If \lim_{x\rightarrow 1}\frac{x^{4}-1}{x-1}=\lim_{x\rightarrow k}\frac{x^{3}-k^{3}}{x^{2}-k^{2}} , then k is:

  • Option 1)

    \frac{8}{3}

  • Option 2)

    \frac{3}{8}

  • Option 3)

    \frac{3}{2}

  • Option 4)

    \frac{4}{3}

 

Answers (1)
V Vakul

\lim_{x\rightarrow 1}\frac{x^{4}-1}{x-1}=\lim_{x\rightarrow 1}4x^{3}=4

\lim_{x\rightarrow k}\frac{x^{3}-k^{3}}{x^{2}-k^{2}}=\lim_{x\rightarrow k}\frac{x^{3}-k^{3}}{x-1}\times \frac{x-1}{x^{2}-k^{2}}=\frac{3k^{2}}{2k}=4

k=\frac{8}{3}

So, option (1) is correct.

 


Option 1)

\frac{8}{3}

Option 2)

\frac{3}{8}

Option 3)

\frac{3}{2}

Option 4)

\frac{4}{3}

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