Find lim when x -> 0 of (1-cosx*cos2x*cos3x)/(1-cosx)

We are tasked with finding the limit:

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

### Step 1: Analyze the denominator
Using the identity for \(\cos x\) around \(x = 0\), we know that:

\[
\cos x \approx 1 - \frac{x^2}{2} \quad \text{as} \quad x \to 0.
\]

Thus, for small \(x\), the denominator can be approximated as:

\[
1 - \cos x \approx \frac{x^2}{2}.
\]

### Step 2: Analyze the numerator
The numerator is \(1 - \cos x \cos 2x \cos 3x\). Since \(\cos x \approx 1\), \(\cos 2x \approx 1\), and \(\cos 3x \approx 1\) as \(x \to 0\), we expect \(\cos x \cos 2x \cos 3x \approx 1\) near zero. Therefore, the numerator behaves like:

\[
1 - \cos x \cos 2x \cos 3x \approx 1 - 1 = 0.
\]

Thus, we need a more detailed expansion for the numerator. Expanding the cosines to second order:

\[
\cos x \approx 1 - \frac{x^2}{2}, \quad \cos 2x \approx 1 - 2x^2, \quad \cos 3x \approx 1 - \frac{9x^2}{2}.
\]

So the product becomes approximately:

\[
\cos x \cos 2x \cos 3x \approx \left(1 - \frac{x^2}{2}\right)\left(1 - 2x^2\right)\left(1 - \frac{9x^2}{2}\right).
\]

Multiplying these gives a series of terms:

\[
\cos x \cos 2x \cos 3x \approx 1 - \left( \frac{x^2}{2} + 2x^2 + \frac{9x^2}{2} \right) + \mathcal{O}(x^4).
\]

This simplifies to:

\[
\cos x \cos 2x \cos 3x \approx 1 - 7x^2 + \mathcal{O}(x^4).
\]

Therefore, the numerator becomes:

\[
1 - \cos x \cos 2x \cos 3x \approx 1 - \left(1 - 7x^2\right) = 7x^2.
\]

### Step 3: Final limit calculation
Now the expression simplifies to:

\[
\frac{1 - \cos x \cos 2x \cos 3x}{1 - \cos x} \approx \frac{7x^2}{\frac{x^2}{2}} = 14.
\]

Thus, the limit is:

\[
\lim_{{x \to 0}} \frac{1 - \cos x \cos 2x \cos 3x}{1 - \cos x} = 14.
\]

### Conclusion:
The value of the limit is:

\[
\boxed{14}.
\]

Would you like further details or have any questions? Here are some related questions to explore this topic:

1. How do Taylor expansions help in finding limits involving trigonometric functions?
2. What is the small-angle approximation for trigonometric functions and how is it derived?
3. How can L'Hopital's Rule be applied to limits involving indeterminate forms?
4. What is the significance of higher-order terms in approximations?
5. How do you determine when to use a series expansion vs another method for limits?

**Tip**: When working with limits of trigonometric functions near zero, using Taylor expansions is a powerful tool for approximating behavior.

This solution demonstrates how to find the limit of (1 - cos(x)cos(2x)cos(3x)) / (1 - cos(x)) as x approaches 0 using Taylor series expansion. Step-by-step explanation of the numerator and denominator simplifications is provided, with a final result of 14.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation