Usage of the standard part function when calculating with infinities

Question

We know that $\lim\limits_{H \to\infty}(\frac{H+1+d}{2H-1+3d})=\frac{1}{2}$.

The concept of limits is something I understand perfectly fine. Though, in non-standard analysis, we have a function called the standard part function, st( ). For example,

$$\text{If }H\text{ infinite and } d\text{ infinitesimal}, \text{then}\;st\left(\frac{H+1+d}{2H-1+3d}\right)=\frac{1}{2}$$

Here you can see that both the standard part function and the calculation with limits yields the same answer. My problem is that while I understand what happens when I'm calculating with limits, I don't understand what is happening when I'm using the standard part function, when calculating with infinities. Could someone help me understand how I should use the standard part function mathematically, particularly when calculating with infinities?

Answer 1

$\DeclareMathOperator{\st}{st}\DeclareMathOperator{\sgn}{sgn}$I will answer in the NSA framework of Nelson known as Internal Set Theory (IST). I'm unusually explicit with how this works here to drive home there's nothing really fancy at work. It's essentially exactly how you would work with limits. This answer is transferrable to other NSA frameworks.

In IST, infinitesimals and "infinite" (unlimited) numbers exist in the reals, $\mathbb{R}$ as usually defined due to IST being a conservative axiomatic extension of ZFC that lets you talk about them.

This system (IST) adds the term standard. All sets (in this case, real numbers) are either standard or nonstandard. The standard numbers correspond to the usual reals (those that are uniquely definable with ZFC alone) while unlimited and nonzero infinitesimals are, e.g., nonstandard. In this framework the standard part function, $\st(x)$, is defined for limited $x$ and returns the unique standard real that differs from $x$ by no more than an infinitesimal.

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I will evaluate your limit using these facts provable from the IST axioms and basic definitions:

If $x,y\in\mathbb{R}$ are limited (i.e., not unlimited):

1. $\st(x+y)=\st(x)+\st(y)$
2. $\st(x\cdot y)=\st(x)\cdot\st(y)$

If $\varepsilon,\delta\in\mathbb{R}$ are infinitesimal, $\ell,k\in\mathbb{R}$ are limited, and $u,v\in\mathbb{R}$ are unlimited:

3. $\st(\varepsilon)=0$
4. $1/u$ is infinitesimal
5. $\varepsilon+\delta$ and $\varepsilon\cdot\delta$ are infinitesimal
6. $u+\ell$ is unlimited
7. $\ell\cdot\varepsilon$ is infinitesimal
8. $\ell+k$ is limited
9. $\sgn(u)=\sgn(v)\implies u+v$ is unlimited

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You ask how to evaluate the following for $H$ unlimited and $d$ infinitesimal. $$\st\left(\frac{H+1+d}{2H-1+3d}\right)$$

\begin{align*} \st\left(\frac{H+1+d}{2H-1+3d}\right) = \st\left(\frac{H}{2H-1+3d}+\frac{1}{2H-1+3d}+\frac{d}{2H-1+3d}\right) \end{align*}

By (9) and repeated application of (6), $u_0=2H-1+3d$ is unlimited. By (4), $\varepsilon_0=1/u_0$ is infinitesimal. \begin{align*} &= \st\left(\frac{H}{u_0}+\varepsilon_0+\varepsilon_0d\right) \end{align*} By (7), $\varepsilon_0d$ is infinitesimal. By (5), $\varepsilon_1=\varepsilon_0+\varepsilon_0d$ is infinitesimal. \begin{align*} &= \st\left(\frac{H}{u_0}+\varepsilon_1\right) \end{align*} If the standard part is defined the expression in the standard part is limited, so by (8), $H/u_0$ is limited. Hence, by (1), we may split the standard part. \begin{align*} &= \st\left(\frac{H}{u_0}\right)+\st\left(\varepsilon_1\right) \end{align*} By (3), the term on the right is $0$. Rewrite as follows (partial fraction decomposition): \begin{align*} &= \st\left(\frac{H}{u_0}\right) \\ &= \st\left(\frac{H}{2H-1+3d}\right) \\ &= \st\left(\frac{1}{2}+\frac{1-3d}{2(2H-1+3d)}\right) \\ &= \st\left(\frac{1}{2}+\frac{1-3d}{2u_0}\right) \end{align*} Again by (8) these are both limited, so we split. \begin{align*} &= \st\left(\frac{1}{2}\right)+\st\left(\frac{1-3d}{2u_0}\right) \end{align*} Since $1/2$ is standard, by the definition of standard part given in the intro, $\st(1/2)=1/2$. \begin{align*} &= \frac{1}{2}+\st\left(\frac{1-3d}{2u_0}\right) \\ &= \frac{1}{2}+\st\left(\frac{1}{2u_0}+\frac{-3d}{2u_0}\right) \\ &= \frac{1}{2}+\st\left(\frac{\varepsilon_0}{2}+\frac{-3d\varepsilon_0}{2}\right) \end{align*} By (7) $\varepsilon_0/2$ is infinitesimal and hence limited. So, again, we can split by (1) and eliminate a term by (3). \begin{align*} &= \frac{1}{2}+\st\left(\frac{\varepsilon_0}{2}\right)+\st\left(\frac{-3d\varepsilon_0}{2}\right) \\ &= \frac{1}{2}+\st\left(\frac{-3d\varepsilon_0}{2}\right) \end{align*} By (5) and (7) the remaining term under the standard part is infinitesimal. Therefore, by (3), it is zero. \begin{align*} &= \frac{1}{2} \end{align*}

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There's other properties and other (probably more efficient) routes to rigorously apply them to get to the same result (e.g., start by dividing top and bottom by $H$ and use (2)), but this is basically how it works. In practice, of course, you'd never do this as you can prove general results that make finding these things easy, just as you do with limits.

If you have any questions about where some of the properties I used come from (or how you would prove them), please leave a comment or ask another question and I'd be happy to clarify.

Answer 2

A simpler answer is that the limit is defined in terms of the standard part function. Thus, $\lim_{x\to0}f(x)$ is by definition the standard part of $f(\epsilon)$ where $\epsilon$ is a nonzero infinitesimal: $\lim_{x\to0}f(x)=\text{st}(\epsilon)$.

More precisely, the limit exists if and only if the standard part has the same value independent of the choice of such $\epsilon\not=0$.