Can we work with $(\forall\epsilon>0)(\mu\in\mathbb R^+)"\mu\epsilon"$ in convergence definition and examples instead of $"\forall\epsilon>0"$? 
There is an example and there are 2 solution($\epsilon$ and $\mu\epsilon$) for that example.If we are able to prove $\mu\epsilon$-style, we will gain time advantages in most $\epsilon-\delta$ proof.
Reader can skip the example, but I insist reader to read brief of main idea.

Brief of main idea: Before listening the story, reader can look at the definition which we are going to talk about.


$$(\forall\epsilon>0)(\exists N\in\mathbb R)(\forall n\in\mathbb N)(n>N\Longrightarrow |x_n-x|<\epsilon)$$


We always say that $\epsilon$ ,which is used in above definition, can be all positive number in $\mathbb R$, why not $\mu\epsilon$ ?
Example;
Addition of limits.
$$(\forall\epsilon>0)(\exists N_1\in\mathbb R)(\forall n\in\mathbb N)(n>N\Longrightarrow |x_n-x|<\epsilon/2) $$  and $$(\forall\epsilon>0)(\exists N_2\in\mathbb R)(\forall n\in\mathbb N)(n>N\Longrightarrow |y_n-y|<\epsilon/2)$$
Hence,for $n>\max\{N_1,N_2\}$
$$|x_n+y_n-(x+y)|\le|x_n-x|+|y_n-x|<\epsilon$$
But if we took $\epsilon$ for each definition, we would find  $<2\epsilon$.(And if reader wants, reader can multiply examples)
I wonder whether we can claim the following equivalent, or not?(I'm not even sure about if I define like I told above or not)



$$(\forall\epsilon>0)(\exists N\in\mathbb R)(\forall n\in\mathbb N)(n>N\Longrightarrow |x_n-x|<\epsilon)\\ \equiv \\ (\forall\epsilon,\mu>0)(\exists N\in\mathbb R)(\forall n\in\mathbb N)(n>N\Longrightarrow |x_n-x|<\mu\epsilon)$$



Eventually, How can we prove that we can also work with $\mu\epsilon$ no matter what or under some conditions? Is it enough for proof?
Note: Yeah, this's kind a very basic situation, but I just wanted to be sure.Thanks in advance.
 A: It is easily seen that the two tests in your third khaki box are equivalent.
Proof. Replace $\epsilon$ by $\epsilon'$ in the upper test. – If a sequence passes the upper test and arbitrary $\epsilon>0$, $\mu>0$ are given then put $\epsilon':=\epsilon \mu$, and realize that the sequence passes for this choice of $\epsilon$ and $\mu$. Since $\epsilon$ and $\mu$ where arbitrary we are done in this direction. The converse is even simpler: If a sequence passes the lower test and an $\epsilon'>0$ is given put $\epsilon:=\epsilon'$, $\mu:=1$, and realize that the sequence passes for this choice of $\epsilon'$. Since $\epsilon'$ was arbitrary we are done.
Now comes the critique: It is true that in some convergence proofs we have to make several quantities small at the same time. In such a situation some people start with ${\epsilon\over4}$ and the like and end up with the relevant difference being $<\epsilon$; others make all parts $<\epsilon$ and end up with the relevant difference being $<4\epsilon$. Both ways are fine, and are easily accepted by the reader. But your proposal would introduce a third dummy variable $\mu$ for every instance of proving some convergence. This would be a $50\%$ increase in mental complexity of each and every such proof, which is definitely not called for.
