Question

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

Answer 1

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

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

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

We know that $\cos 2A=1-2{\sin }^{2}A\Rightarrow 1-\cos 2A=2{\sin }^{2}A$

Using this result we get,

$\Rightarrow \underset{x\to 0}{lim}\frac{8}{x^8}4{\sin }^2\frac{x^2}{8}{\sin }^2\frac{x^2}{4}$

$\Rightarrow \underset{x\to 0}{lim}\frac{32}{64\times 16}\frac{{\sin }^2\frac{x^2}{8}}{\frac{x^4}{64}}\frac{{\sin }^2\frac{x^2}{4}}{\frac{x^4}{16}}$

$\Rightarrow \frac{1}{32}$

Therefore $\underset{x\to 0}{lim}\frac{8}{x^8}[1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}]=\frac{1}{32}$