Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{12}$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{\cos (\sin x)-\cos x}{x^4};\left(\frac{0}{0}\right)$
$\underset{x\to 0}{\lim }\left(\frac{2\cdot \sin \left(\frac{x+\sin x}{2}\right)\cdot \sin \left(\frac{x-\sin x}{2}\right)}{x^4}\right)$
$\underset{x\to 0}{\lim }2\left(\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\right)\left(\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right)(\frac{\frac{x+\sin x}{2})}{x^4}\left(\frac{x-\sin x}{2}\right)$
$\underset{x\to 0}{\lim }\left(\frac{x^2-{\sin }^{2}x}{2x^4}\right):\left(\frac{0}{0}\right)$
Apply L-Hopital Rule :
$\underset{x\to 0}{\lim }\frac{2x-2\sin x\cos x}{2\cdot 4\cdot x^3}$
$\underset{x\to 0}{\lim }\frac{2x-\sin 2x}{8x^3};\frac{0}{0}:$ Again apply L-Hopital rule
$\underset{x\to 0}{\lim }\frac{2-2\cos (2x)}{8(3)x^2}$
$\underset{x\to 0}{\lim }\frac{2(1-\cos (2x))}{24\left(4x^2\right)}\times 4$
$\Rightarrow \frac{2}{24}\times \frac{1}{2}\times 4$
$\Rightarrow \frac{1}{6}$