Answer
$$\dfrac{8}{5}$$
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:
$$ \lim\limits_{x \to 2} \dfrac{(x^3-2x^2+4x-8)}{(x^2+x-6)}=\lim\limits_{x \to 2} \dfrac{(x-2)(x^2+4)}{(x-2)(x+3)} \\=\dfrac{\lim\limits_{x \to 2} (x^2+4)}{\lim\limits_{x \to 2} (x +3)}\\=\dfrac{(2)^2+4}{2+3} \\=\dfrac{4+4}{5} \\=\dfrac{8}{5}$$
