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Question: Answered & Verified by Expert
\(\lim _{x \rightarrow \pi} \frac{1-\sin x / 2}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}=\)
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Options:
  • A \(\frac{3}{\sqrt{2}}\)
  • B \(\frac{-1}{\sqrt{2}}\)
  • C \(\frac{1}{\sqrt{2}}\)
  • D \(\frac{5}{\sqrt{2}}\)
Solution:
1182 Upvotes Verified Answer
The correct answer is: \(\frac{1}{\sqrt{2}}\)
\(\lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}\)
Let \(\quad x=\pi+h, x \rightarrow \pi, h \rightarrow 0\)
\(\begin{gathered}
=\lim _{\substack{x \rightarrow \pi \\
h \rightarrow 0}} \frac{1-\sin \left(\frac{\pi}{2}+\frac{h}{2}\right)}{\cos \left(\frac{\pi}{2}+\frac{h}{2}\right)\left(\cos \left(\frac{\pi+h}{4}\right)-\sin \left(\frac{\pi+h}{4}\right)\right)} \\
=\lim _{h \rightarrow 0} \frac{1-\cos \frac{h}{2}}{-\sin \frac{h}{2} \cdot\left(\cos \frac{\pi}{4} \cos \frac{h}{4}-\sin \frac{\pi}{4} \sin \frac{h}{4}-\sin \frac{\pi}{4} \cdot \cos \frac{h}{4}-\cos \frac{\pi}{4} \cdot \sin \frac{h}{4}\right)}
\end{gathered}\)
\(\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\sqrt{2}\left(1-\cos \frac{h}{2}\right)}{-\sin \frac{h}{2}\left\{\cos \frac{h}{4}-\sin \frac{h}{4}-\cos \frac{h}{4}-\sin \frac{h}{4}\right\}} \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{2}\left(1-\cos \frac{h}{2}\right)}{\sin \frac{h}{2} \cdot\left(-2 \sin \frac{h}{4}\right)} \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{2} \cdot 2 \sin ^2 \frac{h}{4}}{2 \cdot \sin \frac{h}{2} \times \sin \frac{h}{4}} \\ & =\lim _{h \rightarrow 0} \sqrt{2} \frac{\sin \frac{h}{4}}{\sin \frac{h}{2}}=\sqrt{2} \lim _{h \rightarrow 0} \frac{\sin \frac{h}{4}}{4\left(\frac{h}{4}\right)} \cdot \frac{2(h / 2)}{\sin \frac{h}{2}} \\ & =\sqrt{2} \times \frac{2}{4} \cdot \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{4}}{\frac{h}{4}}\right)\left(\frac{\frac{h}{2}}{\sin \frac{h}{2}}\right) \\ & =\sqrt{2} \times \frac{2}{4} \times 1 \times 1=\frac{1}{\sqrt{2}}\end{aligned}\)

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