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

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: Evaluate \( f(0) \)

First, we substitute \( x = 0 \) into the function:

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

Since \( \cos(0) = 1 \), we have:

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

This is an indeterminate form, so we need to use limits.

### Step 2: Apply the Limit

We need to find:

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

### Step 3: Simplify \( 1 - \cos x \)

Using the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \), we can express \( 1 - \cos x \) as:

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

Thus, we have:

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

### Step 4: Use the Cosine Identity Again

Now we apply the cosine identity again:

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

Let \( y = 2 \sin^2\left(\frac{x}{2}\right) \):

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

### Step 5: Substitute Back into the Limit

Now we substitute back into the limit:

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

### Step 6: Use Taylor Series Expansion

For small \( x \), we can use the Taylor series expansion:

\[ \sin z \approx z \quad \text{for small } z \]

Thus, we have:

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

### Step 7: Substitute and Simplify

Now substituting back:

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

### Conclusion

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

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