Evaluate : `lim_(x to 0)(1-cosxcos2xcos3x)/(sin^(2)2x)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x}{\sin^2 2x} \), we can follow these steps:

### Step 1: Simplify the numerator
We start with the expression in the limit:
\[ 1 - \cos x \cos 2x \cos 3x \]
Using the identity \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \), we can rewrite \( \cos x \cos 2x \):
\[ \cos x \cos 2x = \frac{1}{2} (\cos(3x) + \cos(x)) \]
Now, we can express \( \cos x \cos 2x \cos 3x \):
\[ \cos x \cos 2x \cos 3x = \frac{1}{2} \cos 3x \cdot \frac{1}{2} (\cos(3x) + \cos(x)) = \frac{1}{4} (\cos^2 3x + \cos x \cos 3x) \]

### Step 2: Expand using Taylor series
For small values of \( x \), we can use the Taylor series expansions:
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
Thus:
\[ \cos x \cos 2x \cos 3x \approx (1 - \frac{x^2}{2})(1 - \frac{(2x)^2}{2})(1 - \frac{(3x)^2}{2}) = (1 - \frac{x^2}{2})(1 - 2x^2)(1 - \frac{9x^2}{2}) \]
Multiplying these out gives:
\[ 1 - \left(\frac{x^2}{2} + 2x^2 + \frac{9x^2}{2}\right) = 1 - 7x^2 + O(x^4) \]
Thus:
\[ 1 - \cos x \cos 2x \cos 3x \approx 7x^2 \]

### Step 3: Simplify the denominator
Now, we simplify the denominator \( \sin^2 2x \):
Using the Taylor series for \( \sin x \):
\[ \sin 2x \approx 2x \]
Thus:
\[ \sin^2 2x \approx (2x)^2 = 4x^2 \]

### Step 4: Substitute back into the limit
Now substituting back into the limit:
\[ \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x}{\sin^2 2x} \approx \lim_{x \to 0} \frac{7x^2}{4x^2} = \frac{7}{4} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{1 - \cos x \cos 2x \cos 3x}{\sin^2 2x} = \frac{7}{4} \]