Q61 / 150 · MHT CET 2007 · Mathematics
The value of $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to
- A$1 / 5$
- Correct answerB$1 / 6$
- C$1 / 4$
- D$1 / 2$
Solution
$$
\begin{array}{l}
\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}} \\
=\lim _{x \rightarrow 0} \frac{2 \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}} \\
=2 \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)} \times \frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right. \\
\left.\times\left(\frac{x+\sin x}{2 x}\right)\left(\frac{x-\sin x}{2 x^{3}}\right)\right] \\
=2 \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\frac{x+\sin x}{2}} \times \frac{\sin \left(\frac{x-\sin x}{2}\right)}{\frac{x-\sin x}{2}}\right. \\
=\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{x-\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots\right)}{x^{3}} \\
=\lim _{x \rightarrow 0}\left(\frac{1}{3 !}-\frac{x^{2}}{5 !}+\ldots\right)=\frac{1}{6}
\end{array}
$$
Hence, the correct answer is option: 'B'.