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

$\dfrac{3}{2}$
Recall the limit rules: $\ Rule -\ 1 \ : \lim\limits_{x \to a} [A(x)]^n =[\lim\limits_{x \to a} A(x)]^n$ $\ Rule \ - \ 2 : \lim\limits_{x \to a} A(x)=A(a)$ $\ Rule \ - \ 3: \lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$ $\lim\limits_{x\to 2}\dfrac{x^3-8}{x^3--2x^2+4x-8}\\=\lim\limits_{x\to 2} \dfrac{(x-2)(x^2+2x+4)}{(x-2)(x^2+4)}$ Apply $\lim\limits_{x \to a} \dfrac{A(x)}{B(x)}=\dfrac{\lim\limits_{x \to a} A(x)}{\lim\limits_{x \to a} B(x)}$ $\lim\limits_{x\to 2}\dfrac{x^2+2x+4}{x^2+4}=\dfrac{4+4+4}{4+4} \\=\dfrac{3}{2}$