Verified Solution • 2025–26 Session moderate

QJEEM19Ma7676: JEE Main 2019 Mathematics - Class 12 JEEM19Ma7676 Solution: If , then k is :

If lim⁡x→1x4−1x−1=lim⁡x→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}}} , then k is :

Multiple Choice Options

  • A 32{3 \over 2}
  • B 83{8 \over 3}
  • C 43{4 \over 3}
  • D 38{3 \over 8}

Step-by-Step Verified Solution

If lim⁡x→1x4−1x−1=lim⁡x→K(x3−k3x2−k2)\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)

L·H·S·

Ltx→1x4−1x−1=(00form)\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)

⇒Ltx→14x31=4\Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4

Now, lim⁡x→K(x3−k3x2−k2)\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right) = 4

⇒lim⁡x→K3x22x=4\Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4

⇒32k=4⇒k=83\Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}

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