How to solve this limit $\lim\limits_{x \to0}8/x^8 \cdot [1-\cos(x^2/2)-\cos(x^2/4)+\cos(x^2/2)\cos(x^2/4)] $ = ??

Question

Evaluate $$\lim\limits_{x \to 0} \dfrac{8 [1-\cos(x^2/2)-\cos(x^2/4)+\cos(x^2/2)\cos(x^2/4)]}{x^8} $$

I tried using series expansion of $\cos(x)$ and also tried L'Hopital's rule but it comes out to be long in solving.

Answer 1

Hint : $$1-\cos(x^2/2)-\cos(x^2/4)+\cos(x^2/2)\cos(x^2/4) = (1-\cos(x^2/2))(1-\cos(x^2/4))$$ and $$\lim_{x \to 0} \frac{1-\cos(x)}{x^2}=\frac{1}{2}$$

Answer 2

My answer is $\frac{1}{32}$.

Just expand the $\cos(x)$ series up to second summand, eq. : $\cos(x) \approx 1 - \frac{x^2}{2}$ and do the calculation.