Mathematics - Limits Question with Solution | TestHub

Put $x=\frac{\pi }{2}-h$ as $x\to \frac{\pi }{2},h\to 0$ $\therefore$ Given limit $={\lim }_{h\to 0}\frac{1-\tan \left(\frac{\pi }{4}-\frac{\mathrm{h}}{2}\right)}{1+\tan \left(\frac{\pi }{4}-\frac{\mathrm{h}}{2}\right)}\cdot \frac{(1-\cosh )}{(2\mathrm{h}{)}^3}$ $={\lim }_{\mathrm{h}\to 0}\tan \frac{\mathrm{h}}{2}\frac{2{\sin }^2\frac{\mathrm{h}}{2}}{8{\mathrm{h}}^3}$ $={\lim }_{\mathrm{h}\to 0}\frac{1}{4}\cdot \frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}\times 2}{\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)}^2\times \frac{1}{4}$ $={\lim }_{\mathrm{h}\to 0}\frac{1}{32}\cdot \left(\frac{\tan \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}\times 2}\right){\left(\frac{\sin \frac{\mathrm{h}}{2}}{\frac{\mathrm{h}}{2}}\right)}^2=\frac{1}{32}$