If $\lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} = \lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}$, find the value of k.
$\begin{aligned}
\lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} & = \lim\limits_{x \to 1} \dfrac{x^4 -1^4}{x-1} \\
& = 4 \times 1^{4-1} \;\; \left[\text{Note: } \lim\limits_{x \to a} \dfrac{x^n - a^n}{x-a} = na^{n-1}\right] \\
& = 4
\end{aligned}$
$\begin{aligned}
\lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2} & = \lim\limits_{x \to k} \; \dfrac{\dfrac{x^3 - k^3}{x-k}}{\dfrac{x^2 - k^2}{x-k}} \\
& = \dfrac{\lim\limits_{x \to k} \dfrac{x^3-k^3}{x-k}}{\lim\limits_{x \to k} \dfrac{x^2-k^2}{x-k}} = \dfrac{3k^{3-1}}{2k^{2-1}} = \dfrac{3k}{2}
\end{aligned}$
$\therefore$ $\lim\limits_{x \to 1} \dfrac{x^4 -1}{x-1} = \lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2}$ $\implies$ $4=\dfrac{3k}{2}$ $\implies$ $k = \dfrac{8}{3}$