Question

The value of $\underset{x\to 0}{lim}\frac{32}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}.\cos \frac{x^2}{4}\right)$ is

• A. $\frac{1}{8}$
• B. $\frac{3}{8}$
• C. $\frac{7}{8}$
• D. $\frac{9}{8}$

Answer 1

Answer: (A)

$\frac{1}{8}$

$\underset{x\to 0}{lim}$ $\frac{32}{x}(1-\cos \frac{x^2}{2}+\cos \frac{x^2}{2}.\cos \frac{x^2}{4}-\cos \frac{x^2}{4})$

We know that, $1-cos2\theta =2{sin}^2\theta$. Using this identity we can write,

$\underset{x\to 0}{lim}$ $\frac{32}{x^8}(2{\sin }^2\left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right).\cos \left(\frac{x^2}{4}\right)-\cos \left(\frac{x^2}{4}\right))$

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

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

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

$\underset{x\to 0}{lim}$ $\frac{32}{x^8}\left[2{\sin }^2\frac{x^2}{4}.2{\sin }^2\frac{x^2}{8}\right]$

$\underset{x\to 0}{lim}$ $\frac{128}{x^8}[\sin \frac{x^2}{4}.\sin \frac{x^2}{4}\sin \frac{x^2}{8}\sin \frac{x^2}{8}]$

$\underset{x\to 0}{lim}$ $128\frac{\sin \left(x^2/4\right)}{4\left(x^2/4\right)}\frac{\sin \left(x^2/4\right)}{4\left(x^2/4\right)}\frac{\sin \left(x^2/8\right)}{8\left(x^2/8\right)}\frac{\sin \left(x^2/8\right)}{8\left(x^2/8\right)}$

We know that,

$\underset{x\to 0}{lim}$ $\frac{\sin \theta }{\theta }=1$

$\therefore$ $\frac{128}{16\times 64}$ $\underset{x\to 0}{lim}$

$\frac{\sin \left(x^2/4\right)}{\left(x^2/4\right)}.\frac{\sin \left(x^2/4\right)}{\left(x^2/4\right)}.\frac{\sin \left(x^2/8\right)}{\left(x^2/8\right)}.\frac{\sin \left(x^2/8\right)}{\left(x^2/8\right)}$

$=\frac{128}{16\times 64}\times 1\times 1\times 1\times 1$

$=\frac{1}{8}$