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Limits Continuity and Differentiability question

2022 · 29 Jun · Shift 2 · Q26

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JEE Main
Mathematics
Limits Continuity and Differentiability
2022 · 29 Jun · Shift 2 · Q26

Limits Continuity and Differentiability question

2022 · 29 Jun · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

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:

A
${{{\pi ^2}} \over 6}$
B
${{{\pi ^2}} \over 3}$
C
${{{\pi ^2}} \over 2}$
D
$\pi$ 2

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