Question

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

Answer 1

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

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

$\Rightarrow \underset{x\to 0}{lim}\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{16}=2^{-k}$

$\Rightarrow \frac{1}{256}=2^{-k}$
$\Rightarrow 2^{-8}=2^{-k}\Rightarrow k=8$