# What is the limit of  [x^2-5x+6]/ [x^3-8] as x goes to infinity?

$0$
At infinity, the ${x}^{3}$ term in the denominator dominates the function as it grows much quicker than all the other terms.
${\lim}_{x \to \infty} \frac{{x}^{2} - 5 x + 6}{{x}^{3} - 8} = {\lim}_{x \to \infty} {x}^{2} / {x}^{3} = {\lim}_{x \to \infty} \frac{1}{x} = 0$