Answer
$\dfrac{1}{3}$
Work Step by Step
Recall the limit rules:
$ \ Rule -\ 1 \ : \lim\limits_{x \to a} [A(x)]^n =[\lim\limits_{x \to a} A(x)]^n $
$\ Rule \ - \ 2 : \lim\limits_{x \to a} A(x)=A(a)$
$\ Rule \ - \ 3: \lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$
$\lim_{x\to 1}\dfrac{x-1}{x^3-1}\\=\lim_{x\to 1} \dfrac{(x-1)}{(x-1)(x^2+x+1)}$
Apply $\lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$
$\lim_{x\to 1}\dfrac{1}{(x^2+x+1)}\\=\dfrac{1}{(1^2+1+1)}\\=\dfrac{1}{3}$