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Question Number 106771 by bemath last updated on 07/Aug/20

$$@\mathrm{bemath}@$$$$\underset{x\to \pi }{\lim }\left[\frac{4(\mathrm{x}-\pi )\cos {}^2\mathrm{x}}{\pi (\pi -2\mathrm{x})\cos (\mathrm{x}-\frac{\pi }{2})}\right]=?$$

Commented by bemath last updated on 07/Aug/20

Answered by john santu last updated on 07/Aug/20

$${}^{@\mathrm{JS}@}$$$$\underset{x\to \pi }{\lim }\left[\frac{\cos {}^2\mathrm{x}}{\pi (\pi -2\mathrm{x})}\right]\times \underset{x\to \pi }{\lim }\frac{4(\mathrm{x}-\pi )}{\cos (\mathrm{x}-\frac{\pi }{2})}=$$$$\frac{{(-1)}^2}{-{\pi }^2}\times \underset{x\to \pi }{\lim }\frac{4}{-\sin (\mathrm{x}-\frac{\pi }{2})}=$$$$\frac{1}{{\pi }^2}\times 4=\frac{4}{{\pi }^2}.\left(\mathrm{by}{\mathrm{L}}^{\prime }{\mathrm{Hopital}}^{\prime }\mathrm{s}\mathrm{rule}\right)$$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

$$\underset{x\to \pi }{\lim }\frac{4{cos}^{2}x}{\pi (\pi -2x)}\frac{(x-\pi )}{cos(x-\frac{\pi }{2})}$$$$\underset{x\to \pi }{\lim }-\frac{4}{{\pi }^2}\frac{x-\pi }{cos(\frac{\pi }{2}-x)}=\underset{x\to 0}{\lim }\frac{-4}{{\pi }^2}.\frac{x-\pi }{sinx}$$$$=\underset{x\to \pi }{\lim }\frac{4}{{\pi }^2}.\frac{\pi -x}{sin(\pi -x)}=\frac{4}{{\pi }^2}$$

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