Equivalent limit definition Often, when one deals with limits of sequences, and one has to do a proof using the definition, there appears the nasty $< 2 \epsilon$ in the end. Of course, many people are satisfied by it, but I'm not and I think we should prove that this is sufficient. Therefore, consider the following statement. Can someone verify my proof?

Let $k \in \mathbb{R^{+}}$. The following 2 statements are equivalent:
  $$\forall \epsilon>0: \exists N \in \mathbb{N}: \left(n >N \implies d(u_n,L) <\epsilon\right)$$
  $$\forall \epsilon>0: \exists N \in \mathbb{N}: \left(n >N \implies d(u_n,L) <k\epsilon\right)$$

Proof: $\boxed{\Rightarrow}$ Let $\epsilon > 0$. Then $k \epsilon >0$ and by assumption, there exists a positive integer $N$ such that $n > N$ implies $d(u_n,L) <k\epsilon$
$\boxed{\Leftarrow}$ Let $\epsilon > 0$, then $\epsilon/k >0$ and by assumption, there exists a positive integer $N$ such that $n>N$ implies $d(u_n,L) < k \epsilon/k = \epsilon$
QED
 A: Your proof is indeed correct, but to me it illustrates why people usually aren't bothered by the "$2\varepsilon$"; it's really easy to formally prove they are equivalent, and intuitively, the distance still gets 'arbitrarily small'.
A: Your proof is correct, and the reason that this works as you'd want it is that $x\mapsto kx$ (for positive $k$) is bijection of $(0,+\infty)$ onto itself.
Thus, let me write your proof again, but replacing $\epsilon\mapsto k\epsilon$ with $f\colon (0,+\infty)\to (0+\infty)$ bijective:

Proof: $\boxed{\Rightarrow}$ Let $\epsilon > 0$. Then $f( \epsilon) >0$ and by assumption, there exists a positive integer $N$ such that $n > N$ implies $d(u_n,L) <f( \epsilon).$
$\boxed{\Leftarrow}$ Let $\epsilon > 0$, then $f^{-1}(\epsilon) >0$ and by
  assumption, there exists a positive integer $N$ such that $n>N$
  implies $d(u_n,L) < f(f^{-1}(\epsilon)) = \epsilon$
QED

But, if you examine it closer, you will notice that you don't actually need that $f$ is bijective, just that it has right inverse. Assuming axiom of choice, this is equivalent to $f$ being surjective. But, then again, if you are doing $\epsilon -\delta$ proof and have to resort to some surjection that is not bijection to complete your proof, you are probably doing something wrong.
A: I don't see a real need to provide a proof. You can formally trade $\epsilon$ for $k\epsilon$ and write
$$\forall k\epsilon>0: \exists N \in \mathbb{N}: \left(n >N \implies d(u_n,L) <k\epsilon\right).$$
Now the equivalence of $k\epsilon>0$ and $\epsilon>0$ is obvious.
A: Everything is O.K. Your proof is fine !
