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Question Number 98450 by bemath last updated on 14/Jun/20

$$\mathrm{Find}\mathrm{the}\mathrm{supremum}\mathrm{and}\mathrm{the}$$$$\mathrm{infimum}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\sin \mathrm{x}},\mathrm{x}\in (0,\frac{\pi }{2}]$$

Commented by bobhans last updated on 14/Jun/20

$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{\sin \mathrm{x}-\mathrm{xcos}\mathrm{x}}{\sin {}^2\mathrm{x}}\mathrm{\_}\mathrm{\_}\mathrm{\_}\left(1\right)$$$$\mathrm{take}\mathrm{g}\left(\mathrm{x}\right)=\sin \mathrm{x}-\mathrm{xcos}\mathrm{x};\mathrm{x}\in [0,\frac{\pi }{2}]$$$$\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)=\mathrm{xsin}\mathrm{x}>0\mathrm{on}[0,\frac{\pi }{2}].\mathrm{hence}\mathrm{g}\left(\mathrm{x}\right)$$$$\mathrm{increasing}\mathrm{in}[0,\frac{\pi }{2}].\mathrm{g}\left(0\right)<\mathrm{g}\left(\mathrm{x}\right)$$$$\therefore \mathrm{g}\left(\mathrm{x}\right)=\sin \mathrm{x}-\mathrm{xcos}\mathrm{x}>0$$$$\mathrm{from}\left(1\right){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)>0\mathrm{for}\mathrm{x}\in (0,\frac{\pi }{2}]$$$$\therefore \mathrm{infimum}=\underset{x\to 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{x\to 0}{\lim }\frac{\mathrm{x}}{\sin \mathrm{x}}=1$$$$\mathrm{supremum}=\mathrm{f}\left(\frac{\pi }{2}\right)=\frac{\frac{\pi }{2}}{\sin \frac{\pi }{2}}=\frac{\pi }{2}◼$$

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