If the function $$f(x) = \frac{\cos(\sin x) - \cos x}{x^4}$$ is continuous at each point in its domain and $$f(0) = \frac{1}{k}$$, then $$k$$ is ________.

We need $$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$$ so that $$f(0) = \frac{1}{k}$$.

We use Taylor series. Recall $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ and $$\cos u = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \cdots$$

First, expand $$\cos x$$ up to $$x^4$$: $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ $$-(1)$$

Next, expand $$\cos(\sin x)$$. We have $$\sin x = x - \frac{x^3}{6} + \cdots$$, so $$(\sin x)^2 = \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + \cdots$$ (keeping terms up to $$x^4$$).

Also $$(\sin x)^4 = x^4 + \cdots$$ (higher-order terms are $$x^6$$ and above).

Therefore $$\cos(\sin x) = 1 - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{24} - \cdots = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} - \cdots$$

$$= 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} + \cdots = 1 - \frac{x^2}{2} + \frac{4x^4 + x^4}{24} + \cdots = 1 - \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$$ $$-(2)$$

Subtracting $$(1)$$ from $$(2)$$: $$\cos(\sin x) - \cos x = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) + \cdots = \frac{5x^4}{24} - \frac{x^4}{24} + \cdots = \frac{4x^4}{24} + \cdots = \frac{x^4}{6} + \cdots$$

Therefore $$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} = \frac{1}{6}$$.

Since $$f(0) = \frac{1}{k} = \frac{1}{6}$$, we get $$k = 6$$.