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 find the value of $k$.

• A. $\frac{5}{3}$
• B. $-\frac{8}{3}$
• C. $\frac{8}{3}$
• D. $Noneofthese$

Answer 1

Answer: (C)

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

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

$=(1+1)(1+1)$

$=\left(2\right)\left(2\right)$

$=4$ ...(1)

Now $\underset{x\to 1}{lim}\frac{x^3-k^3}{x^2-k^2}$

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

$=x^2+\frac{k^2+k^2+k^2}{2k}$

$=\frac{3k^2}{2k}=\frac{3k}{2}$...(2)

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

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