Answer
$$0$$
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:
$$ \lim\limits_{x \to -1} \dfrac{(x+1)^2}{x^2-1}=\lim\limits_{x \to -1} \dfrac{(x+1)(x+1)}{(x+1)(x-1)} \\=\dfrac{\lim\limits_{x \to -1} (x+1)}{\lim\limits_{x \to -1} x-1} \\=\dfrac{-1+1}{-1-1} \\=0$$
