Question

If : $\displaystyle\lim _{x \rightarrow 0} \frac{8}{x^{8}}\left[1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2} \cos \frac{x^{2}}{4}\right]$ is equal to $\frac{\lambda}{k}$ then find $k-\lambda$

Answer 1

$\displaystyle\lim _{x \rightarrow 0} \frac{8}{x^{8}}\left(1\left(1-\cos \frac{x^{2}}{2}\right)-\cos \frac{x^{2}}{4}\left(1-\cos \frac{x^{2}}{2}\right)\right)$
$\displaystyle\lim _{x \rightarrow 0} 8 \frac{\left(1-\cos \frac{x^{2}}{2}\right)\left(1-\cos \frac{x^{2}}{4}\right)}{x^{8}}$
$\displaystyle\lim _{x \rightarrow 0} 8 \frac{2 \sin ^{2} \frac{x^{2}}{4} \cdot 2 \sin ^{2} \frac{x^{2}}{8}}{x^{8}}$
$\displaystyle\lim _{x \rightarrow 0} 8 \frac{2\left(\frac{\sin \frac{x^{2}}{4}}{\frac{x^{2}}{4}}\right)^{2} \cdot 2\left(\frac{\sin \frac{x^{2}}{8}}{\frac{x^{2}}{8}}\right)^{2} \cdot\left(\frac{x^{2}}{4}\right)^{2}\left(\frac{x^{2}}{8}\right)^{2}}{x^{8}}$
$=8 \cdot \frac{1}{256}=\frac{1}{32}$