# Is there a “functional” definition of a limit?

Say we have a convergent sequence $(x_n)$ where $x_n \in E$ for all $n \in \mathbb{N}$ and $E$ is a subset of a metric space $(X,d)$.

With this setup, we usually define it's limit as a point $x \in X$ such that for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N$ implies $d(x_n,x) < \epsilon$.

In some sense, I think that the above definition of the limit effectively mean that given any "$\epsilon > 0$, there is some kind of $N$ that we can use to get the sequence within $\epsilon$ of the limit $x$.

I am wondering whether this can be reformulated as follows: there is a function $f: \mathbb{R} \rightarrow \mathbb{N}$, so that $f(\epsilon) = N$ and $n > N$ implies $d(x_n,x) < \epsilon$. If so, the function $f$ would have some nice properties (it would be onto, and monotonically decreasing in $\epsilon$ for instance).

Is there any use to thinking about functions in this way / has it been introduced in this way?

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Note that your $f$ need not be onto. – Hagen von Eitzen Dec 19 '12 at 21:14

This concept is often known as the modulus of convergence, though for recursion-theoretic purposes it's often given as a function $g(n): \mathbb{N}\to\mathbb{N}$, where (in terms of your definition for $f()$) $g(n)$ would be defined as $g(n) = f(\frac{1}{n})$ or $g(n) = f(2^{-n})$. The concept is important in recursion theory because, for instance, we can compute the digits of the limit recursively in $g$ (i.e., using a Turing machine for $g$). Note that the function $g$ doesn't have to be computable even if the series converges; see, for instance, http://mathoverflow.net/questions/51794/simple-example-of-a-sequence-without-computable-modulus-of-convergence .
To obtain a well-defined function you might define $f(\epsilon)$ to be the smallest $N$ such that $d(x_n,x)<\epsilon$ for all $n>N$. Convergence of the sequence $x_n$ guarantees existence of such $N$'s and one of them has to be the smallest.
Conversely, if there exists a function $f$ such that, for every $\epsilon$, $d(x,x_n)<\epsilon$ for all $n>f(\epsilon)$, then $x_n$ is convergent with limit $x$, because you may take $N=f(\epsilon)$.