Question

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

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

Answer 1

The correct option is C

$\frac{1}{32}$

${lim}_{x\to 0}\frac{8}{x^8}[1-cos\frac{x^2}{2}cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}]={lim}_{x\to 0}frac8x^8[(1-cos\frac{x^2}{4})-cos\frac{x^2}{4}(1-cos\frac{x^2}{4})]$
$={lim}_{x\to 0}\frac{8}{x^8}\left[(1-cos\frac{x^2}{4})(1-cos\frac{x^2}{2})\right]={lim}_{x\to 0}\frac{8}{x^8}\left[\left(2si{n}^2\frac{x^2}{8}\right)\left(2si{n}^2\frac{x^2}{4}\right)\right]$
$={lim}_{x\to 0}4\times 8\frac{\left(si{n}^2\frac{x^2}{8}\right)\left(si{n}^2\frac{x^2}{4}\right)}{(64\times \frac{x^2}{64})16\left(\frac{x^2}{16}\right)}=\frac{32}{64\times 16}=\frac{1}{32}$