find the value of k if lim x tends to 1 x^4-1/x-1 = lim x tends to k x^3-k^3/x^2-k^2

In the given question, we need to find the value of k if $\underset{x\to 1}{\lim }\frac{x^4-1}{x-1}=\underset{x\to k}{\lim }\frac{x^3-k^3}{x^2-k^2}$
Here, let us first evaluate $\underset{x\to 1}{\lim }\frac{x^4-1}{x-1}$. So,
$$\begin{gathered} \underset{x\to 1}{\lim }\frac{x^4-1}{x-1}=\underset{x\to 1}{\lim }\frac{{\left(x^2\right)}^2-1^2}{x-1} \\ =\underset{x\to 1}{\lim }\frac{\left(x^2+1\right)\left(x^2-1\right)}{x-1} \left[\mathit{Using}\mathit{ }{\mathit{a}}^{\mathit{2}}\mathit{-}{\mathit{b}}^{\mathit{2}}\mathit{=}\left(\mathit{a}\mathit{+}\mathit{b}\right)\left(\mathit{a}\mathit{-}\mathit{b}\right)\right] \\ =\underset{x\to 1}{\lim }\frac{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{x-1} \\ =\underset{x\to 1}{\lim }\left(x^2+1\right)\left(x+1\right) \\ =\left(1^2+1\right)\left(1+1\right) \\ =\left(2\right)\left(2\right) \\ =4 \end{gathered}$$
Next, evaluate $\underset{x\to k}{\lim }\frac{x^3-k^3}{x^2-k^2}$. So,
$$\begin{gathered} \underset{x\to k}{\lim }\frac{x^3-k^3}{x^2-k^2}=\underset{x\to k}{\lim }\frac{\left(x-k\right)\left(x^2+xk+k^2\right)}{x^2-k^2} \left[\mathrm{Using} a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ \underset{ x\to k}{ =\lim }\frac{\left(x-k\right)\left(x^2+xk+k^2\right)}{\left(x+k\right)\left(x-k\right)} \left[\mathrm{Using} a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ =\underset{x\to k}{\lim }\frac{x^2+xk+k^2}{x+k} \\ =\frac{k^2+\left(k\right)\left(k\right)+k^2}{k+k} \\ =\frac{3k^2}{2k} \end{gathered}$$
Now, as it is given that $\underset{x\to 1}{\lim }\frac{x^4-1}{x-1}=\underset{x\to k}{\lim }\frac{x^3-k^3}{x^2-k^2}$. So,
$$\begin{gathered} 4=\frac{3k^2}{2k} \\ \left(4\right)\left(2k\right)=3k^2 \\ 8k=3k^2 \\ 3k^2-8k=0 \\ k\left(3k-8\right)=0 \\ \mathrm{So}, \\ k=0 \\ \mathrm{Also}, \\ 3k-8=0 \\ 3k=8 \\ k=\frac{8}{3} \end{gathered}$$
Substituting k = 0 in $\frac{3k^2}{2k}$. We get
$$\begin{gathered} \frac{3k^2}{2k}=\frac{3{\left(0\right)}^2}{2\left(0\right)} \\ =0 \end{gathered}$$
Next substitute $k=\frac{8}{3}$in $\frac{3k^2}{2k}$. We get
$$\begin{gathered} \frac{3k^2}{2k}=\frac{3{\left({\displaystyle \frac{8}{3}}\right)}^2}{2\left({\displaystyle \frac{8}{3}}\right)} \\ =\frac{3\left({\displaystyle \frac{64}{9}}\right)}{{\displaystyle \frac{16}{3}}} \\ =\frac{{\displaystyle \frac{64}{3}}}{{\displaystyle \frac{16}{3}}} \\ =4 \end{gathered}$$
Therefore, k = 8/3