Mathematics - Limits Question with Solution | TestHub

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

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

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

$=\underset{x\to 0}{\lim }\frac{\left(1-\cos \frac{x^2}{2}\right)}{4\times \left({\displaystyle \frac{x^4}{4}}\right)}\times \frac{\left(1-\cos \frac{x^2}{4}\right)}{16\times \left({\displaystyle \frac{x^4}{16}}\right)}$

Using $\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2}$

$=\frac{1}{2}\times \frac{1}{4}\times \frac{1}{2}\times \frac{1}{16}=\frac{1}{256}=2^{-8}$

$\Rightarrow 2^{-8}=2^{-k} \Rightarrow k=8$