JEE Main 201910 Apr, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002If limx→1x4−1x−1=limx→kx3−k3x2−k2\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to k} {{{x^3} - {k^3}} \over {{x^2} - {k^2}}}x→1limx−1x4−1=x→klimx2−k2x3−k3, then k is :AOption A. 32{3 \over 2}23BOption B. 83{8 \over 3}38COption C. 43{4 \over 3}34DOption D. 38{3 \over 8}83CheckSkip · show solution
