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?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =\lim_{x\to \infty}\frac{x^3\left(2+0-0\right)}{x^3\left(3-0-0+0\right)} \\
\ =\frac{2}{3}$$
 A: Yes, your calculations look correct. However, for future reference, there is an easier way to do this. If you have a rational function with a polynomial on the top and bottom and  you are taking the limit as $x \to \infty$:


*

*If the polynomials are of the same degree, then the limit is the ratio of their leading coefficients

*If the polynomial in the numerator has a higher degree, the limit is $\infty$.

*If the polynomial in the denominator has a higher degree, the limit is $0$.

A: 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.
A: Exactly correct! In the future you can also check your work on Wolfram Alpha.
A: Put $y=1/x $ and compute
$$\lim_ {y\to 0^+}\frac { 2+y^2-2y^3 }{3-y -y^2  +y^3}=2/3$$
