Question

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

Answer 1

Answer: 8

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