the value of k is___________.
Question:

If $\lim _{x \rightarrow 0}\left\{\frac{1}{x^{8}}\left(1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}\right)\right\}=2^{-k}$, then the value of $k$ is___________.

Solution:

$\lim _{x \rightarrow 0} \frac{\left(1-\cos \frac{x^{2}}{2}\right)}{x^{4}} \frac{\left(1-\cos \frac{x^{2}}{4}\right)}{x^{4}}=2^{-k}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x^{2}}{4}}{\frac{x^{4}}{16} \times 16} \times \frac{2 \sin ^{2} \frac{x^{2}}{8}}{\frac{x^{4}}{64} \times 64}=2^{-k}$

$\Rightarrow \frac{4}{16 \times 64}=2^{-8}=2^{-k}$  $\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$

$\therefore k=8$