MCQ
$\lim\limits _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to :
  • A
    $\frac{1}{3}$
  • B
    $\frac{1}{4}$
  • $\frac{1}{6}$
  • D
    $\frac{1}{12}$

Answer

Correct option: C.
$\frac{1}{6}$
c
$\lim\limits _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}} ;\left(\frac{0}{0}\right)$

$\lim \limits_{x \rightarrow 0}\left(\frac{2 \cdot \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}}\right)$

$\lim\limits _{x \rightarrow 0} 2\left(\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\left(\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right)\left(\frac{\left.\frac{x+\sin x}{2}\right)}{x^{4}}\left(\frac{x-\sin x}{2}\right)\right.\right.$

$\lim\limits _{x \rightarrow 0}\left(\frac{x^{2}-\sin ^{2} x}{2 x^{4}}\right):\left(\frac{0}{0}\right)$

Apply $L-Hopital\; Rule$ :

$\lim \limits_{x \rightarrow 0} \frac{2 x-2 \sin x \cos x}{2.4 \cdot x^{3}}$

$\lim\limits _{x \rightarrow 0} \frac{2 x-\sin 2 x}{8 x^{3}} ; \frac{0}{0}:$ Again apply $L-Hopital \;rule$

$\lim \limits_{x \rightarrow 0} \frac{2-2 \cos (2 x)}{8(3) x^{2}}$

$\lim \limits_{x \rightarrow 0} \frac{2(1-\cos (2 x))}{24\left(4 x^{2}\right)} \times 4 \Rightarrow \frac{2}{24} \times \frac{1}{2} \times 4 \Rightarrow \frac{1}{6}$

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