What is the limit

$${{\lim }_{x\to\infty}}x^\epsilon$$

for an infinitesimal $\epsilon$? Does it give zero or infinity?

Note that I'm considering the infinitesimals described in



Since I was asked to show some of my own thoughts on the subject, I'd like to contribute the following:

$$\lim_{x\to\infty} x^\epsilon = \lim_{x\to\infty} \exp(\log (x^\epsilon))= \lim_{x\to\infty} \exp(\epsilon\log(x))\\= \lim_{x\to\infty}(1+\epsilon\log(x))= \lim_{x\to\infty}(1+\epsilon)=1+\epsilon$$

Where we used a series expansion of $\exp(y)$ and the properties of infinitesimals $\epsilon^2=0$ and $a\epsilon=\epsilon$ for any number $a$. However, I am not sure if this is some sort of cheating my way around the limit at hand or not. Help would be appreciated!

  • 3
    $\begingroup$ What's an infinitesimal? Are you doing nonstandard analysis? $\endgroup$
    – egreg
    Commented Feb 17, 2015 at 23:51
  • $\begingroup$ I am doing smooth infinitesimal analysis. As described in en.wikipedia.org/wiki/Smooth_infinitesimal_analysis $\endgroup$
    – Kagaratsch
    Commented Feb 17, 2015 at 23:52
  • $\begingroup$ I wasn't “voicing my frustration”, but just asking for what setting you're in. There are several theories dealing with infinitesimals. Can you show some of your thoughts about the subject? You'll increase the chances to receive an answer. $\endgroup$
    – egreg
    Commented Feb 17, 2015 at 23:56
  • $\begingroup$ I am sorry, I misunderstood. For the sake of flexibility, I am interested to learn the answer in any arbitrary but meaningful theory dealing with infinitesimals you prefer. $\endgroup$
    – Kagaratsch
    Commented Feb 17, 2015 at 23:59
  • $\begingroup$ I have shown my thoughts on the problem in the edit. Does it make any sense? $\endgroup$
    – Kagaratsch
    Commented Feb 18, 2015 at 0:09

1 Answer 1


In the Robinson framework, we can still define $x^\varepsilon$ to be $\exp(\varepsilon \log(x))$ when $\varepsilon$ is a positive infinitesimal and $x$ is a positive hyperreal number. We identify $\lim_{x \to \infty} f(x)$ (as it is formulated in the standard framework) with the standard part of $f(M)$ where $M$ is a positive infinite number, provided that the standard part does not depend on the choice of $M$. (Alternately, if it is always positive and infinite, we say the limit is $+\infty$, and if it is always negative and infinite, we say the limit is $-\infty$).

So here we substitute in an infinite number $M$, obtaining $M^\varepsilon = \exp(\varepsilon \log(M))$. This should immediately look like a problem, because $\varepsilon$ is infinitesimal while $\log(M)$ is infinite. Consequently $\varepsilon \log(M)$ might be infinite, if $M$ is for instance $\exp(\exp(1/\varepsilon))$. It might be finite, if $M$ is for instance $\exp(1/\varepsilon)$. Or it might be infinitesimal, if $M$ is for example $1/\varepsilon$.

We get a similar problem in the standard framework: if $f,g > 0$, $\lim_{x \to \infty} f(x)=\infty$ and $\lim_{x \to \infty} g(x)=0$ then $\lim_{x \to \infty} f(x)^{g(x)}$ depends on how fast $f$ grows and how fast $g$ decays.

In this respect I would think that the limit simply can't be said to exist. Please note that I am not an expert on this subject, though; most of my knowledge is confined to proving basic standard theorems (e.g. Heine-Cantor) by nonstandard methods.

  • $\begingroup$ It still feels like a purely standard framework answer, where we only can say anything about the limit once we know the exact functional dependence of the entering elements on the limiting variable. Basically, thinking of an "infinitesimal" as something going to zero asymptotically, this might also be considered an order of limits problem. Oh, well. I guess this will have to suffice. $\endgroup$
    – Kagaratsch
    Commented Feb 18, 2015 at 0:41
  • 1
    $\begingroup$ @Kagaratsch My understanding is that the actual nonstandard theorems of nonstandard analysis, that is, the ones that are defined purely in the nonstandard language, are not really of any particular interest to analysts. $\endgroup$
    – Ian
    Commented Feb 18, 2015 at 1:04

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