Question

Find K where
$\underset{x\to k}{lim}\left[\frac{x^4-1}{x-1}\right]=\underset{x\to k}{lim}\left(\frac{x^3-k^3}{x^2-k^2}\right)$

Answer 1

$\underset{x\to 1}{Lt}\frac{x^4-1}{x-1}=\underset{x\to 1}{Lt}\frac{(x^2{)}^2-1^2}{x-1}=\underset{x\to 1}{Lt}\frac{(x^2+1)(x^2-1)}{x-1}=\underset{x\to 1}{Lt}(x^2+1)(x+1)$

$=(1^2+1)(1+1)=\left(2\right)\left(2\right)=4$

$\underset{x\to k}{Lt}\frac{x^3-k^3}{x^2-k^2}=\underset{x\to k}{Lt}\frac{(x-k)(x^2+xk+k^2)}{x^2-k^2}=\underset{x\to k}{Lt}\frac{(x-k)(x^2+xk+k^2)}{(x+k)(x-k)}$

$=\underset{x\to k}{Lt}\frac{x^2+xk+k^2}{x+k}=\frac{k^2+\left(k\right)\left(k\right)+k^2}{k+k}=\frac{3k^2}{2k}$

$\underset{x\to 1}{Lt}\frac{x^4-1}{x-1}=\underset{x\to 1}{Lt}\frac{x^3-k^3}{x^2-k^2}$ so

$4=\frac{3k^2}{2k}$

$4\left(2k\right)=3k^2$

$8k=3k^2$

$\Rightarrow 3k^2-8k=0$

$\Rightarrow k(3k-8)=0$

$k=\frac{8}{3}$ & $0$

If $k=0$ in $\frac{3k^2}{2k}$ we get

$\frac{3k^2}{2k}=\frac{3(0{)}^2}{2\left(0\right)}=0$

If $k=\frac{8}{3}$ in $\frac{3k^2}{2k}$ we get

$\frac{3k^2}{2k}=\frac{3{\left(\frac{8}{3}\right)}^2}{2\left(\frac{8}{3}\right)}=\frac{3\left(\frac{64}{9}\right)}{\frac{16}{3}}=\frac{\frac{64}{3}}{\frac{16}{3}}=4$

$k=\frac{8}{3}$.