Question

If $\underset{x\to 0}{lim}\frac{1}{x^8}[1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}]=2^{-k},$ then the value of $k$ is​​​​​​​

Answer 1

$\underset{x\to 0}{lim}\frac{1}{x^8}[(1-\cos \frac{x^2}{2})-\cos \frac{x^2}{4}(1-\cos \frac{x^2}{2})]$

$=\underset{x\to 0}{lim}\frac{1}{x^8}\left[(1-\cos \frac{x^2}{2})(1-\cos \frac{x^2}{4})\right]$

$=\underset{x\to 0}{lim}\frac{1}{x^8}\left[2{\sin }^2\frac{x^2}{4}2{\sin }^2\frac{x^2}{8}\right]$

$=\underset{x\to 0}{lim}\frac{1}{256}{\left(\frac{\sin \frac{x^2}{4}}{\left(\frac{x^2}{4}\right)}\right)}^2{\left(\frac{\sin \frac{x^2}{8}}{\left(\frac{x^2}{8}\right)}\right)}^2=\frac{1}{256}=2^{-8}$

$\therefore k=8$