Question

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

• A. $1/5$
• B. $1/6$
• C. $1/4$
• D. $1/2$

Answer 1

Answer: (B)

$\frac{2\cos \left(\frac{\sin x+x}{2}\right)\cdot \sin \left(\frac{x-\sin x}{2}\right)}{x^4}$
$=\underset{x\to 0}{\lim }2\left[\frac{\sin \left(\frac{\sin x+x}{2}\right)}{\left(\frac{\sin x+x}{2}\right)}\right]\left[\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right]$
$\times \left[\frac{1}{2\left(\frac{1}{\frac{\sin x}{x}+1}\right)\cdot 3\frac{x^3}{(x-\sin x)}}\right]$
$=\frac{1}{6}$