Q268 / 270 · NDA 2020 · Mathematics

If $\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$, where $k \neq 0$, then what is the value of $k$ ?

  • A
    $\frac{2}{3}$
  • B
    $\frac{4}{3}$
  • Correct answer
    C
    $\frac{8}{3}$
  • D
    4

Solution

$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^4-1}{x-1} & =\lim _{x \rightarrow 1} \frac{(x-1)(x+1)\left(x^2+1\right)}{(x-1)} \\ & =\lim _{x \rightarrow 1}(x+1)\left(x^2+1\right)=(1+1)(1+1)=4 \end{aligned}$ Now, $\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}=4$ $\lim _{x \rightarrow k} \frac{(x-k)\left(x^2+k^2+x k\right)}{(x-k)(x+k)}=4$ $\begin{aligned} & \lim _{x \rightarrow k} \frac{\left(x^2+k^2+x k\right)}{(x+k)}=4 \\ & \frac{k^2+k^2+k \cdot k}{k+k}=4 \Rightarrow \frac{3}{2} k=4 \Rightarrow k=\frac{8}{3}\end{aligned}$

Hence, the correct answer is option: 'C'.

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