Answer
$\dfrac{8}{5}$.
Work Step by Step
As we can see from the attached graph, as $x$ gets closer and closer to $2$ from the left and the right, the value of the function (the $y$-value) gets closer and closer to $\frac{8}{5}$. Thus the limit is
$\displaystyle \lim_{x\to2}\frac{x^3-2x^2+4x-8}{x^2+x-6}=\frac{8}{5}$
