Question

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

Answer 1

$\displaystyle\lim _{x \rightarrow 0}\left\{\frac{1}{x^{8}}\left(1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}\right)\right\}=2^{-k}$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{\left(1-\cos \frac{x^{2}}{2}\right)}{4\left(\frac{x^{2}}{2}\right)^{2}} \frac{\left(1-\cos \frac{x^{2}}{4}\right)}{16\left(\frac{x^{2}}{4}\right)^{2}}$
$=\frac{1}{8} \times \frac{1}{32}=2^{-k}$
$\Rightarrow 2^{-8}=2^{-k}$
$\Rightarrow k=8$