If `f(x)=(cos(sinx)-cosx)/x^4` is continuous over the domain and `f(0)=1/k` , then `k=?`

Answer

To find the value of \( k \) such that \( f(0) = \frac{1}{k} \) for the function

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

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). This requires us to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-step Solution:

1. **Evaluate the Limit**:
We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} \]

2. **Apply the Cosine Difference Formula**:
We can use the identity \( \cos a - \cos b = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \):
Let \( a = \sin x \) and \( b = x \):
\[ \cos(\sin x) - \cos x = -2 \sin\left(\frac{\sin x + x}{2}\right) \sin\left(\frac{\sin x - x}{2}\right) \]

3. **Substituting into the Limit**:
Thus, we rewrite the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-2 \sin\left(\frac{\sin x + x}{2}\right) \sin\left(\frac{\sin x - x}{2}\right)}{x^4} \]

4. **Using Taylor Series Expansion**:
We can expand \( \sin x \) around 0:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
Therefore,
\[ \sin x - x = -\frac{x^3}{6} + O(x^5) \]
and
\[ \sin x + x = 2x - \frac{x^3}{6} + O(x^5) \]

5. **Evaluate the Sine Terms**:
Substituting back:
\[ \sin\left(\frac{\sin x + x}{2}\right) \approx \sin\left(x - \frac{x^3}{12}\right) \approx x - \frac{x^3}{12} \]
and
\[ \sin\left(\frac{\sin x - x}{2}\right) \approx \sin\left(-\frac{x^3}{12}\right) \approx -\frac{x^3}{12} \]

6. **Combining the Results**:
Now substituting these approximations into the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-2 \left(x - \frac{x^3}{12}\right) \left(-\frac{x^3}{12}\right)}{x^4} \]
Simplifying gives:
\[ = \lim_{x \to 0} \frac{2 \left(x - \frac{x^3}{12}\right) \frac{x^3}{12}}{x^4} = \lim_{x \to 0} \frac{2 \left(x^4 - \frac{x^6}{12}\right)}{12x^4} = \lim_{x \to 0} \frac{2}{12} = \frac{1}{6} \]

7. **Setting the Limit Equal to \( f(0) \)**:
Since \( f(0) = \frac{1}{k} \), we have:
\[ \frac{1}{k} = \frac{1}{6} \]
Therefore, solving for \( k \):
\[ k = 6 \]

### Final Answer:
\[ k = 6 \]