lim x--> 0 tan4x/(xcos2x) use limit rule involving sin to solve

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.

Explore how to solve the trigonometric limit lim x → 0 tan(4x)/(x cos(2x)) by applying the sine limit rule. This problem involves expressing tangent in terms of sine and cosine and using key trigonometric limits. Perfect for Grades 11-12 and early college students.

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