Limit when x approaches a number $\ne \infty$ I was rushing through a analysis task book when I suddenly encountered 
$$\text{Find }~\lim_{x \to 3}\frac{x^2-9}{x-3}~\text{ by using the definition of limits.}$$
I thought, as the term equals to x+3, the limit should be 6.
Here's my beginning: I have to show, that $$\forall \epsilon >0~\exists ~X\in \mathbb N :\forall x \gt X : |a_x-a|<\epsilon$$
So $|x+3-6|=|x-3|<\epsilon$ is true because as x is approaching 3, |x-3| is approaching 0.
I just don't know how to wrap it up formally.
Can I say: $|x-3|=|3-x|\overset{as ~x~ is~ positive}{=}3-x<\epsilon,$
so choose $X=3-\epsilon$ ?

Since you want to prove the limit, $L=6$, you need to show $\forall\epsilon>0,\ \exists\delta>0$ such that $|x-3|<\delta\implies \left|\dfrac{x^2-9}{x-3}-6\right|<\epsilon$.  It would be a good idea to start with the right side of the $\implies$ and work backwards, to show you can actually find such a $\delta$ that makes this statement true.  After you find it, then write out the proof.  Now, to find a suitable $\delta$:
Since $\left|\dfrac{x^2-9}{x-3}-6\right|=\left|\dfrac{x^2-9-6(x-3)}{x-3}\right|=\left|\dfrac{x^2-6x+18}{x-3}\right|=\left|\dfrac{(x-3)^2}{x-3}\right|=|x-3|<\epsilon.$ Since $|x-3|<\delta$ by assumption, then just let $\delta=\epsilon.$
Now to start the proof:
Given $\epsilon>0$, let $\delta=\epsilon$.  Then $$|x-3|<\epsilon\implies\left|\dfrac{(x-3)^2}{x-3}\right|=\left|\dfrac{x^2-6x+9}{x-3}\right|=\left|\dfrac{x^2-6x-9+18}{x-3}\right|=\left|\dfrac{x^2-9-6(x-3)}{x-3}\right|=\left|\dfrac{x^2-9}{x-3}-6\frac{(x-3)}{x-3}\right|=\left|\dfrac{x^2-9}{x-3}-6\right|<\epsilon$$ which proves that $\lim\limits_{x\to 3}\frac{x^2-9}{x-3}=6$ and we are done.
