How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$? 
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.

I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator of the fraction. The numerator can be factorized as $(x-1)(x+1)(2x^2-1)$ and the numerator can be factorized as $x(6x^3+x^2-3x)$. So, we can rewrite the problem as follows: $$\lim\limits_{x\to\infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}=\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}$$ But this doesn't help as there is no common factor in the numerator and denominator. I've also tried the following: $$\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}=\lim\limits_{x\to\infty}\frac{x-1}{x}\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}=1\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}$$ Here I used $\frac{x-1} x=1$ as $x$ approaches infinity. Yet this does not help.
The answer is $\frac 1 3$ in the book but the book does not include the solution.

So, how to solve the problem?

Divide the numerator and denominator of the expression by $x^4$:$$\lim\limits_{x\to \infty}\frac {2-3/x^2+1/x^4}{6+1/x-3/x^3}$$and use the fact that $\lim\limits_{x\to\infty}\frac{1}{x^n} = 0$ for any positive $n$
