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 $k=?$

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{8}{3}$
• D. none

Answer 1

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

$\underset{x\to 1}{lim}\frac{x^4-1}{x-1}={4.1}^{4-1}=4$

$\underset{x\to k}{lim}\frac{x^3-k^3}{x^2-k^2}=\underset{x\to k}{lim}\frac{x^3-k^3}{x-k}.\underset{x\to k}{lim}\frac{x-k}{x^2-k^2}$

$=3k^{3-1}\frac{1}{2k^{2-1}}=\frac{3}{2}k$

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

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