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I am trying to figure out whether the following is true: a function $f$ has limit $L$ at the point $p$ if and only if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $$ s,t \in \mathbb{B}_{\delta}(p) \implies |f(s) - f(t)| < \epsilon. $$ I believe that it is true because by definition of limit $$ |t - p| < \delta \implies |f(t) - L| < \frac{\epsilon}{2} $$ and $$ |s - p| < \delta \implies |f(s) - L| < \frac{\epsilon}{2} $$ which can be combined to give $$ s,t \in \mathbb{B}_{\delta}(p) \implies |f(t) - L| + |f(s) - L| < \epsilon \implies |f(s) - f(t)| < \epsilon $$ Since every step is reversible then this proves the statment is true. Is this the right way to think about this? This is the first time I've seen the limit characterized in this way and it makes intuitive sense. So, am I on the right track here?

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We often (at least in the US) require $s$ and $t$ to be different from $p$, to allow the function to be undefined at $p$. – Arturo Magidin Mar 8 '12 at 20:07
You may want to read something about Cauchy convergence criterion for functions, which states that a function $f:A\to \mathbb{R}$ converges in $p$ iff for each $\varepsilon >0$ there exists $\delta >0$ s.t. $\forall s,t\in A\cap \mathbb{B}_\delta (p)\setminus \{p\},\ |f(s)-f(t)|<\varepsilon$ – Pacciu Mar 8 '12 at 20:08
There is similar definition for Cauchy sequence. You are mostly right, but in fact point $p$ maybe absent from space (the space is not a complete metric space then). – dtldarek Mar 8 '12 at 20:12
Not every step is reversible, and your definition makes no mention of $L$, so it cannot be a definition for "the limit is $L$"; it's just a definition of "the limit exists" (provided you are working over $\mathbb{R}$ or some similar). The step $|f(t)-L|+|f(s)-L|\lt\epsilon\implies |f(s)-f(t)\lt\epsilon$ is not reversible. – Arturo Magidin Mar 8 '12 at 20:16
@Pacciu When you say the function "coverges" at $p$ do you just mean it has a limit at P? – AFX Mar 8 '12 at 20:16
up vote 3 down vote accepted

Note that your proposed definition makes no mention of $L$. That means that, using your definition, I could conclude that the limit of $f(x)=x$ as $x\to 0$ is $1$, or $2$, or $3$, or whatever I want. That's a bit of a problem.

Rather, what you give is a definition for the existence of the limit (provided you disallow $s,t=p$). So you would have:

Definition. Let $f$ be a function defined on a punctured neighborhood of $p$. Then the limit of $f(x)$ as $x\to p$ exists if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $s,t\in \mathbb{B}_{\delta}(p)-\{p\}$, then $|f(s)-f(t)|\lt \epsilon$.

The idea is fine, and works over the real numbers; but the presentation leaves something to be desired. It is not true that "every step is reversible": for example while it is true that if $|f(t)-L|+|f(s)-L|\lt \epsilon$, then $|f(s)-f(t)|\lt \epsilon$, that implication is not reversible in general.

Your characterization is just a "Cauchy-like" definition; you ask the values to approach one another instead of a specific number. So the definition doesn't tell you what the limit is, it just tells you that a limit exists (provided you are working over the reals, or some other Cauchy-complete field).

Now, your argument shows that if the limit exists in the usual sense, then it exists in this sense. But it does not establish the converse! For example, every function $\mathbb{Q}\to\mathbb{Q}$ that has a limit will satisfy your definition, but there are functions $\mathbb{Q}\to\mathbb{Q}$ that satisfy your definition and don't have a limit (because the limit is not a rational).

In order to show that your definition implies the limit exists in the usual sense, you need to invoke completeness of the real numbers. The argument is similar as for proving that every Cauchy sequence converges.

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+1. One concrete counterexample would be: $$f(x)=\max\{nx\mid n\in\mathbb Z\land (nx)^2<2\}$$ This takes rationals to rationals, and satisfies the Cauchy-like condition for $x\to 0$ (if we require that $s,t\ne p$), but does not have a rational limit for $x\to 0$. – Henning Makholm Mar 8 '12 at 20:28

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