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The image presents three problems where the limits of functions are to be calculated as \( x \) approaches 3. Let's solve each one by determining the left-hand limit (as \( x \) approaches 3 from the left) and the right-hand limit (as \( x \) approaches 3 from the right).

### Problem 24:
\[
f(x) = \frac{(x+2)(x-6)}{x-3}
\]
We need to find the left and right limits as \( x \to 3 \).

- **Left-hand limit ( \( \lim_{x \to 3^-} f(x) \) ):** When \( x \) approaches 3 from the left, the denominator \( x-3 \) becomes slightly negative. We will examine the behavior of the numerator and denominator:
    - Numerator: \( (x+2)(x-6) \), which remains continuous.
    - Denominator: \( x - 3 \) becomes negative (approaching zero from below).
    \[
    \lim_{x \to 3^-} f(x) = \frac{(5)(-3)}{-0^-} = -\infty
    \]
    So, \( \lim_{x \to 3^-} f(x) = -\infty \).

- **Right-hand limit ( \( \lim_{x \to 3^+} f(x) \) ):** When \( x \) approaches 3 from the right, \( x - 3 \) becomes slightly positive.
    \[
    \lim_{x \to 3^+} f(x) = \frac{(5)(-3)}{0^+} = \infty
    \]
    So, \( \lim_{x \to 3^+} f(x) = \infty \).

### Problem 25:
\[
f(x) = \frac{(x-3)(x+3)}{x(x-3)}
\]
Here, the function simplifies except at \( x = 3 \) where a potential discontinuity arises.

- **Left-hand limit ( \( \lim_{x \to 3^-} f(x) \) ):** Simplify the function:
    \[
    f(x) = \frac{(x+3)}{x}, \quad \text{for} \, x \neq 3
    \]
    As \( x \to 3^- \), we simply substitute into the simplified expression:
    \[
    \lim_{x \to 3^-} f(x) = \frac{6}{3} = 2
    \]
    So, \( \lim_{x \to 3^-} f(x) = 2 \).

- **Right-hand limit ( \( \lim_{x \to 3^+} f(x) \) ):** Similarly, the right-hand limit:
    \[
    \lim_{x \to 3^+} f(x) = \frac{6}{3} = 2
    \]
    So, \( \lim_{x \to 3^+} f(x) = 2 \).

### Problem 26:
\[
f(x) = \frac{-2}{(x-3)^2}
\]
This is a rational function where the denominator is squared, leading to symmetry around \( x = 3 \).

- **Left-hand limit ( \( \lim_{x \to 3^-} f(x) \) ):** As \( x \to 3^- \), \( (x-3)^2 \) is always positive, and the function becomes:
    \[
    \lim_{x \to 3^-} f(x) = \frac{-2}{0^+} = -\infty
    \]
    So, \( \lim_{x \to 3^-} f(x) = -\infty \).

- **Right-hand limit ( \( \lim_{x \to 3^+} f(x) \) ):** Similarly, on the right:
    \[
    \lim_{x \to 3^+} f(x) = \frac{-2}{0^+} = -\infty
    \]
    So, \( \lim_{x \to 3^+} f(x) = -\infty \).

### Summary of limits:

1. **Problem 24:**
   \[
   \lim_{x \to 3^-} f(x) = -\infty, \quad \lim_{x \to 3^+} f(x) = \infty
   \]
   
2. **Problem 25:**
   \[
   \lim_{x \to 3^-} f(x) = 2, \quad \lim_{x \to 3^+} f(x) = 2
   \]

3. **Problem 26:**
   \[
   \lim_{x \to 3^-} f(x) = -\infty, \quad \lim_{x \to 3^+} f(x) = -\infty
   \]

Let me know if you would like more details on any step!

### Five related questions:
1. What happens to a rational function when the denominator approaches zero?
2. How do left-hand and right-hand limits affect continuity at a point?
3. Can a function have different left and right limits at the same point?
4. How do you simplify rational functions with removable discontinuities?
5. What role do squares in the denominator play in limit behavior?

**Tip**: Always check for simplifications before calculating limits, especially when there are factors common to both the numerator and denominator!

For each of the following, write the left and right limit statements for f(x) as x approaches 3.

Learn how to solve left and right limit problems for rational functions. This guide covers limits as x approaches 3, focusing on asymptotic behavior, rational function properties, and continuity. Suitable for high school students in Grades 11-12 or early Calculus learners.

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