Find the value of ′k ′ if lim(x→1)(x^4-1)/(x-1)=lim(x→k)(x^3-k^3)/x^2-k^2)

Question

Find the value of ′k ′ if :

\(\lim\limits_{x \to 1}\frac{x^4-1}{x-1}=\lim\limits_{x \to k}\frac{x^3-k^3}{x^2-k^2}\)

Answer 1

Given,

\(\lim\limits_{x \to 1}\frac{x^4-1}{x-1}=\lim\limits_{x \to k}\frac{x^3-k^3}{x^2-k^2}\) 

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

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

⇒ \(\lim\limits_{x \to 1}{(x+1)(x^2+1)}=\frac{(k^2+k^2+k^2)}{(k+k)}\) 

⇒ (1 +1)(1²+1) = \(\frac{3k^2}{2k}\) 

⇒ 4 = \(\frac{3k}{2}\) 

⇒ k = \(\frac{8}{3}\)