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

Solution:

we have,

$\underset{x\to 1}{lim} \frac{x^4-1}{x-1}=\underset{x\to 1}{lim} \frac{x^4-1^4}{x-1}=4(1{)}^{4-1}=4$ 

and, 

$\begin{array}{l}\underset{x\to k}{lim} \frac{x^3-k^3}{x^2-k^2}\\ =\underset{x\to k}{lim} \frac{x^3-k^3}{x-k}\times \frac{x-k}{x^2-k^2}\\ =\underset{x\to k}{lim} \frac{x^3-k^3}{x-k}+\frac{x^2-k^2}{x-k}\\ =\underset{x\to k}{lim} \frac{x^3-k^3}{x-k}÷\underset{x\to k}{lim} \frac{x^2-k^2}{x-k}=\frac{3k^2}{2k}=\frac{3}{2}k\end{array}$ 

$\therefore \underset{x\to 1}{lim} \frac{x^4-1}{x-1}=\underset{x\to k}{lim} \frac{x^3-k^3}{x^2-k^2}\Rightarrow 4=\frac{3k}{2}\Rightarrow k=\frac{8}{3}$