Question

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

(a) $\frac{1}{16}$

(b) $-\frac{1}{16}$

(c) $\frac{1}{32}$

(d) $-\frac{1}{32}$

Answer 1

(c) $\frac{1}{32}$
$$\begin{gathered} \underset{x\to 0}{\lim }\frac{8}{x^8}\left[1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right] \\ =\underset{x\to 0}{\lim }\frac{8}{x^8}\left[\left(1-\cos \frac{x^2}{4}\right)-\cos \frac{x^2}{2}\left(1-\cos \frac{x^2}{4}\right)\right] \\ =\underset{x\to 0}{\lim }\frac{8}{x^8}\left[\left(1-\cos \frac{x^2}{4}\right)\left(1-\cos \frac{x^2}{2}\right)\right] \\ =\underset{x\to 0}{\lim }\frac{8}{x^8}\left[\left(2{\sin }^2\frac{x^2}{8}\right)\left(2{\sin }^2\frac{x^2}{4}\right)\right] \\ =\underset{x\to 0}{\lim }4\times 8\frac{\left({\sin }^2{\displaystyle \frac{x^2}{8}}\right)}{\left(64\times {\displaystyle \frac{x^4}{64}}\right)}\frac{\left({\sin }^2{\displaystyle \frac{x^2}{4}}\right)}{16\left({\displaystyle \frac{x^4}{16}}\right)} \\ =\frac{32}{64\times 16} \\ =\frac{1}{32} \end{gathered}$$