Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =\lim_{x\to \infty}\frac{x^3\left(2+0-0\right)}{x^3\left(3-0-0+0\right)} \\
\ =\frac{2}{3}$$

You are correct, if you have the ratio of two polynomial of the same degree $n$ then
$$\lim_{x\to +\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots +a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots +b_0}=\lim_{x\to +\infty}\frac{x^n(a_n+\frac{a_{n-1}}{x}+\dots +\frac{a_0}{x^n})}{x^n(b_n+\frac{b_{n-1}}{x}+\dots +\frac{b_0}{x^n})}\\=\lim_{x\to +\infty}\frac{a_n+\frac{a_{n-1}}{x}+\dots +\frac{a_0}{x^n}}{b_n+\frac{b_{n-1}}{x}+\dots +\frac{b_0}{x^n}}=\frac{a_n+0+\dots +0}{b_n+0+\dots +0}=\frac{a_n}{b_n}$$
where $a_n$ and $b_n$ are different from zero.
