Question Number 98450 by bemath last updated on 14/Jun/20
![Find the supremum and the infimum of f(x) = (x/(sin x)) ,x∈ (0,(π/2) ]](https://www.tinkutara.com/question/Q98450.png)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{supremum}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{infimum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\:,\mathrm{x}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right] \\ $$
Commented by bobhans last updated on 14/Jun/20
![f ′(x)=((sin x−xcos x)/(sin^2 x)) ___(1) take g(x) = sin x−xcos x ; x∈ [0,(π/2) ] g ′(x)= xsin x > 0 on [0 ,(π/2) ] . hence g(x) increasing in [ 0,(π/2) ]. g(0) < g(x) ∴ g(x) =sin x−xcos x > 0 from (1) f′(x) > 0 for x∈(0,(π/2) ] ∴ infimum = lim_(x→0) f(x) = lim_(x→0) (x/(sin x)) = 1 supremum = f((π/2)) = ((π/2)/(sin (π/2))) = (π/2) ■](https://www.tinkutara.com/question/Q98451.png)
$$\mathrm{f}\:'\left(\mathrm{x}\right)=\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{xcos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:\:\_\_\_\left(\mathrm{1}\right) \\ $$$$\mathrm{take}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{xcos}\:\mathrm{x}\:;\:\mathrm{x}\in\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right]\: \\ $$$$\mathrm{g}\:'\left(\mathrm{x}\right)=\:\mathrm{xsin}\:\mathrm{x}\:>\:\mathrm{0}\:\mathrm{on}\:\left[\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\:\right]\:.\:\mathrm{hence}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{increasing}\:\mathrm{in}\:\left[\:\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right].\:\mathrm{g}\left(\mathrm{0}\right)\:<\:\mathrm{g}\left(\mathrm{x}\right)\: \\ $$$$\therefore\:\mathrm{g}\left(\mathrm{x}\right)\:=\mathrm{sin}\:\mathrm{x}−\mathrm{xcos}\:\mathrm{x}\:>\:\mathrm{0}\: \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\:\mathrm{f}'\left(\mathrm{x}\right)\:>\:\mathrm{0}\:\mathrm{for}\:\mathrm{x}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\:\right]\: \\ $$$$\therefore\:\mathrm{infimum}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\:=\:\mathrm{1} \\ $$$$\mathrm{supremum}\:=\:\mathrm{f}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\frac{\frac{\pi}{\mathrm{2}}}{\mathrm{sin}\:\frac{\pi}{\mathrm{2}}}\:=\:\frac{\pi}{\mathrm{2}}\:\blacksquare \\ $$