Answer
$\frac{8}{5}$
Work Step by Step
Factor each polynomial:
$$\lim_{x\to 2}\frac{x^3-2x^2+4x-8}{x^2+x-6}\\=\lim_{x\to 2}\frac{(x-2)(x^2+4)}{(x-2)(x+3)}.$$
Cancel the common factors: $$\lim_{x\to 2}\frac{(x^2+4)}{(x+3)}\\=\frac{(2^2+4)}{(2+3)}\\=\frac{(4+4)}{5}\\=\frac{8}{5}$$