#### Answer

The solution is
$$\lim_{x\to\infty}\frac{3x^4-x^2}{6x^4+12}=\frac{1}{2}.$$

#### Work Step by Step

To solve this limit follow the steps below:
$$\lim_{x\to\infty}\frac{3x^4-x^2}{6x^4+12}=\lim_{x\to\infty}\frac{x^4(3-\frac{x^2}{x^4})}{x^4(6+\frac{12}{x^4})}=\lim_{x\to\infty}\frac{3-\frac{1}{x^2}}{6+\frac{12}{x^4}}=\left[\frac{3-\frac{1}{\infty^2}}{6+\frac{12}{\infty^4}}\right]=\frac{3}{6}=\frac{1}{2}.$$