Find-the-supremum-and-the-infimum-of-f-x-x-sin-x-x-0-pi-2-

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

Find the supremum and the

infimum of f (x) = \frac{x}{\sin x}, x \in (0, \frac{π}{2}]

Commented by bobhans last updated on 14/Jun/20

f^{'} (x) = \frac{\sin x - xcos x}{\sin^{2} x}___(1)

take g (x) = \sin x - xcos x; x \in [0, \frac{π}{2}]

g^{'} (x) = xsin x > 0 on [0, \frac{π}{2}] . hence g (x)

increasing in [0, \frac{π}{2}] . g (0) < g (x)

∴ g (x) = \sin x - xcos x > 0

from (1) f^{'} (x) > 0 for x \in (0, \frac{π}{2}]

∴ infimum = \lim_{x \to 0} f (x) = \lim_{x \to 0} \frac{x}{\sin x} = 1

supremum = f (\frac{π}{2}) = \frac{\frac{π}{2}}{\sin \frac{π}{2}} = \frac{π}{2}

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