JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 3)

The value of $$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}$$ is equal to:
$${{{\pi ^2}} \over 6}$$
$${{{\pi ^2}} \over 3}$$
$${{{\pi ^2}} \over 2}$$
$$\pi$$2

Explanation

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

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

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

Let $$x = 1 + h$$

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

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

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

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

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

$$ = {\pi ^2}$$

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