Question about using arbitrary $\epsilon$ in real analysis proofs I've notice that in a lot of the proofs that are assigned in an undergraduate analysis course, we are often trying to show that some quantity is bounded by an arbitrary epsilon. 
For example, if I want to show that 
$$\lim_{n\to\infty}\frac{2}{\sqrt{n}}+\frac{1}{n}+3 = 3$$ 
I could try to show that for any $\epsilon > 0$ I can find an index $N$ such that 
$$\left| \frac{2}{\sqrt{n}}+\frac{1}{n} \right| < \epsilon$$
I would try to do something like using the Archimedean principle to show that if $\epsilon = \epsilon'^2/9$ (where $\epsilon'$ is dependent on the choice of $\epsilon$) that I can get find an index such that...
My real question is that how would I justify that any arbitrary $\epsilon$ can be written as $\epsilon = \epsilon'^2/9$? How do I justify that this equality would always have a solution? Would I appeal to density of the real numbers somehow?
Thanks
P.S. I think some of the answers are based on helping me prove the sequence converges, but I was mainly talking about how given some $\epsilon >0$, how I can assert that I can always find another real number $k$ such that $\epsilon = k^2/9$. Writing it this way makes the proof easier to read in my opinion, but I don't want to rely on something that seems obvious but I can't rigidly justify.
 A: $x\mapsto \frac{x^2}{9}$ is a bijection of $\Bbb{R}^+$. So if you show that a statement holds for every $\epsilon$ "of the form" $\frac{\epsilon'^2}{9}$, you show that it holds for ALL $\epsilon>0$.
A: In a limit case you have to prove that $\forall \epsilon >0$ something holds. In general this would mean that you would have to prove it for every epsilon and you are correctly bothered by only choosing what appear to be some epsilon of a particular shape.
But there are two things to notice. First of all (as someone already pointed out) since $\epsilon>0$ you can always write it as ${\epsilon^\prime}^2/9$ for some $\epsilon^\prime$, and second of all in limit cases what you are proving is that $\forall \epsilon \exists \delta .... \implies |f(x)-L|<\epsilon$. This means that if you choose an $\epsilon^\prime$ smaller then the $\epsilon$ you start with you are fine since if $|f(x)-L|<\epsilon^\prime<\epsilon$ then $|f(x)-L|<\epsilon$. This is a particular case of a more general approach where when you need to prove something for all numbers it might suffice to prove it for some nice numbers which imply it then holds for all the other numbers by some extension. (For example for limits if you can prove that you can get a $\delta$ for each $\epsilon=\frac{1}{n}$ where $n\in\mathbb{N}$ then you have proved it for all $\epsilon$ by an "easy" "density" argument.
A: Hint:
Since $\epsilon>0$ we can allways find $\epsilon'=3\sqrt{\epsilon} \in \mathbb{R}$
A: Generally from the analysis point of view when we say that $a=b$, we generally mean $|a-b|<\epsilon \quad \forall \epsilon >0$. The way of looking at this is, since the distance between $a$ and $b$ is fixed it cannot be varied arbitrarily. If you are able to do it indefinitely by a small positive quantity which is this $\epsilon$, it means that they are equal.
In your statement, when you say $\epsilon'^2/9$, you are limiting your proof to this specific form. In general it should hold for all $\epsilon>0$.
A: A certain tolerance  $\epsilon>0$ is  given to us. We have to find an $n_0$ such that 
$$0<r_n:=\left|{2\over\sqrt{n}}+{1\over n}\right|<\epsilon$$
for all $n\geq n_0$, but we are not required to find the smallest $n_0$ that does the job.
You already have remarked that $$0<r_n\leq {3\over\sqrt{n}}\qquad(n\geq1)\ .$$
If $n$ is large enough to make $${3\over\sqrt{n}}<\epsilon\tag{1}$$ then $r_n<\epsilon$ is guaranteed as well. Now $(1)$ is equivalent with
$$n>{9\over\epsilon^2}\ ,$$
and this is fulfilled for all $n$ with
$$n\geq n_0:=\left\lfloor{9\over\epsilon^2}\right\rfloor+1\ .$$
Note that we didn't have to construct any new real $\epsilon'$ or similar, but we implicitly have used the Archimedean principle in the definition of $n_0$.
