## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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