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Question: Answered & Verified by Expert
$$
\lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)=
$$
MathematicsLimitsTS EAMCETTS EAMCET 2022 (19 Jul Shift 1)
Options:
  • A 8
  • B $\frac{1}{6}$
  • C $\frac{1}{24}$
  • D $\frac{2}{3}$
Solution:
1598 Upvotes Verified Answer
The correct answer is: $\frac{1}{24}$
$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right) \\ & \lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}\right)\left(1-\cos \frac{x^2}{4}\right)\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{4 !}{x^8}\left(1-1+2 \sin ^2 \frac{x^2}{6}\right)\left(1-1+2-\sin ^2 \frac{x^2}{8}\right)\right) \\ & \left.=4 !\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x^2}{6}}{\frac{x^2}{6}}\right) \times \frac{2}{36} \times\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}\right)^2 \times \frac{2}{64}\right] \\ & =4 \times 3 \times 2\left[1 \times \frac{2}{36} \times 1 \times \frac{2}{64}\right]=\frac{1}{24}\end{aligned}$

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