Let `f(x) = cos2x.cos4x.cos6x.cos8x.cos10x` then `lim_(x ->0) (1 - (f(x))^3)/(55sin^2x)` equals

Explanation

To solve the limit problem, we start with the function:

\[ f(x) = \cos(2x) \cos(4x) \cos(6x) \cos(8x) \cos(10x) \]

We need to evaluate:

\[ \lim_{x \to 0} \frac{1 - (f(x))^3}{55 \sin^2 x} \]

### Step 1: Evaluate \( f(0) \)

First, we find \( f(0) \):

\[ f(0) = \cos(0) \cos(0) \cos(0) \cos(0) \cos(0) = 1 \]

### Step 2: Rewrite the limit

Since \( f(0) = 1 \), we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{1 - (f(x))^3}{55 \sin^2 x} = \lim_{x \to 0} \frac{1 - f(x)^3}{55 \sin^2 x} \]

### Step 3: Use the identity for cubes

We can use the identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \):

\[ 1 - f(x)^3 = (1 - f(x))(1 + f(x) + f(x)^2) \]

Thus, we can rewrite the limit:

\[ \lim_{x \to 0} \frac{(1 - f(x))(1 + f(x) + f(x)^2)}{55 \sin^2 x} \]

### Step 4: Substitute \( \sin^2 x \)

We know that \( \sin^2 x \) can be approximated as \( x^2 \) near \( x = 0 \):

\[ \sin^2 x \approx x^2 \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{(1 - f(x))(1 + f(x) + f(x)^2)}{55 x^2} \]

### Step 5: Evaluate \( 1 - f(x) \)

Next, we need to evaluate \( 1 - f(x) \):

Using the approximation \( 1 - \cos \theta \approx \frac{\theta^2}{2} \) for small \( \theta \):

\[ 1 - f(x) = 1 - \left(\cos(2x) \cos(4x) \cos(6x) \cos(8x) \cos(10x)\right) \]

We can expand \( f(x) \) using the small angle approximation for each cosine term:

\[ \cos(kx) \approx 1 - \frac{(kx)^2}{2} \]

Thus:

\[ f(x) \approx \left(1 - \frac{(2x)^2}{2}\right)\left(1 - \frac{(4x)^2}{2}\right)\left(1 - \frac{(6x)^2}{2}\right)\left(1 - \frac{(8x)^2}{2}\right)\left(1 - \frac{(10x)^2}{2}\right) \]

### Step 6: Calculate \( 1 - f(x) \)

Multiplying these approximations, we find that:

\[ f(x) \approx 1 - \left(\frac{2^2 + 4^2 + 6^2 + 8^2 + 10^2}{2}\right)x^2 \]

Calculating the sum of squares:

\[ 2^2 + 4^2 + 6^2 + 8^2 + 10^2 = 4 + 16 + 36 + 64 + 100 = 220 \]

Thus:

\[ f(x) \approx 1 - \frac{220}{2} x^2 = 1 - 110 x^2 \]

So:

\[ 1 - f(x) \approx 110 x^2 \]

### Step 7: Substitute back into the limit

Now substituting back into our limit:

\[ \lim_{x \to 0} \frac{110 x^2 (1 + f(x) + f(x)^2)}{55 x^2} \]

The \( x^2 \) terms cancel out:

\[ \lim_{x \to 0} \frac{110 (1 + f(x) + f(x)^2)}{55} \]

### Step 8: Evaluate the limit

As \( x \to 0 \), \( f(x) \to 1 \):

\[ 1 + f(x) + f(x)^2 \to 1 + 1 + 1 = 3 \]

Thus, the limit becomes:

\[ \frac{110 \cdot 3}{55} = \frac{330}{55} = 6 \]

### Final Answer

The final answer is:

\[ \boxed{6} \]