If the function f(x) = ( x ) - x x ^ 4 is continuous at each point...

Given $\lim_{x \to 0}\dfrac{\cos ( \sin x ) - \cos x}{{x}^{4}} = f\left(0\right)$

$⇒ \lim_{x \to 0}\dfrac{2\sin \left(\dfrac{\sin x + x}{2}\right)\sin \left(\dfrac{x - \sin x}{2}\right)}{{x}^{4}} = \dfrac{1}{k}$

$⇒ \lim_{x \to 0}\dfrac{2\sin \left(\dfrac{\sin x + x}{2}\right)\sin \left(\dfrac{x - \sin x}{2}\right)}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{x - \sin x}{2}\right){x}^{4}} \times \left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{x - \sin x}{2}\right) = \dfrac{1}{k}$

$⇒ \lim_{x \to 0}2\left(\dfrac{\sin x + x}{2x}\right)\left(\dfrac{x - \sin x}{2{x}^{3}}\right) = \dfrac{1}{k}, \left(∵ \lim_{x \to 0}\dfrac{\sin x}{x} = 1\right)$

$⇒ 2\left(\lim_{x \to 0}\left(\dfrac{\sin x + x}{2x}\right)\right)\left(\lim_{x \to 0}\left(\dfrac{x - \sin x}{2{x}^{3}}\right)\right) = \dfrac{1}{k}$

Apply L'Hospital rule for an indeterminate form of $\dfrac{0}{0}$, we get,

$⇒ 2 \times \dfrac{( 1 + 1 )}{2} \times \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{k}$

$⇒ k = 6$