Math Problem Statement

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

Solution

The expression you've uploaded is:

limx01cos2xcos2xcos3xx2\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 1cos(2x)cos(2x)cos(3x)1 - \cos(2x) \cdot \cos(2x) \cdot \cos(3x). First, note that this is essentially 1cos2(2x)cos(3x)1 - \cos^2(2x) \cdot \cos(3x).

  2. Use a small angle approximation: For small values of xx, cos(kx)1(kx)22\cos(kx) \approx 1 - \frac{(kx)^2}{2}. We can apply this approximation to cos(2x)\cos(2x) and cos(3x)\cos(3x), to simplify the terms near x=0x = 0.

    • For cos(2x)\cos(2x), using the approximation: cos(2x)1(2x)22=12x2\cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2
    • For cos(3x)\cos(3x), using the same approximation: cos(3x)1(3x)22=19x22\cos(3x) \approx 1 - \frac{(3x)^2}{2} = 1 - \frac{9x^2}{2}
  3. Multiply the approximations: We can now multiply cos2(2x)\cos^2(2x) with cos(3x)\cos(3x) using their approximations: cos2(2x)(12x2)2=14x2+O(x4)\cos^2(2x) \approx \left(1 - 2x^2\right)^2 = 1 - 4x^2 + O(x^4) and cos(3x)19x22\cos(3x) \approx 1 - \frac{9x^2}{2} Thus, the product is approximately: cos2(2x)cos(3x)(14x2)(19x22)=117x22+O(x4)\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: 1(117x22)x2=17x22x2=172\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:

limx01cos2xcos2xcos3xx2=172\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?


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 cos2x\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=0x = 0, small-angle approximations (cosx1x2/2\cos x \approx 1 - x^2/2) are very useful!

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Math Problem Analysis

Mathematical Concepts

Limits
Trigonometric Functions
Small Angle Approximation

Formulas

cos(kx) ≈ 1 - (kx)^2 / 2 for small x
Limit of trigonometric functions as x → 0

Theorems

Small Angle Approximation
Taylor Series Approximation

Suitable Grade Level

Undergraduate Calculus