Question

The value of the function $\underset{x\to 0}{lim}\frac{8}{x^8}(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}.\cos \frac{x^2}{4})$ is

• A. $\frac{1}{16}$
• B. $\frac{1}{15}$
• C. $\frac{1}{32}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{32}$

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

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

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

$=\underset{x\to 0}{lim}\frac{8\times 4}{x^8}\left({\sin }^2\frac{x^2}{4}\right)\left({\sin }^2\frac{x^2}{8}\right)$

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