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 &gt;0, \exists  \delta &gt;0$ s.t. if $0&lt; | x-1 | &lt; \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | &lt; \epsilon $.

$|  \frac{3x^2 - x - 2}{3(x+2)} | =  |  \frac{(3x+2)(x-1)}{3x+6}|&lt; \epsilon $

Let $\delta=\frac{1}{3}$. Then, $|x-1|&lt;\frac{1}{3} \Rightarrow |x|&lt;\frac{4}{3}, |3x+2|&lt;6$..

Then, $|  \frac{(3x+2)(x-1)}{3x+6}|&lt;|  \frac{6(x-1)}{6}|&lt;\epsilon $

$|  x-1|&lt;\epsilon $, so  $\delta$ =min{1/3,  $\epsilon$ }

i think this feels correct, but to be sure... i thank in advance!
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 &gt;0, \exists  \delta &gt;0$ s.t. if $0&lt; | x-1 | &lt; \delta$ , then  $|  \frac{x^2}{x+2} - \frac{1}{3}  | &lt; \epsilon $.


I personally would have used more English words (“We have to show that for each $\epsilon&gt;0$ 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 $|x-1|&lt;\frac{1}{3}$ we have $0&lt;|x|&lt;\frac{4}{3}$ and hence $|3x+2|&lt;6$ and $|  \frac{(3x+2)(x-1)}{3x+6}|&lt;|  \frac{6(x-1)}{6}|=\lvert x-1\rvert$.


I have added $0&lt;\lvert x\rvert$, because you are using this for $\lvert 3x+6\rvert&gt;6$.

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 $\epsilon&gt;0$ be given. We set $\delta =\mathrm{min}\{1/3,  \epsilon \}$. Then for all $x$ with $\lvert x-1\rvert&lt;\delta$ we have 
  \begin{align*}
\left\lvert\frac{x^2}{x+2} - \frac{1}{3}\right\rvert
&amp;&lt;\lvert x-1\rvert&amp;&amp;\text{because $|x-1|&lt;\frac{1}{3}$}
\\&amp;&lt;\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.
There is a slight mistake you made 3x+2 &lt; 6 and 3x+6&lt;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:$$\frac{\lim_{x \to 1} x^2}{\lim_{x \to 1} x + \lim_{x \to 1}2} = \frac{1}{1+2}=\frac13$$