Answer
$\lim\limits_{n \to \infty} \frac{3n^3-1}{2n^3+1} = \frac{3}{2}$
Work Step by Step
$\lim\limits_{n \to \infty} \frac{3n^3-1}{2n^3+1} = \lim\limits_{n \to \infty} \frac{\frac{3n^3}{n^3}-\frac{1}{n^3}}{\frac{2n^3}{n^3}+\frac{1}{n^3}} = \lim\limits_{n \to \infty} \frac{3-\frac{1}{n^3}}{2+\frac{1}{n^3}} = \frac{3-0}{2+0} = \frac{3}{2}$
