Answer
$$\dfrac{1}{2}$$
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.
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Thus, we have:
$$ \lim\limits_{x \to -1} \dfrac{x^2+x}{x^2-1}=\lim\limits_{x \to -1} \dfrac{x(x+1)}{(x-1)(x+1)} \\=\dfrac{\lim\limits_{x \to -1} x}{\lim\limits_{x \to -1} x-1} \\=\dfrac{-1}{-1-1}\\=\dfrac{1}{2}$$
