Find the Limit

The given expression can be rewritten as $\dfrac{1}{32}\times \dfrac{\left (\sin \dfrac{x^2}{4}\right ) ^2}{\left (\dfrac{x^2}{4}\right) ^2}\times \dfrac{\left (\sin \dfrac{x^2}{8}\right ) ^2}{\left (\dfrac{x^2}{8}\right )^2}$ . Hence the limiting value of the expression is $\dfrac{1}{32}=\boxed {0.03125}$ .

$\begin{aligned} L & = \lim_{x \to 0} \frac 8{x^8}\left(1 - \cos \frac {x^2}2 - \cos \frac {x^2}4 + \cos \frac {x^2}2 \cdot \cos \frac {x^2}4 \right) & \small \blue{\text{Let }u = \frac {x^2}4} \\ & = \lim_{u \to 0} \frac {1 - \cos 2u - \cos u + \cos 2u \cdot \cos u}{32u^4} & \small \blue{\text{By Maclaurin series}} \\ & = \lim_{u \to 0} \frac {1-\left(1 - \frac {4u^2}{2!} + \frac {16u^4}{4!} - \cdots \right) - \left(1 - \frac {u^2}{2!} + \frac {u^4}{4!} - \cdots \right) + \left(1 - \frac {4u^2}{2!} + \frac {16u^4}{4!} - \cdots \right) \left(1 - \frac {u^2}{2!} + \frac {u^4}{4!} - \cdots \right)}{32u^4} \\ & = \lim_{u \to 0} \frac {-1 + \frac {5u^2}2 - \frac {17u^4}{24} - \cdots + \left(1 - \frac {5u^2}2 + \left(\frac {17}{24}+1\right)u^2 - \cdots \right)}{32u^4} \\ & = \lim_{u \to 0} \frac {u^4 + O(u^6)}{32u^4} & \small \blue{\text{Divide up and down by }u^4} \\ & = \lim_{u \to 0} \frac {1 + O(u^2)}{32} = \frac 1{32} = \boxed{0.03125} \end{aligned}$

Reference: Maclaurin series