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 $0<d(x, a)<\delta$. 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.


2 Answers 2


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 \begin{equation} (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]. \end{equation} 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 $|x-c|<d$ 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 \begin{equation} (2) \quad (\exists\delta\in\mathbb R^{*+})(\forall x\in D^*_f)\big[|x-c|<\delta \; \Rightarrow \; |f^*(x)-L|<\epsilon\big] \end{equation} 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.

  • $\begingroup$ 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. $\endgroup$
    – Alice
    Commented May 24 at 12:36
  • 1
    $\begingroup$ @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. $\endgroup$ Commented May 26 at 6:27

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