What is the limit ${{\lim }_{x\to\infty}}x^\epsilon$ for an infinitesimal $\epsilon$?

Question

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

http://en.wikipedia.org/wiki/Smooth_infinitesimal_analysis

EDIT:

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!

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.