Mathematics - Limits Question with Solution | TestHub

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

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

$=\underset{h\to 0}{\lim }\frac{\tan h-\sin h}{{\left(\pi -2\times \frac{\pi }{2}+2h\right)}^3}$

$=\underset{h\to 0}{\lim }\left[\frac{\tan h-\sin h}{8h^3}\right]$

$= \frac{1}{8} \underset{h\to 0}{\lim }\frac{\sin h}{h}\left[\frac{\sec h-1}{h^2}\right]$

$=\frac{1}{8} \left[\underset{h\to 0}{\lim }\left(\frac{\sin h}{h}\right)\right]\left[\underset{h\to 0}{\lim }\left(\frac{1-\cos h}{\left(\cos h\right) h^2}\right)\right]$

$=\frac{1}{8} \times 1\times \left[\underset{h\to 0}{\lim }\left(\frac{1-\cos h}{\left(\cos h\right) h^2}\right)\right]$

$=\frac{1}{8} \underset{h\to 0}{\lim }\left[\frac{2{\sin }^2\left({\displaystyle \frac{h}{2}}\right)}{\left(\cos h\right) h^2}\right]$

$=\frac{1}{8} \underset{h\to 0}{\lim }\left[\frac{2{\sin }^2\left(\frac{h}{2}\right)}{\left(\frac{h^2}{4}\right)\times 4\times \cos h}\right]$

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