Answer
$$\dfrac{5}{4}$$
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:
$$ \lim\limits_{x \to -3} \dfrac{x^2+x-6}{x^2+2x-3}=\lim\limits_{x \to -3} \dfrac{(x+3)(x-2)}{(x+3)(x-1)} \\=\dfrac{\lim\limits_{x \to -3} x-2}{\lim\limits_{x \to -3} x-1} \\=\dfrac{-3-2}{-3-1}\\=\dfrac{5}{4}$$
