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.