Question

The value of
$ \displaystyle\lim_{x \to 0} \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

• A. $\frac{1}{8}$
• B. $\frac{1}{16}$
• C. $\frac{1}{32}$
• D. $\frac{1}{64}$

Answer 1

Answer: (C)

Given expression is
$ \displaystyle\lim_{x \to 0} \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\}$
$= \displaystyle\lim _{x \to 0} \frac{8}{x^{8}} \left\{1-cos \frac{x^{2}}{2} - cos \frac{x^{2}}{4}\left(1-cos \frac{x^{2}}{2}\right) \right\}$
$= \displaystyle\lim _{x \to 0} \frac{8}{x^{8}} \left\{\left(1-cos \frac{x^{2}}{2}\right)\left(1-cos \frac{x^{2}}{4}\right)\right\}$
$= \displaystyle\lim _{x \to 0} \frac{8}{x^{8}} \left\{2\,sin^{2} \left(\frac{x^{2}}{4}\right) 2\, sin^{2}\left(\frac{x^{2}}{8}\right)\right\}$
(using $cos \,2\theta = 1- 2\,sin^{2} \,\theta$)
$= \displaystyle\lim _{x \to 0}\left[32 \frac{sin^{2} \left(\frac{x^{2}}{4}\right)}{x^{4}}\times\frac{sin^{2} \left(\frac{x^{2}}{8}\right)}{x^{4}}\right]$
$= 32.\displaystyle\lim _{x \to 0}\left[\frac{sin \frac{x^{4}}{16}}{16. \frac{x^{4}}{16}}\times\frac{sin \frac{x^{4}}{64}}{64. \frac{x^{4}}{64}}\right]$
$= 32\times \frac{1}{16}\times \frac{1}{64} = \frac{1}{32}$ (Using $\displaystyle\lim _{\theta \to 0 \frac{sin\,\theta}{\theta}} = 1$)