Question

If $\displaystyle\lim_{x\to1} \frac{x^{4}-1}{x-1} =\lim_{x\to k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}} $, then k is :

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

Answer 1

Answer: (D)

$\displaystyle \lim_{x \to1} \frac{x^{4}-1}{x-1} =\lim_{x\to k} \frac{x^{3}-k^{3}}{x^{2} -k^{2}} $
$\displaystyle \Rightarrow \lim_{x\to1} \left(x+1\right)\left(x^{2}+1\right) = \frac{k^{2}+k^{2}+k^{2}}{2k } $
$ \Rightarrow k = 8/3 $