Answer
$3$
Work Step by Step
In order to simplify the given expression, we will use the following rules.
$(a) \lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)} \\ (b) \lim\limits_{x \to a} k(x)=k(a)$ ;
where $a$ is a constant.
---
Thus, we have:
$$ \lim\limits_{x \to 1} \dfrac{x^3-1}{x-1}=\lim\limits_{x \to 1} \dfrac{(x-1)(x^2+x+1)}{(x-1)} \\=\lim\limits_{x \to 1} x^2+x+1 \\=(1)^2+1+1 \\=3$$
