Mathematics - Limits Question with Solution | TestHub

$\underset{x\to 1^{+}}{\lim }\frac{\left(1-\left|x\right|+\sin \left|1-x\right|\right)\sin \left(\left[1-x\right]\frac{\mathrm{\pi }}{2}\right)}{\left|1-x\right|\left[1-x\right]}$

$=\underset{h\to 0}{\lim }\frac{\left(1-\left|\left(1+h\right)\right|+\sin \left|1-\left(1+h\right)\right|\right)\sin \left(\frac{\pi }{2}\left[1-\left(1+h\right)\right]\right)}{\left|1-\left(1+h\right)\right|\left[1-\left(1+h\right)\right]}$

$=\underset{h\to 0}{\lim }\frac{\left(1-h-1+\sin \left|\left(1-h-1\right)\right|\right)\sin \left(\frac{\mathrm{\pi }}{2}\left[1-1-h\right]\right)}{\left|\left(1-1-h\right)\right|\left[1-1-h\right]}$

$=\underset{h\to 0}{\lim }\frac{\left(-h+\sin h\right)\sin \left(\frac{\mathrm{\pi }}{2}\left[-h\right]\right)}{\left|-h\right|\left[-h\right]}$

$=\underset{h\to 0}{\lim }\frac{\left(-h+\sin h\right)\sin \left(-\frac{\mathrm{\pi }}{2}\right)}{h\left(-1\right)}=\underset{h\to 0}{\lim }\frac{\left(-h+\sin h\right)\left(-1\right)}{-h}$

$=1-1=0$.