Evaluate the limit as $x$ approaches $3$ : $\displaystyle\frac{x^3 -6x+2}{x^2+2x-3}$ Evaluate the limit x approaches 3 : $\displaystyle\frac{x^3 -6x+2}{x^2+2x-3}$
I am currently doing a limit problem, will be doing many in the coming months. I was wondering how do I factor a cube such as $x^3 -6x+2$? Is there a special rule? 
The ones I am familiar with are  $$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$ 
$$a^3 – b^3 = (a – b)(a^2 + ab + b^2)$$
but that does not work here. I also can not factor anything out.
 A: If $f(x) = x^{3} -6x +2$ and $g(x) = x^{2} +2x-3$
You have $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$ = $\frac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}$ with $a\in \mathbb{R}$ and obviously $g(x)\ne {0}$ when $x\rightarrow a$
Here $a=3$
so for $x \rightarrow 3$ :
$f(x) \rightarrow 11$
and $g(x) \rightarrow 12$
The final limit is : $\frac{11}{12}$
A: If it can be factored, a cubic polynomial will have two factors whose degrees add up to 3.  So if we disregard constant factors, it would be the product of a first degree and a second degree factor. 
In this particular example, $x^3 -6x +2$, it cannot be done with polynomials having integer coefficients.  We say the polynomial is irreducible over the integers because it cannot be factored, and the easy way to show this is applying Eisenstein's criterion with the prime $p=2$.
A: There is no need to factor numerator or denominator because at $x\to3$ numerator and denominator are defined and denominator isn't $0$, so just write:
$$\lim_{x\to3}\frac{x^3-6x+2}{x^2+2x-3}=\frac{3^3-6\cdot3+2}{3^2+2\cdot3-3}=\frac{11}{12}$$
A: Your need to factor the fraction should only arise when it is tending to an indeterminate form $0\over 0$. Here you can simply put 3 and evaluate the limit which comes out to $11\over12$.
