Using the epsilon-delta to prove $\lim_{x \rightarrow 1} \frac{x^2}{x+2} = \frac{1}{3} $ I'm trying to use the epsilon-delta definition to prove 
$\lim_{x \rightarrow 1} \frac{x^2}{x+2} = \frac{1}{3} $.
Could someone please check if i got this right?
$ \forall  \epsilon >0, \exists  \delta >0$ s.t. if $0< | x-1 | < \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | < \epsilon $.
$|  \frac{3x^2 - x - 2}{3(x+2)} | =  |  \frac{(3x+2)(x-1)}{3x+6}|< \epsilon $
Let $\delta=\frac{1}{3}$. Then, $|x-1|<\frac{1}{3} \Rightarrow |x|<\frac{4}{3}, |3x+2|<6$..
Then, $|  \frac{(3x+2)(x-1)}{3x+6}|<|  \frac{6(x-1)}{6}|<\epsilon $
$|  x-1|<\epsilon $, so  $\delta$ =min{1/3,  $\epsilon$ }
i think this feels correct, but to be sure... i thank in advance!
 A: Your calculations are correct, but that alone does not make a correct solution. It should be clear what the connection between the different lines is. The first line is what you want to prove, so say that.

We have to show: $ \forall  \epsilon >0, \exists  \delta >0$ s.t. if $0< | x-1 | < \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | < \epsilon $.

I personally would have used more English words (“We have to show that for each $\epsilon>0$ there is a $\delta$ such that...”), but that is a matter of taste.
If I was grading this as homework, then regarding your next line I would ask: Where does the $\epsilon$ come from? So far you have not introduced an epsilon. Of course, I understand that you want that inequality to hold, but then you have to say that. I would prefer to just state what you are actually doing, namely a quick calculation:

We have $|  \frac{x^2}{x+2} - \frac{1}{3}  | = |  \frac{3x^2 - x - 2}{3(x+2)} | =  |  \frac{(3x+2)(x-1)}{3x+6}|$.

You begin the next line with “let $\delta=\frac13$”. This is not good, because actually you want to make a statement that holds whenever $\delta\le\frac13$. But so far there is no need to introduce $\delta$, just state what you are claiming.

We note that for $|x-1|<\frac{1}{3}$ we have $0<|x|<\frac{4}{3}$ and hence $|3x+2|<6$ and $|  \frac{(3x+2)(x-1)}{3x+6}|<|  \frac{6(x-1)}{6}|=\lvert x-1\rvert$.

I have added $0<\lvert x\rvert$, because you are using this for $\lvert 3x+6\rvert>6$.
Now we can wrap things up. What I do not like about your formulation is the “so  $\delta =\mathrm{min}\{1/3,  \epsilon \}$”. Since $\delta$ has not yet been defined, you cannot claim that it equals something else. What you mean is that setting $\delta$ to this value will work. So let's just do this.

Now let $\epsilon>0$ be given. We set $\delta =\mathrm{min}\{1/3,  \epsilon \}$. Then for all $x$ with $\lvert x-1\rvert<\delta$ we have 
  \begin{align*}
\left\lvert\frac{x^2}{x+2} - \frac{1}{3}\right\rvert
&<\lvert x-1\rvert&&\text{because $|x-1|<\frac{1}{3}$}
\\&<\delta\le\epsilon.
\end{align*}
  This is what we had to show.

Now I am sure that you can still improve some of these formulations. The important thing is that you really ask yourself what your claims are and that you express these clearly. 
A: There is a slight mistake you made 3x+2 < 6 and 3x+6<6
All in all this is a correct proof except that slight mistake.
But if I were you I would use some theorems about limits to handle this much more easier.Here is how I would have done that.
It is true that if $$\lim_{x \to a} f(x) = l$$ $$and$$ $$\lim_{x \to a}g(x) = m$$ then following are also true $$\lim_{x \to a } f(x)g(x) = lm$$ and $$\lim_{x \to a}(f+g)(x) = l + m $$
Using these facts we can then transform your limit into something like this $$\frac{\lim_{x \to 1} x^2}{\lim_{x \to 1} x + \lim_{x \to 1}2}$$
Since 2 would be constant function its limit will be 2.
Since x approaches 1 near one so does sqaure of x
Thus we can conclude that:$$\frac{\lim_{x \to 1} x^2}{\lim_{x \to 1} x + \lim_{x \to 1}2} = \frac{1}{1+2}=\frac13$$
