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We are tasked with solving the following limit step by step:

\[
\lim_{z \to 0} \frac{z \sin^2(z)}{\tan^3(2z)}
\]

### Step 1: Evaluate the behavior of the function as \( z \to 0 \)

First, let's check the individual components of the function as \( z \) approaches zero:

1. \( \sin^2(z) \to \sin^2(0) = 0^2 = 0 \)
2. \( \tan(2z) \to \tan(0) = 0 \)
3. \( z \to 0 \)

Thus, as \( z \to 0 \), the limit approaches an indeterminate form of type \( \frac{0}{0} \), which suggests we can apply **L'Hopital's Rule**. L'Hopital's Rule is used for limits of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), and it involves taking derivatives of the numerator and denominator.

### Step 2: Apply L'Hopital's Rule

To apply L'Hopital's Rule, we need to differentiate the numerator and denominator separately.

#### Numerator: \( z \sin^2(z) \)

The derivative of \( z \sin^2(z) \) using the product rule:

\[
\frac{d}{dz} \left(z \sin^2(z)\right) = \sin^2(z) + z \cdot 2\sin(z)\cos(z)
\]

This simplifies to:

\[
\sin^2(z) + 2z\sin(z)\cos(z)
\]

#### Denominator: \( \tan^3(2z) \)

The derivative of \( \tan^3(2z) \) using the chain rule:

\[
\frac{d}{dz} \left(\tan^3(2z)\right) = 3\tan^2(2z) \cdot \frac{d}{dz}(\tan(2z)) = 3\tan^2(2z) \cdot 2\sec^2(2z)
\]

This simplifies to:

\[
6\tan^2(2z)\sec^2(2z)
\]

### Step 3: Re-evaluate the limit

Now, we re-evaluate the limit with the derivatives:

\[
\lim_{z \to 0} \frac{\sin^2(z) + 2z\sin(z)\cos(z)}{6\tan^2(2z)\sec^2(2z)}
\]

As \( z \to 0 \):

- \( \sin^2(z) \to 0 \)
- \( z\sin(z)\cos(z) \to 0 \)
- \( \tan^2(2z) \to 0 \)
- \( \sec^2(2z) \to 1 \)

Thus, the limit still results in an indeterminate form \( \frac{0}{0} \). To proceed, we could apply L'Hopital's Rule again or use series expansions.

### Step 4: Use Series Expansions

Let’s use series expansions around \( z = 0 \) for the numerator and denominator:

- \( \sin(z) \approx z - \frac{z^3}{6} \)
- \( \tan(2z) \approx 2z + \frac{8z^3}{3} \)

Substitute these approximations into the limit:

\[
\lim_{z \to 0} \frac{z \left( z^2 - \frac{z^4}{3} \right)}{(2z + \frac{8z^3}{3})^3}
\]

Simplifying this expression gives a result of \( \frac{1}{8} \).

### Final Answer:

\[
\lim_{z \to 0} \frac{z \sin^2(z)}{\tan^3(2z)} = \frac{1}{8}
\]

---

Would you like further details or explanations? Here are 5 related questions to expand on this concept:

1. How does L'Hopital's Rule help resolve indeterminate forms?
2. What are the Taylor series expansions of \( \sin(z) \) and \( \tan(z) \) near \( z = 0 \)?
3. How does the chain rule apply when differentiating trigonometric functions?
4. Can we use a different approach (e.g., squeeze theorem) to solve this limit?
5. How does this limit change if the powers of sine or tangent are altered?

**Tip:** When dealing with indeterminate forms, always consider both L'Hopital's Rule and series expansions as possible tools.

Step-by-step solution of lim (z → 0) z * sin^2(z) / tan^3(2z). This problem explores the use of L'Hopital's Rule and series expansions for solving limits involving trigonometric functions and indeterminate forms. Suitable for Grades 11-12 or Early College students.

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