Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q70
If lim_x → 1 (x^4-1)/(x-1)= lim_x → k (x^3- k^3)/(x^2- k^2), then k is:
If $\displaystyle \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow \mathrm{k}} \frac{x^3-\mathrm{k}^3}{x^2-\mathrm{k}^2}$, then k is :
Official answer
(2)
$\displaystyle \frac{8}{3}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.