Question

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

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

Answer 1

Answer: (C)

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