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Let's evaluate the given limits step by step.

### Problem (b)
\[
\lim_{y \to 2} \frac{|2y - 4|}{4y - 8 - y^3 + 2y^2}
\]

1. **Simplify the expression:**

   Start by simplifying the absolute value in the numerator and the expression in the denominator:
   \[
   |2y - 4| = |2(y - 2)|
   \]
   Since we are taking the limit as \(y \to 2\), the absolute value can be simplified as follows:
   \[
   |2(y - 2)| = 2|y - 2|
   \]

   Now let's simplify the denominator:
   \[
   4y - 8 - y^3 + 2y^2 = 4(y - 2) - y^3 + 2y^2
   \]
   Factor \(y - 2\) from the terms:
   \[
   4(y - 2) - y^3 + 2y^2 = (y - 2)(4 - (y^2 + 2y))
   \]
   The expression inside the parenthesis simplifies as:
   \[
   4 - (y^2 + 2y) = 4 - y(y + 2)
   \]

2. **Evaluate the limit:**

   Since \(y \to 2\), substitute \(y = 2\) into the simplified expression:
   \[
   \lim_{y \to 2} \frac{2|y - 2|}{(y - 2)(4 - y(y + 2))}
   \]

   Since \(y - 2\) is a common factor in the numerator and denominator, cancel it out:
   \[
   \lim_{y \to 2} \frac{2}{4 - y(y + 2)}
   \]

   Substitute \(y = 2\) into the remaining expression:
   \[
   \frac{2}{4 - 2(2 + 2)} = \frac{2}{4 - 8} = \frac{2}{-4} = -\frac{1}{2}
   \]

   **Final Answer for (b):**
   \[
   \lim_{y \to 2} \frac{|2y - 4|}{4y - 8 - y^3 + 2y^2} = -\frac{1}{2}
   \]

### Problem (c)
\[
\lim_{z \to -\infty} \frac{\sqrt{9z^2 - 1}}{z + 3}
\]

1. **Simplify the expression:**

   First, notice that as \(z \to -\infty\), the term \(z + 3\) behaves like \(z\), so:
   \[
   \frac{\sqrt{9z^2 - 1}}{z + 3} \approx \frac{\sqrt{9z^2}}{z} = \frac{3|z|}{z}
   \]
   Since \(z \to -\infty\), \(|z| = -z\), so:
   \[
   \frac{3|z|}{z} = \frac{3(-z)}{z} = -3
   \]

2. **Evaluate the limit:**

   The expression tends towards \(-3\) as \(z \to -\infty\).

   **Final Answer for (c):**
   \[
   \lim_{z \to -\infty} \frac{\sqrt{9z^2 - 1}}{z + 3} = -3
   \]

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Would you like further details or have any questions about these solutions?

Here are some related questions that might help you explore further:
1. How do you evaluate limits when the function involves absolute values?
2. How can the behavior of functions at infinity influence the result of a limit?
3. What is the importance of simplifying the expression before evaluating a limit?
4. How do you handle indeterminate forms when calculating limits?
5. What techniques can be used to evaluate limits involving square roots?

**Tip:** When evaluating limits at infinity, always consider the highest degree terms as they dominate the behavior of the function.

Learn how to evaluate limits involving absolute values and square roots with detailed step-by-step solutions. This includes examples of limits approaching specific values and behavior at infinity. Suitable for college-level students and those preparing for calculus courses.

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