Answer
$\lim\limits_{n \to \infty}\frac{n^3}{n^4 + 1} = 0$
Work Step by Step
$\lim\limits_{n \to \infty}\frac{n^3}{n^4 + 1} = \lim\limits_{n \to \infty}\frac{\frac{n^3}{n^4}}{\frac{n^4}{n^4} + \frac{1}{n^4}} = \frac{\frac{1}{\infty}}{1+\frac{1}{\infty}}= \frac{0}{1+0} = 0$