Is $\delta$ always a function of $\epsilon$? In the epsilon-delta definition of limit, $\delta$ seems to rely on $\epsilon$. Of course, $\delta$ can be a constant, and it doesn't have to look like a continuous. But I'm not sure what the problem is with the expression $\delta:\mathbb{R}^{>0}\to\mathbb{R}^{>0}$. Is there any reason why this statements are not common?
 A: Functions are a concept which are useful, but not universally so. Many things can technically be considered functions, but we only introduce this viewpoint when it helps to describe or understand a certain object, i.e. when it seems convenient, which it isn't here. In order to think of $\delta(\epsilon)$ as a function we must first pick such a $\delta$ given $\epsilon$ (since it is not uniquely defined), coming down to a more-or-less arbitrary choice, and after we've gone through the effort of doing so we do not gain anything. This view of $\delta$ as a map $\mathbb R_{>0}\to\mathbb R_{>0}$ did not make theorems regarding limits easier to prove, or the definition easier to understand, or allowed us to gain anything at all in terms of analysis. That's why this view, while correct, is not at all common.
A: For the $\epsilon$-$\delta$ definition of $\lim_{x\to a} f(x) = L$ we have a unique function $\epsilon \mapsto \delta_{\text{max}}(\epsilon),$ where $\delta_{\text{max}}(\epsilon)$ is the maximal valid $\delta>0$ for a given $\epsilon>0$.
But we usually do not take $\delta$ be the maximum possible (this confuses students first learning $\epsilon$-$\delta$ proofs). Instead we make a choice $\delta(\epsilon) < \delta_{\text{max}}(\epsilon)$. The choice is a function. It is not unique, though; several choices are possible. The choice is taken so that it's easy to show that $|f(x) - L| < \epsilon$ whenever $|x - a| < \delta.$
Thus we have functions, $\delta_{\text{max}}(\epsilon)$ and the choice $\delta(\epsilon)$, but as YiFan writes, it doesn't gain us much thinking of them as functions.
A: It's not the point. The point is that, given an arbitrary $\varepsilon>0$, it should be possible to find a $\delta>0$ such that blah, blah, blah. You should see $\varepsilon$-$\delta$ definitions as a "challenge": if you are challenged with a $\varepsilon$, you should be able to find a $\delta$ to answer the challenge. 
