Question about hyperreal numbers

Question

I recently did a question about infinitesimals, which I now know them as hyperreal numbers, and it was not very clear what I was asking. Let me try again.

Let $\epsilon$ be a positive number smaller than every positive real. Formally, $0<\epsilon<|x|,\enspace x\in\mathbb{R}-\{0\}$.

$$\large\mathbb{H^-}=\{a\epsilon^k\enspace|\enspace a\in\mathbb{R},\enspace k>0\}$$ $$\large\mathbb{R}=\{a\epsilon^k\enspace|\enspace a\in\mathbb{R},\enspace k=0\}$$ $$\large\mathbb{H^+}=\{a\epsilon^k\enspace|\enspace a\in\mathbb{R}-\{0\},\enspace k<0\}$$

$$\large\forall\epsilon_0\in\mathbb{H^-},\forall\epsilon_1\in\mathbb{H^+},\forall x\in\mathbb{R}-\{0\}:\epsilon_0<|x|<\epsilon_1$$

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Define an equivalence relation $$\large\leftrightarrow\enspace:=\enspace \{\left\enspace|\enspace x=y,\enspace x,y\in\mathbb{R},\enspace \epsilon_0,\epsilon_1\in\mathbb{H^-}\} \cup\enspace \{\left\enspace|\enspace x,y\in\mathbb{R},\enspace \epsilon_0,\epsilon_1\in\mathbb{H^+}\}$$

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To calculate the limits of, say, $f(x)=\frac{x}{x-1}$ at infinity (more or less $\epsilon^{-1}$) just substitute for it.

$$f(\epsilon^{-1})=\frac{\epsilon^{-1}}{\epsilon^{-1}-1}=\frac{1}{\epsilon(\epsilon^{-1}-1)}=\frac{1}{\epsilon(\frac{1-\epsilon}{\epsilon})}=\frac{1}{1-\epsilon}\leftrightarrow\frac{1}{1}=1$$

For the limits of $g(x)=\frac{1}{x^2}+4$ at $0^-$:

$$g(-\epsilon)=\frac{1}{(-\epsilon)^2}+4=\frac{1}{\epsilon^2}+4\leftrightarrow\epsilon^{-1}\approx\infty$$

and at $-\infty$:

$$g(-\epsilon^{-1})=\frac{1}{(-\epsilon^{-1})^2}+4=\frac{1}{\frac{1}{\epsilon^2}}+4=\epsilon^2+4\leftrightarrow4$$

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Is this system useful or even valuable to calculate limits?

Answer 1

Making your observations effective

Your calculations are very similar to the way calculus is usually done with the hyperreals, but with a few significant issues.

1. The handling of $\pm\infty$

Note that your definition of $\mathbb{H}^{+}$ makes it so that (by choosing $a=\pm1$), $\epsilon^{-1},-\epsilon^{-1}\in\mathbb{H}^{+}$. But that means that under your equivalence relation $\leftrightarrow$, we have $\epsilon^{-1}\leftrightarrow-\epsilon^{-1}$. This means you can't use your $\leftrightarrow$ idea to tell the difference between the limit of $g\left(x\right)$ at $0$ being $\infty$ or being $-\infty$. If you separate out $\mathbb{H}^{+}$ into two sets: the positive infinities and the negative infinities, then you can change the definition of $\leftrightarrow$ to fix this problem.

2. The choice of $\epsilon$

You made it sound like you can use this method to calculate limits by just using any particular infinitesimal $\epsilon$. This doesn't quite always work. For example, when you want to investigate ${\displaystyle \lim_{x\to0^{+}}}\sin\dfrac{1}{x}$. If you choose one particular infinitesimal, you might get $\sin\dfrac{1}{\epsilon}=1$ (or any other particular value between $1$ and $-1$), but the limit isn't $1$. To fix this problem, you really have to look at all infinitesimals: if they all give you the same real number (up to your $\leftrightarrow$ equivalence), then that's the limit.

3. The range of $k $ in $\mathbb H^\pm$

Thanks to Hurkyl for pointing this out. The range of $k $ in the definition of $\mathbb H^\pm$ is a problem.

If it's all hyperreals, then $k=\pm\dfrac{\log 2}{\log \epsilon}$ would make $2^{\mp1}\epsilon=1$ so that $\mathbb H ^- $ would have noninfinitesimals in it and $\mathbb H ^+ $ would have finite numbers in it.

On the other hand, if it's just positive (finite) integers or reals, then you can't get to infinitesimals significantly larger than $\epsilon$ in $\mathbb H^- $ or infinities significantly smaller than $\epsilon^{-1} $ in $\mathbb H^+$, which could be a problem for this to work properly for some limits.

Useful/Valuable?

With the above changes, and proper care, you can use these methods to calculate all of the same limits from Calculus. In that sense it's useful/valuable. However, it doesn't really let you do any Calculus that couldn't also be done without the hyperreals. That said, mathematicians may find it a little easier to phrase some proofs in terms of this sort of language, for technical reasons. Here is a very technical blog post about this.

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Standard terminology

To help clear up your intuition/make this easier to think about, I'll summarize how this stuff is usually phrased.

Firstly, $\mathbb{H}^{-}$ should be the set of infinitesimals: hyperreals between $-\left|a\right|$ and$ \left|a\right|$ for a nonzero real number $a$. $\mathbb{H}^{+}$ is the set of both positive and negative infinite hyperreals. Reciprocals of infinite hyperreals are indeed infinitesimal, etc.

Your equivalence relation $\leftrightarrow$ could be phrased “two hyperreals numbers are equivalent if they're both infinitely close to the same real number or they're both infinite”. The most common equivalence relation when doing this sort of work is $a\approx b$ ($a$ and $b$ are “infinitely close”) if and only if $a-b$ is infinitesimal (whether or not $a$ and $b$ are finite). Another relavent notation is the “standard part”: If a hyperreal number $a$ is not infinite, then the real number it's infinitely close to is denoted $\mathrm{st}\left(a\right)$.

Limit calculations

We would say that since $f\left(H\right)\approx1$ for any positive infinite hyperreal $H$, ${\displaystyle \lim_{x\to\infty}}f\left(x\right)=1$. Similarly, since $g\left(-\epsilon\right)$ is positive infinite for any positive infinitesimal $\epsilon$, ${\displaystyle \lim_{x\to0^{-}}}g\left(x\right)=\infty$. And since $g\left(-H\right)\approx4$ for any positive infinite hyperreal $H$, ${\displaystyle \lim_{x\to-\infty}}g\left(x\right)=4$.

I would strongly recommend reading through sections 1.5, 1.6, and 5.1 of Elementary Calculus: An Infinitesimal Approach for a good treatment of this material.