Помогите пожалуйста решить предел

Решите задачу:

$\displaystyle\lim_{x\to +\infty}\Bigg(\frac{6x^3-x^2+1}{2x^2-6x}-\frac{3x^2}{x-3}\Bigg)=\lim_{x\to +\infty}\Bigg(\frac{6x^3-x^2+1}{2x(x-3)}-\frac{3x^2}{x-3}\frac{2x}{2x}\Bigg)=$
$\displaystyle =\lim_{x\to +\infty}\Bigg(\frac{6x^3-x^2+1-6x^3}{2x(x-3)}\Bigg)=\lim_{x\to +\infty}\bigg(\frac{1-x^2}{2x^2-6x}\bigg)=$
$\displaystyle =\lim_{x\to +\infty}\bigg(\frac{d(1-x^2)}{d(2x^2-6x)}\bigg)=\lim_{x\to +\infty}\bigg(\frac{2x}{6-4x}\bigg)=\lim_{x\to +\infty}\bigg(\frac{x}{3-2x}\bigg)=$
$\displaystyle =\lim_{x\to +\infty}\bigg(\frac{dx}{d(3-2x)}\bigg)=\boxed{-\frac{1}{2}}\phantom{.}.$