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?
 A: Essentially, divide the numerator and denominator by the largest  power of numerator:$$\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x} = \lim\limits_{x\to \infty}\frac {\frac{2x^4-3x^2+1}{x^4}}{\frac{6x^4+x^3-3x}{x^4}} = \lim\limits_{x\to \infty}\frac {2-\frac{3}{x^2} + \frac{1}{x^4}}{6 + \frac{1}{x} - \frac{3}{x^3}} = \frac{2 - 0 + 0}{6 + 0 - 0}= \frac13$$

Factorization usually works when you have $0/0$ expressions. For instance,
$$\lim_{x\to2} \frac{x^2-5x + 6}{x^2 - 3x + 2}$$
Note that both numerator and denominator are equal to zero when $x = 2$. From this, we can suspect that one of the roots of those expressions is $2$. Then, let's factorize and cancel that out (note that we can cancel since $x \to 2$, not $x = 2$)
$$\lim_{x\to2} \frac{x^2-5x + 6}{x^2 - 3x + 2} = \lim_{x\to2}\frac{(x-2)(x-3)}{(x-2)(x-1)} = \lim_{x\to2}\frac{x-3}{x-1}$$
which is fine at $x = 2$ so we can just go ahead and plug this in to get
$$\lim_{x\to2}\frac{x-3}{x-1} = \frac{2-3}{2-1} = -1$$
Note: We can plug in $x=2$ because rational functions are continuous.

A: 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 $n$
A: When $x$ is very large the dominant term in Num is $2x^4$ and in den it is $6x^4$. So the required limit is $$\lim_{x \to \infty} \frac{2x^4}{6x^4}=\frac{1}{3}.$$.
