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.