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The limit you are dealing with is:

\[
\lim_{x \to 0} \frac{4 \sin^2(2x)}{x \tan(2x)}
\]

### Steps to solve this:

1. **Simplify the expression:**

   We can rewrite the given expression as:

   \[
   \lim_{x \to 0} 4 \cdot \frac{\sin^2(2x)}{x \tan(2x)}
   \]

   We will handle the limit involving \( \frac{\sin(2x)}{\tan(2x)} \) using known standard limits and trigonometric identities.

2. **Trigonometric identity:**  
   
   Recall that:
   \[
   \tan(2x) = \frac{\sin(2x)}{\cos(2x)}
   \]
   Therefore, the expression becomes:
   \[
   \lim_{x \to 0} 4 \cdot \frac{\sin^2(2x)}{x \cdot \frac{\sin(2x)}{\cos(2x)}} = \lim_{x \to 0} 4 \cdot \frac{\sin(2x) \cdot \cos(2x)}{x}
   \]

3. **Use the standard limit:**  
   
   The standard limit for small \( x \) is:
   \[
   \lim_{x \to 0} \frac{\sin(kx)}{x} = k
   \]
   In our case, \( k = 2 \) for \( \sin(2x) \), so:
   \[
   \lim_{x \to 0} \frac{\sin(2x)}{x} = 2
   \]
   
   Therefore, the expression becomes:
   \[
   4 \cdot 2 \cdot \cos(0) = 4 \cdot 2 \cdot 1 = 8
   \]

### Final Answer:
\[
\lim_{x \to 0} \frac{4 \sin^2(2x)}{x \tan(2x)} = 8
\]

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. What is \( \lim_{x \to 0} \frac{\sin x}{x} \)?
2. How do you compute \( \lim_{x \to 0} \frac{\tan(3x)}{x} \)?
3. Can you solve \( \lim_{x \to 0} \frac{x^2 - \sin^2(x)}{x^4} \)?
4. What is the limit of \( \frac{\sin(5x)}{x} \) as \( x \to 0 \)?
5. How can L'Hopital's rule be applied to trigonometric limits?

**Tip**: When working with trigonometric limits, remember that the standard limit \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \) is a crucial tool.

What is the limit of (4 * sin^2(2x)) / (x * tan(2x)) as x approaches 0?

Solve the limit of (4 * sin^2(2x)) / (x * tan(2x)) as x approaches 0 with step-by-step explanations, using trigonometric identities and standard limits. Learn the techniques for handling trigonometric limits and standard limit forms.

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