JEE Main 2019 — Mathematics Calculus

$$\underset{x\rightarrow {1}^{+}}{\mathrm{lim}}\frac{(1-|x|+\mathrm{sin}|1-x|)\mathrm{sin}([1-x]\frac{\pi }{2})}{|1-x|[1-x]}$$x→1+lim​∣1−x∣[1−x](1−∣x∣+sin∣1−x∣)sin([1−x]2π​)​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(1-|(1+h)|+\mathrm{sin}|1-(1+h)|)\mathrm{sin}(\frac{\pi }{2}[1-(1+h)])}{|1-(1+h)|[1-(1+h)]}$$=h→0lim​∣1−(1+h)∣[1−(1+h)](1−∣(1+h)∣+sin∣1−(1+h)∣)sin(2π​[1−(1+h)])​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(1-h-1+\mathrm{sin}|(1-h-1)|)\mathrm{sin}(\frac{\pi }{2}[1-1-h])}{|(1-1-h)|[1-1-h]}$$=h→0lim​∣(1−1−h)∣[1−1−h](1−h−1+sin∣(1−h−1)∣)sin(2π​[1−1−h])​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(-h+\mathrm{sin}h)\mathrm{sin}(\frac{\pi }{2}[-h])}{|-h|[-h]}$$=h→0lim​∣−h∣[−h](−h+sinh)sin(2π​[−h])​

$$=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(-h+\mathrm{sin}h)\mathrm{sin}(-\frac{\pi }{2})}{h(-1)}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{(-h+\mathrm{sin}h)(-1)}{-h}$$=h→0lim​h(−1)(−h+sinh)sin(−2π​)​=h→0lim​−h(−h+sinh)(−1)​

$=1-1=0$=1−1=0.