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We are tasked with evaluating the following limit:

\[
\lim_{y \to 0} \frac{5y^3 + 8y^2}{3y^4 - 16y^2}
\]

### Step 1: Factor both the numerator and denominator
The numerator is \( 5y^3 + 8y^2 \). We can factor out \( y^2 \):

\[
5y^3 + 8y^2 = y^2(5y + 8)
\]

The denominator is \( 3y^4 - 16y^2 \). Similarly, we can factor out \( y^2 \):

\[
3y^4 - 16y^2 = y^2(3y^2 - 16)
\]

### Step 2: Simplify the expression
Now substitute these factored forms back into the limit expression:

\[
\lim_{y \to 0} \frac{y^2(5y + 8)}{y^2(3y^2 - 16)}
\]

We can cancel \( y^2 \) from both the numerator and the denominator, provided \( y \neq 0 \):

\[
\lim_{y \to 0} \frac{5y + 8}{3y^2 - 16}
\]

### Step 3: Evaluate the limit
Now, substitute \( y = 0 \) directly into the simplified expression:

\[
\frac{5(0) + 8}{3(0)^2 - 16} = \frac{8}{-16} = -\frac{1}{2}
\]

### Final Answer:
The value of the limit is:

\[
\boxed{-\frac{1}{2}}
\]

Would you like further details on any part of the solution? Here are some related questions:

1. What happens if we couldn't cancel the \( y^2 \) terms?
2. How can we determine when a limit tends to infinity?
3. Can L'Hôpital's Rule be applied in this case?
4. How would the result change if we approach infinity instead of zero?
5. What does this limit represent graphically?

**Tip:** Always check for common factors when evaluating limits involving polynomials; it simplifies the process.

Evaluate the limit: lim (y→0) (5y^3 + 8y^2) / (3y^4 - 16y^2)

This detailed solution walks you through how to evaluate the limit of (5y^3 + 8y^2) / (3y^4 - 16y^2) as y approaches 0. The process includes factoring, simplifying rational expressions, and applying limit laws.

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