$$\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$$ is equal to

We need to evaluate $$\displaystyle\lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}$$.

Recall that $$\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).$$

Let $$A = \sin x$$ and $$B = x$$, so that $$\cos(\sin x) - \cos x = -2\sin\left(\frac{\sin x + x}{2}\right)\sin\left(\frac{\sin x - x}{2}\right).$$

Using the Taylor series expansion for small x, $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots.$$

It follows that $$\sin x + x = 2x - \frac{x^3}{6} + O(x^5).$$

Therefore, $$\frac{\sin x + x}{2} = x - \frac{x^3}{12} + O(x^5).$$

Similarly, $$\sin x - x = -\frac{x^3}{6} + O(x^5),$$ so $$\frac{\sin x - x}{2} = -\frac{x^3}{12} + O(x^5).$$

As x → 0, both arguments of sine are small, so one may approximate $$\sin u \approx u$$ for each.

Thus $$\sin\left(\frac{\sin x + x}{2}\right) \approx x - \frac{x^3}{12},$$ whose leading term is x.

Also, $$\sin\left(\frac{\sin x - x}{2}\right) \approx -\frac{x^3}{12}.$$

Substituting these approximations into the difference of cosines gives $$\cos(\sin x) - \cos x \approx -2 \left(x - \frac{x^3}{12}\right)\left(-\frac{x^3}{12}\right).$$

Retaining terms up to order $$x^4$$ yields $$-2x\left(-\frac{x^3}{12}\right) + O(x^6) = \frac{2x^4}{12} = \frac{x^4}{6}.$$

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

Thus the limit is $$\dfrac{1}{6}$$.