L'Hospital's Rule

Relevant wiki: Maclaurin Series

$\begin{aligned} L & = \lim_{x \to 0} \frac {1-\cos(1-\cos x)}{\sin^4 x} \\ & =\lim_{x \to 0} \frac {1-\cos(1-\cos x)}{(1-\cos^2 x)^2} \\ & =\lim_{x \to 0} \frac {1-\cos(1-\cos x)}{(1-\cos x)^2(1+\cos x)^2} & \small \color{#3D99F6} \text{Let }u = 1-\cos x \\ & =\lim_{x \to 0} \frac {1-\color{#3D99F6} \cos u}{u^2(2-u)^2} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \frac {1-\color{#3D99F6} \left(1 - \frac {u^2}{2!}+\frac {u^4}{4!}-\cdots \right)}{4u^2-4u^3+u^4} \\ & = \lim_{x \to 0} \frac {\frac {u^2}{2!}-\frac {u^4}{4!}+\frac {u^6}{6!} -\cdots}{4u^2-4u^3+u^4} & \small \color{#3D99F6} \text{Divide up and down by }u^2 \\ & = \lim_{x \to 0} \frac {\frac 1{2!}-\frac {u^2}{4!}+\frac {u^4}{6!} -\cdots}{4-4u+u^2} \\ & = \frac 18 = \boxed{0.125} \end{aligned}$

Simplifying the numerator as $1 - \cos \,(1- \cos x) = 1 - \cos \left( 2\sin ^2 \frac{x}{2}\right) = 2 \sin ^2 \left( \frac{2 \sin ^2 \frac{x}{2}}{2}\right) = 2 \sin ^2 \left(\sin ^2 \frac{x}{2}\right)$ Further $\begin{aligned} \lim_{x\to 0} \left( \dfrac{1 - \cos \,(1- \cos x) }{\sin ^4 x} \right) & = \lim_{x\to 0 }\left( \dfrac{2 \sin ^2 \left(\sin ^2 \frac{x}{2}\right) }{\sin ^4 x}\right) \\ & = 2 \lim_{x\to 0} \left(\dfrac{\frac{\sin \left(\sin ^2\frac{x}{2}\right)}{\sin ^2 \frac{x}{2} }\cdot \sin ^2\frac{x}{2}}{\left(\frac{\sin x}{x}\cdot x \right)^2 }\right) ^2 \\ & = 2\lim_{x\to 0} \left( \dfrac{\sin\frac{x}{2}}{\frac{x}{2}\cdot 2}\right) ^4 \\& = \dfrac{2}{2^4} = \dfrac{1}{8}= 0.125\end{aligned}$

Using L-Hopital's rule $\lim_{x\to 0} \dfrac{1-\cos \,(1 -\cos x)}{\sin ^4 x}= \lim_{x\to 0}\dfrac{\sin x \cdot \sin\,(1-\cos x)}{4\sin ^3 x \cdot \cos x} = \dfrac{1}{4}\lim_{x\to 0}\left(\dfrac{\sin x \cos (1-\cos x)}{\sin x \,(2\cos ^2x - \sin x)} \right) =\dfrac{1}{4\cdot 2}=\dfrac{1}{8}$