Question

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

Answer 1

Answer: 31.00

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