If \lim_{x \to 1} \frac{x^4-1}{x-1}=\lim_{x \to k} \frac{x^3-k^3}{x^2-k^2} then k=

NIMCET 2023 Question Paper with solutions

Question : 32 of 120

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\lim ↙{x → 1} \;{x^4-1}/{x-1}=\lim ↙{x → k} \;{x^3-k^3}/{x^2-k^2}

then

k=

Solution:

Using L'Hospital rule,

\lim ↙{x → 1} \;{4 x^3}/{1}\;=\lim ↙{x → k} \;{3 x^2}/{2 x}

⇒ 4\;=\;{3}/{2} k

k\;=\;{8}/{3} .

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