# Equivalence between definitions of limit

In non-standard analysis, it is possible to define a limit as follows:

$$\displaystyle \left[\lim_{x \to a} f(x)=L \right] :=x\approx a\implies f(x)\approx L$$

($$a\approx b$$ denotes that the difference $$a-b$$ is infinitesimal or zero)

I would like to show that this definition of the limit is equivalent to the standard $$\epsilon$$-$$\delta$$ definition. I think I have some idea of how one may show it, but I am unsure of if it holds. It goes something like this:

Suppose that the limit of $$f(x)$$ does exist at some point $$x=a$$, where $$\epsilon, \delta \in\mathbb{R}$$. Then:

$$\displaystyle \forall \epsilon>0, \exists\delta>0$$ such that, $$0<|x-a|<\delta \implies |f(x)-L|<\epsilon$$

Now assume we choose $$x \approx a$$. Then, obviously, $$0<|x-a| <\delta$$. So we know that for $$x \approx a$$, regardless of our choice of $$\delta$$, we will always have $$0. Shouldn't that then imply that $$d(f(x), L) <\epsilon$$? So assuming the limit exists at $$x=a$$, the $$\epsilon$$-$$\delta$$-criterion seems to be equivalent to the logical statement at the very top.

But I suspect I am way, way out of my depth here. Any help or motivations for why these definitions are equivalent would be appreciated.

Like Sassatelli Giulio says, you should stipulate $$x \ne a$$ in the nonstandard definition of limit. Otherwise this is mostly fine, with a couple of notes:

1. You should be clear about the quantification of $$x$$. It is not a free variable, and it should range only over standard real numbers in the standard definition, but also nonstandard real numbers in nonstandard. Suppose $$f : U \to \mathbb R$$ for some $$U \subseteq \mathbb R$$. You should write something like $$\forall x\in {}^*U.\:x\approx a\text{ and }x\ne a \implies f(x) \approx L$$ and $$\forall\epsilon>0.\,\exists\delta>0.\,\forall x\in U.\:x\ne a\text{ and }|x - a| < \delta \implies |f(x) - L| < \epsilon.$$ If we were being really pedantic we would also put a $$*$$ on $$f$$ in the nonstandard definition.

2. Your proof that standard definition $$\implies$$ nonstandard definition is missing some details. In the standard definition of limit, $$x$$ is only ranging only over standard reals. The way to remedy this is simply to apply the Transfer principle; because $$\epsilon, \delta, f$$ are standard we can Transfer just the $$\forall x\ldots$$ part, and this lets us choose $$a\ne x\approx a$$. Now you need to conclude that for any such $$x$$ that $$|f(x) - L| < \epsilon$$ for all standard $$\epsilon>0$$. By definition of $$\approx$$, this means $$f(x) \approx L$$.

3. You also need to prove nonstandard definition $$\implies$$ standard definition. This will also involve an application of Transfer.

Here is a proof of the opposite direction in the context of continuity (a minor adjustment will turn this into a proof for limits).

Let $$L=f(c)$$. Recall that a real function $$f$$ is continuous at $$c$$ in the $$\epsilon,\delta$$ sense if $$$$(1) \quad (\forall \epsilon\in\mathbb R^+)(\exists\delta\in\mathbb R^+) (\forall x\in D_f)\big[|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon \big].$$$$ Assume that $$f^*$$ is microcontinuous at $$c$$, so that $$(\forall x \in D^*_f)\quad \left[x\approx c \; \Rightarrow \;f^*(x)\approx L\right].$$ Let us prove that $$f$$ is continuous in the sense of formula (1). Choose a real number $$\epsilon>0$$ as in the leftmost quantifier in (1). Let $$d>0$$ be infinitesimal. If $$|x-c| then in particular $$x\approx c$$. By microcontinuity of $$f^*$$ we necessarily have $$f^*(x)\approx L,$$ and in particular $$|f^*(x)-L|<\epsilon$$ since $$\epsilon$$ is appreciable. Then the value $$\delta=d$$ is witness to the truth of the existence claim expressed by the formula $$$$(2) \quad (\exists\delta\in\mathbb R^{*+})(\forall x\in D^*_f)\big[|x-c|<\delta \; \Rightarrow \; |f^*(x)-L|<\epsilon\big]$$$$ where our chosen $$\epsilon$$ is a fixed parameter in formula (2) (unlike formula (1) which quantifies over $$\epsilon$$). We now apply downward transfer to formula (2) to obtain $$(\exists\delta\in\mathbb R^+)(\forall x\in D_{f})\big[|x-c|<\delta\;\Rightarrow \; |f(x)-L|<\epsilon\big].$$ We conclude that there exists a real $$\delta>0$$ as required.

• Out of curiosity, is it possible to motivate the jump from (2) to the last logical statement without use of the transfer principle? For example, since $D_f \subseteq D_f^*$, obviously the statement could be simplified to $(\exists\delta\in\mathbb{R}^{*+})(\forall x\in D_f)...$, but the range of $\delta$ would still cause issues. Commented May 24 at 12:36
• @naytte2, one can't avoid using transfer here. This is the essential part of the proof. Transfer guarantees that if there is a nonstandard solution, then there is also a standard solution. The availability of transfer is what distinguishes nonstandard analysis from other modern theories of infinitesimals, which are too weak to handle infinitesimal analysis. Commented May 26 at 6:27