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The expression you've uploaded is:

\[
\lim_{x \to 0} \frac{1 - \cos 2x \cdot \cos 2x \cdot \cos 3x}{x^2}
\]

We can simplify this limit step by step.

### Step-by-Step Solution:

1. **Simplify the numerator:**
   The numerator is \( 1 - \cos(2x) \cdot \cos(2x) \cdot \cos(3x) \). First, note that this is essentially \( 1 - \cos^2(2x) \cdot \cos(3x) \).

2. **Use a small angle approximation:**
   For small values of \( x \), \(\cos(kx) \approx 1 - \frac{(kx)^2}{2} \). We can apply this approximation to \( \cos(2x) \) and \( \cos(3x) \), to simplify the terms near \( x = 0 \).

   - For \( \cos(2x) \), using the approximation:
     \[
     \cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2
     \]
   - For \( \cos(3x) \), using the same approximation:
     \[
     \cos(3x) \approx 1 - \frac{(3x)^2}{2} = 1 - \frac{9x^2}{2}
     \]

3. **Multiply the approximations:**
   We can now multiply \( \cos^2(2x) \) with \( \cos(3x) \) using their approximations:
   \[
   \cos^2(2x) \approx \left(1 - 2x^2\right)^2 = 1 - 4x^2 + O(x^4)
   \]
   and
   \[
   \cos(3x) \approx 1 - \frac{9x^2}{2}
   \]
   Thus, the product is approximately:
   \[
   \cos^2(2x) \cdot \cos(3x) \approx (1 - 4x^2) \cdot (1 - \frac{9x^2}{2}) = 1 - \frac{17x^2}{2} + O(x^4)
   \]

4. **Substitute into the original limit:**
   Now, substituting this approximation into the original expression:
   \[
   \frac{1 - \left(1 - \frac{17x^2}{2}\right)}{x^2} = \frac{\frac{17x^2}{2}}{x^2} = \frac{17}{2}
   \]

Thus, the limit evaluates to:

\[
\lim_{x \to 0} \frac{1 - \cos 2x \cdot \cos 2x \cdot \cos 3x}{x^2} = \frac{17}{2}
\]

Would you like more details or explanations on any specific step?

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Here are some related questions to further explore:

1. What other trigonometric limits are important to understand for calculus?
2. How would the answer change if \( \cos 2x \) was replaced with another trigonometric function?
3. How can we use Taylor expansions to approximate other functions in limits?
4. What are small angle approximations, and why are they useful in limits?
5. How would L'Hopital's Rule apply if we directly differentiated the numerator and denominator?

**Tip:** For limits involving trigonometric functions near \( x = 0 \), small-angle approximations (\( \cos x \approx 1 - x^2/2 \)) are very useful!

lim(x→0) (1 - cos(2x)cos(2x)cos(3x)) / x^2

This solution demonstrates how to evaluate the limit of a trigonometric function using small-angle approximation techniques. The problem simplifies lim(x→0) (1 - cos(2x)cos(2x)cos(3x)) / x^2 and includes detailed step-by-step instructions. Suitable for undergraduate calculus students.

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