Question

The value of $\underset{x\to 0}{lim}\frac{\cos \left(\sin x\right)-\cos x}{x^4}=$

• A. $1$
• B. $6$
• C. $-\frac{1}{6}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D $\frac{1}{6}$

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

Rewriting the difference in the numerator as a product

$\cos \theta -\cos \varphi \cdot 2\sin \frac{\theta +\varphi }{2}\sin \frac{\varphi -\theta }{2}$

$\underset{x\to 0}{lim}\frac{2\sin \frac{x+\sin x}{2}\sin \frac{x-\sin x}{2}}{x^4}$

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

$2\underset{x\to 0}{lim}[\frac{1}{x^4}\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\times \frac{x+\sin x}{2}\times \frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\times \frac{x-\sin x}{2}$

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

$2\times 1\times 1{lim}_{x\to 0}[\frac{1}{x^4}\times \frac{x+\sin x}{2}\times \frac{x-\sin x}{2}]$

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

Replacing \sin x by Maclarian series:-

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}\cdot \cdot \cdot$

$\Rightarrow \sin x=x^2-\frac{2x^4}{3!}+p\left(x\right)$

We do not need exact expression of p(x)

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

$\Rightarrow \frac{1}{2}\underset{x\to 0}{lim}\left[\frac{x^2-x^2-\frac{2x^4}{3!}+p\left(x\right)}{x^4}\right]$

$\frac{1}{2}\underset{x\to 0}{lim}\left[\frac{\frac{2x^4}{3!}-p\left(x\right)}{x^4}\right]$

$\therefore p^{{}^{\prime }}\left(x\right)$ contains $x$\,power $>4$

$\therefore p\left(x\right)\Rightarrow 0$

$\Rightarrow \frac{1}{2}\underset{x\to 0}{lim}[\frac{2}{3!}-0]=\frac{1}{6}$