Question

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

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

Answer 1

Answer: (C)

Given expression is
$\underset{x\to 0}{\lim }\frac{8}{x^8}\left\{1-cos\frac{x^2}{2}-cos\frac{x^2}{4}+cos\frac{x^2}{2}cos\frac{x^2}{4}\right\}$
$=\underset{x\to 0}{\lim }\frac{8}{x^8}\left\{1-cos\frac{x^2}{2}-cos\frac{x^2}{4}\left(1-cos\frac{x^2}{2}\right)\right\}$
$=\underset{x\to 0}{\lim }\frac{8}{x^8}\left\{\left(1-cos\frac{x^2}{2}\right)\left(1-cos\frac{x^2}{4}\right)\right\}$
$=\underset{x\to 0}{\lim }\frac{8}{x^8}\left\{2si{n}^2\left(\frac{x^2}{4}\right)2si{n}^2\left(\frac{x^2}{8}\right)\right\}$
(using $cos2\theta =1-2si{n}^2\theta$)
$=\underset{x\to 0}{\lim }\left[32\frac{si{n}^2\left(\frac{x^2}{4}\right)}{x^4}\times \frac{si{n}^2\left(\frac{x^2}{8}\right)}{x^4}\right]$
$=32.\underset{x\to 0}{\lim }\left[\frac{sin\frac{x^4}{16}}{16.\frac{x^4}{16}}\times \frac{sin\frac{x^4}{64}}{64.\frac{x^4}{64}}\right]$
$=32\times \frac{1}{16}\times \frac{1}{64}=\frac{1}{32}$ (Using $\underset{\theta \to 0\frac{sin\theta }{\theta }}{\lim }=1$)