To solve $\lim_{x \to 0} \frac{\tan(4x)}{x \cos(2x)}$ using the limit rule involving $\sin$, we will start by expressing $\tan(4x)$ in terms of sine and cosine.

Step-by-step solution:

1. Recall that $\tan(4x) = \frac{\sin(4x)}{\cos(4x)}$.

   So the given limit becomes: $\lim_{x \to 0} \frac{\sin(4x)}{x \cos(4x) \cos(2x)}$

2. Now, use the standard limit result: $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$ where $k$ is a constant.

3. To use this, rewrite the expression as: $\lim_{x \to 0} \frac{4 \sin(4x)}{4x} \cdot \frac{1}{\cos(4x) \cos(2x)}$

4. The limit of $\frac{\sin(4x)}{4x}$ as $x \to 0$ is 1: $\lim_{x \to 0} \frac{\sin(4x)}{4x} = 1$

5. Now, evaluate the limits of the cosine terms at $x = 0$: $\cos(4x) \to \cos(0) = 1 \quad \text{and} \quad \cos(2x) \to \cos(0) = 1$

6. Therefore, the overall limit becomes: $\lim_{x \to 0} \frac{4 \cdot 1}{1 \cdot 1} = 4$

Final answer:

$\lim_{x \to 0} \frac{\tan(4x)}{x \cos(2x)} = 4$

Would you like more details on any of the steps?

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Here are five related questions to explore further:

1. What is the limit of $\frac{\sin(5x)}{x}$ as $x \to 0$?
2. How would you solve $\lim_{x \to 0} \frac{\tan(3x)}{x}$?
3. Can you derive the limit $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$?
4. How does the small angle approximation for sine and cosine help in limit problems?
5. What is the limit of $\frac{\tan(2x)}{x \cos(3x)}$ as $x \to 0$?

Tip: When solving trigonometric limits, using known standard limits like $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ can significantly simplify the process.