Question

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

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

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{\cos (\sin x)-\cos x}{x^4}$
$=\underset{x\to 0}{\lim }\frac{2\sin \left(\frac{x+\sin x}{2}\right)\sin \left(\frac{x-\sin x}{2}\right)}{x^4}$
$=2\underset{x\to 0}{\lim }[\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\times \frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}$
$\times \left(\frac{x+\sin x}{2x}\right)\left(\frac{x-\sin x}{2x^3}\right)]$
$=2\underset{x\to 0}{\lim }[\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\frac{x+\sin x}{2}}\times \frac{\sin \left(\frac{x-\sin x}{2}\right)}{\frac{x-\sin x}{2}}$
$\times \left(\frac{1}{2}+\frac{\sin x}{2x}\right)\left(\frac{x-\sin x}{2x^3}\right)]$
$=2\times 1\times 1\times \left(\frac{1}{2}+\frac{1}{2}\right){\lim }_{x\to 0}\frac{x-\sin x}{2x^3}$
$=\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}=\underset{x\to 0}{\lim }\frac{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \right)}{x^3}$
$=\underset{x\to 0}{\lim }\left(\frac{1}{3!}-\frac{x^2}{5!}+\dots \right)=\frac{1}{6}$