Answer
$\lim\limits_{n \to \infty} \frac{n^{12}}{3n^{12}+4}=\frac{1}{3}$
Work Step by Step
$\lim\limits_{n \to \infty} \frac{n^{12}}{3n^{12}+4} = \lim\limits_{n \to \infty} \frac{\frac{n^{12}}{n^{12}}}{\frac{3n^{12}}{n^{12}}+\frac{4}{n^{12}}} = \lim\limits_{n \to \infty}\frac{1}{3 + \frac{4}{n^{12}}} = \frac{1}{3+0} = \frac{1}{3}$
