Question

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 $k$ is :

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

Answer 1

The correct option is B $\frac{8}{3}$

$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}=\underset{x\to 1}{lim}\frac{(x-1)x^3+x^2+x+1}{(x-1)}=\underset{x\to 1}{lim}\frac{x^3+x^2+x+1}{1}=4$

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