Q106 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics
$$
\lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)=
$$
- A8
- B$\frac{1}{6}$
- Correct answerC$\frac{1}{24}$
- D$\frac{2}{3}$
Solution
$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right) \\ & \lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}\right)\left(1-\cos \frac{x^2}{4}\right)\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-1+2 \sin ^2 \frac{x^2}{6}\right)\left(1-1+2-\sin ^2 \frac{x^2}{8}\right)\right) \\ & \left.=4 !\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x^2}{6}}{\frac{x^2}{6}}\right) \times \frac{2}{36} \times\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}\right)^2 \times \frac{2}{64}\right] \\ & =4 \times 3 \times 2\left[1 \times \frac{2}{36} \times 1 \times \frac{2}{64}\right]=\frac{1}{24}\end{aligned}$
Hence, the correct answer is option: 'C'.