_ x → 0 ( x ) - x x ^ 4 is equal to

Consider $\lim_{x \to 0}\dfrac{\cos ( \sin x ) - \cos x}{{x}^{4}}\left(\dfrac{0}{0}\text{form}\right)$

$\lim_{x \to 0}\dfrac{2\sin \left(\dfrac{\sin x + x}{2}\right) \cdot \sin \left(\dfrac{x - \sin x}{2}\right)}{{x}^{4}}$

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

$\lim_{x \to 0}2\left[\dfrac{\sin \left(\dfrac{\sin x + x}{2}\right)}{\left(\dfrac{\sin x + x}{2}\right)}\right]\left[\dfrac{\sin \left(\dfrac{x - \sin x}{2}\right)}{\left(\dfrac{x - \sin x}{2}\right)}\right] \times \left(\dfrac{{x}^{2} - \sin^{2}x}{4{x}^{4}}\right)$

$\lim_{x \to 0}2 \times \left(\dfrac{{x}^{2} - \sin^{2}x}{4{x}^{4}}\right)\left(\dfrac{0}{0}\text{form}\right)\left(∵ \lim_{t \to 0}\dfrac{\sin t}{t} = 1\right)$

Applying L'Hospital's Rule,

$\lim_{x \to 0}2 \times \left(\dfrac{2x - 2\sin x \cos x}{4 \cdot 4{x}^{3}}\right) = \lim_{x \to 0}\left(\dfrac{2x - \sin 2x}{8{x}^{3}}\right)\left(\dfrac{0}{0}\text{form}\right)$

$\lim_{x \to 0}\left(\dfrac{2 - 2\cos 2x}{24{x}^{2}}\right)\left(\dfrac{0}{0}\text{form}\right)$

$\lim_{x \to 0}\left(\dfrac{4\sin 2x}{48x}\right) = \lim_{x \to 0}\left(\dfrac{\sin 2x}{2x}\right) = \dfrac{1}{6}$