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Question: Answered & Verified by Expert
\(\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\) \(\left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\}=\)
MathematicsLimitsJEE Main
Options:
  • A \(\frac{1}{16}\)
  • B \(\frac{1}{32}\)
  • C \(\frac{1}{64}\)
  • D \(\frac{1}{8}\)
Solution:
1059 Upvotes Verified Answer
The correct answer is: \(\frac{1}{32}\)
\(\begin{aligned}
\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left\{1-\cos \left(\frac{x^2}{2}\right)-\right. & \cos \left(\frac{x^2}{4}\right) \\
& \left.+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right\}
\end{aligned}\)
\(\begin{gathered}\Rightarrow \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left\{\left[1-\cos \left(\frac{x^2}{2}\right)\right]\left[-\cos \left(\frac{x^2}{4}\right)\right]\left[1-\cos ^2\left(\frac{x^2}{2}\right)\right]\right\} \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left[1-\cos \left(\frac{x^2}{2}\right)\right]\left(1-\cos \left(\frac{x^2}{4}\right)\right) \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left(2 \sin ^2\left(\frac{x^2}{4}\right)\right)\left(2 \sin ^2\left(\frac{x^2}{8}\right)\right) \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{32 \sin ^2\left(\frac{x^2}{4}\right) \cdot \sin ^2\left(\frac{x^2}{8}\right)}{\sin ^8 x} \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{32 \cdot \sin ^2\left(\frac{x^2}{4}\right) \cdot \sin ^2\left(\frac{x^2}{8}\right) \times x^8}{\left(\frac{x^2}{4} \times \frac{x^2}{8}\right)^2 \cdot \sin ^8 x \times(4 \times 8)^2}=\frac{1}{32}\end{gathered}\)

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