We have to find the following limit.
$$\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$$
In this I thought to use Lhopital . But using that it will become too long . Is there ny short method .
HINT:
$$1-a-b+ab=(1-a)(1-b)$$
$$\lim_{h\to0}\frac{1-\cos2h}{h^2}=2\lim_{h\to0}\left(\frac{\sin h}h\right)^2 = \text{?}$$
Beside the good hint by lab bhattacharjee, as Henry W commented, Taylor series make the problem quite simple.
Starting with $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ and replacing $y$ successively by $\frac {x^2} 2$ and $\frac {x^2} 4$ $$\cos \left(\frac{x^2}{2}\right)=1-\frac{x^4}{8}+\frac{x^8}{384}+O\left(x^{12}\right)$$ $$\cos \left(\frac{x^2}{4}\right)=1-\frac{x^4}{32}+\frac{x^8}{6144}+O\left(x^{12}\right)$$ $$1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right)\cos \left(\frac{x^2}{4}\right)=\frac{x^8}{256}+O\left(x^{12}\right)$$
