JEE Main 202229 Jun, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002The value of limx→1(x2−1)sin2(πx)x4−2x3+2x−1\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^3} + 2x - 1}}x→1limx4−2x3+2x−1(x2−1)sin2(πx) is equal to:AOption A. π26{{{\pi ^2}} \over 6}6π2BOption B. π23{{{\pi ^2}} \over 3}3π2COption C. π22{{{\pi ^2}} \over 2}2π2DOption D. π\piπ²CheckSkip · show solution
