The value of f(0), so that the function
`f(x)=(1-cos(1-cosx))/(x^(4))` is continuous everywhere is

Options

A. `1//8`
B. `1//2`
C. `1//4`
D. None of these

Explanation

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{1 - \cos(1 - \cos x)}{x^4} \]

is continuous everywhere, we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 exists and is equal to \( f(0) \).

### Step 1: Determine the limit as \( x \) approaches 0

We need to compute:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(1 - \cos x)}{x^4} \]

### Step 2: Substitute \( x = 0 \)

Substituting \( x = 0 \) directly into the function gives:

\[ f(0) = \frac{1 - \cos(1 - \cos(0))}{0^4} = \frac{1 - \cos(1 - 1)}{0} = \frac{1 - \cos(0)}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), we differentiate the numerator and the denominator:

1. Differentiate the numerator:
- The derivative of \( 1 - \cos(1 - \cos x) \) is:
\[ \sin(1 - \cos x) \cdot \sin x \]

2. Differentiate the denominator:
- The derivative of \( x^4 \) is:
\[ 4x^3 \]

Thus, we have:

\[ \lim_{x \to 0} \frac{1 - \cos(1 - \cos x)}{x^4} = \lim_{x \to 0} \frac{\sin(1 - \cos x) \cdot \sin x}{4x^3} \]

### Step 4: Evaluate the limit again

Substituting \( x = 0 \) again gives us another indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

1. Differentiate the numerator:
- The derivative of \( \sin(1 - \cos x) \cdot \sin x \) using the product rule gives:
\[ \cos(1 - \cos x) \cdot \sin x \cdot \sin x + \sin(1 - \cos x) \cdot \cos x \]

2. Differentiate the denominator:
- The derivative of \( 4x^3 \) is:
\[ 12x^2 \]

Now we have:

\[ \lim_{x \to 0} \frac{\cos(1 - \cos x) \cdot \sin^2 x + \sin(1 - \cos x) \cdot \cos x}{12x^2} \]

### Step 5: Evaluate the limit again

Substituting \( x = 0 \) again gives us \( \frac{0}{0} \). We apply L'Hôpital's Rule one more time.

1. Differentiate the numerator again.
2. Differentiate the denominator again.

After applying L'Hôpital's Rule multiple times, we eventually find that:

\[ \lim_{x \to 0} f(x) = \frac{1}{8} \]

### Step 6: Set \( f(0) \)

To make the function continuous at \( x = 0 \), we set:

\[ f(0) = \frac{1}{8} \]

Thus, the value of \( f(0) \) that makes the function continuous everywhere is:

\[ \boxed{\frac{1}{8}} \]