similar limit behavior?

**Tip:** Always check for removable discontinuities by simplifying the function before finding limits!

Considere la función: f(x) = (x^2 - 9) / (3 - x). ¿Cuál de las siguientes opciones corresponde a los valores de la función cuando x → 3⁻?

Learn how to calculate the limit of the function f(x) = (x^2 - 9) / (3 - x) as x approaches 3 from the left. This problem involves factoring and simplifying rational functions, suitable for students in Grades 10-12.

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