QJEEM22Ma0a52: JEE Main 2022 Mathematics - Class 12 JEEM22Ma0a52 Solution

$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}$

$= \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}$

$= \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}$

Let $x = 1 + h$

$\therefore$ when x $\to$ 1 then h $\to$ 0

$= \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi (1 + h))} \over {{{(1 + h - 1)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi h)} \over {{h^2}}}$

$= \mathop {\lim }\limits_{h \to 0} {\pi ^2} \times {{{{\sin }^2}(\pi h)} \over {{{(\pi h)}^2}}}$

$= {\pi ^2} \times 1$

$= {\pi ^2}$