_ x → (π)/(2) x - x (π - 2x) ^ 3 equals

Put, $x = \dfrac{\pi}{2} - h$ in the given limit, we get

$= \lim_{h \to 0}\left[\dfrac{\cot \left(\dfrac{\pi}{2} - h\right) - \cos \left(\dfrac{\pi}{2} - h\right)}{{\left(\pi - 2 \times \left(\dfrac{\pi}{2} - h\right)\right)}^{3}}\right]$

$= \lim_{h \to 0}\dfrac{tanh - sinh}{{\left(\pi - 2 \times \dfrac{\pi}{2} + 2h\right)}^{3}}$

$= \lim_{h \to 0}\left[\dfrac{tanh - sinh}{8{h}^{3}}\right]$

$= \dfrac{1}{8}\lim_{h \to 0}\dfrac{sinh}{h}\left[\dfrac{sech - 1}{{h}^{2}}\right]$

$= \dfrac{1}{8}\left[\lim_{h \to 0}\left(\dfrac{sinh}{h}\right)\right]\left[\lim_{h \to 0}\left(\dfrac{1 - cosh}{\left(cosh\right){h}^{2}}\right)\right]$

$= \dfrac{1}{8} \times 1 \times \left[\lim_{h \to 0}\left(\dfrac{1 - cosh}{\left(cosh\right){h}^{2}}\right)\right]$

$= \dfrac{1}{8}\lim_{h \to 0}\left[\dfrac{2\sin^{2}\left(\dfrac{h}{2}\right)}{\left(cosh\right){h}^{2}}\right]$

$= \dfrac{1}{8}\lim_{h \to 0}\left[\dfrac{2\sin^{2}\left(\dfrac{h}{2}\right)}{\left(\dfrac{{h}^{2}}{4}\right) \times 4 \times cosh}\right]$

$= \dfrac{1}{16}$.