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

Answer

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 evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Rewrite the function using trigonometric identities
We know that \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \). Therefore, we can rewrite \( 1 - \cos(1 - \cos x) \) as:

\[ 1 - \cos(1 - \cos x) = 1 - \cos\left(2 \sin^2\left(\frac{x}{2}\right)\right) \]

### Step 2: Apply the limit
To find \( f(0) \), we need to compute:

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

### Step 3: Use Taylor series expansion
Using the Taylor series expansion for \( \cos \) around 0, we have:

\[ \cos y \approx 1 - \frac{y^2}{2} \quad \text{for small } y \]

Substituting \( y = 1 - \cos x \):

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

Now substituting \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \):

\[ 1 - \cos(1 - \cos x) \approx \frac{(2 \sin^2\left(\frac{x}{2}\right))^2}{2} = 2 \sin^4\left(\frac{x}{2}\right) \]

### Step 4: Substitute back into the limit
Now we can rewrite our limit:

\[ \lim_{x \to 0} \frac{2 \sin^4\left(\frac{x}{2}\right)}{x^4} \]

### Step 5: Simplify using the small angle approximation
Using the small angle approximation \( \sin y \approx y \):

\[ \sin\left(\frac{x}{2}\right) \approx \frac{x}{2} \]

Thus,

\[ \sin^4\left(\frac{x}{2}\right) \approx \left(\frac{x}{2}\right)^4 = \frac{x^4}{16} \]

### Step 6: Substitute this back into the limit
Now substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{2 \cdot \frac{x^4}{16}}{x^4} = \lim_{x \to 0} \frac{2}{16} = \frac{1}{8} \]

### Conclusion
Thus, for the function to be continuous at \( x = 0 \), we need:

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