# Boxing Pythagoras

## Classical Limits vs. Non-Standard Limits

One of the most important and fundamental concepts taught in modern Calculus classes is that of the Limit. I have discussed this idea once before, but I thought I would revisit it, here. In that first article, I noted that the classical definition for a Limit is fairly complex and that we can utilize a more intuitive notion of infinitesimals to accomplish the same task, insofar as derivatives are concerned. However, there are other uses and purposes for limits, in mathematics, so we would not want to simply omit them entirely, even using a non-standard approach to Calculus lessons.

Thankfully, even the very difficult and complex definition of “limit” can be simplified and made easier to understand by use of non-standard infinitesimals.

# The ε-δ Method

First, let us review the classical definition of a limit and how to evaluate it. As you’ll recall, there are times when we might want to discuss the value of a particular function, $f(x)$, at a particular value, when $x=c$. The definition for a limit is then given as follows:

Let $f(x)$ be a function defined over an interval containing $x=c$, except perhaps at $x=c$. Then we say that,

$lim_{x\rightarrow c} f(x)=L$

if for every number $\epsilon >0$ there exists some number $\delta >0$ such that,

$|f(x)-L|< \epsilon$ whenever $0<|x-c|< \delta$

However, the astute among you might recognize that this definition is not yet complete. There are other ways in which a limit can be used for which this definition is inadequate. Particularly, this definition is not enough to describe what happens to a function in the case that $x$ becomes arbitrarily large. If you remember your Calculus lessons, you may recall this being written as $lim_{x\rightarrow \infty} f(x)$.

The lemniscate symbol, $\infty$, is not a number. As such, we cannot substitute it in for $c$ in our definition, above. So, this case requires a slightly different definition.

Let $f(x)$ be a function defined on $x>k$ for some $k$. Then we say that,

$lim_{x\rightarrow \infty} f(x)=L$

if for every number $\epsilon >0$ there is some number $m>0$ such that

$|f(x)-L|<\epsilon$ whenever $x>m$

Of course, this still isn’t the whole story. Just as we can look arbitrarily far in the positive direction, we can also look arbitrarily far in the negative direction, so we also need a definition to handle that case:

Let $f(x)$ be a function defined on $x for some $k$. Then we say that,

$lim_{x\rightarrow -\infty} f(x)=L$

if for every number $\epsilon >0$ there is some number $n<0$ such that

$|f(x)-L|<\epsilon$ whenever $x

So now we have three different definitions in place for the concept of a limit which converges to a particular Real-valued number. However, things are still even more complex than that. The limit of a function doesn’t necessarily converge to a Real-valued number. It may diverge infinitely in either the positive or the negative direction. So, for each of these three definitions, we can add two more to describe the behavior of divergent functions. I will save space by refraining from listing them here (this page has an excellent overview with examples), but we are now up to 9 separate definitions which are necessary to make the concept of a limit work!

So, now, let’s look at an example. Perhaps we want to solve $lim_{x\rightarrow 5} 6x+1$. Intuitively, we feel like it should be easy to prove that this limit is equal to 31, since we can just plug 5 into the function and evaluate. Unfortunately, the proof isn’t quite so simple.

Given our definition, above, to prove that $lim_{x\rightarrow 5} 6x+1=31$ we need to show that for any arbitrary number $\epsilon >0$ we can find a number $\delta >0$ such that $|(6x+1)-31|<\epsilon$ whenever $0<|x-5|<\delta$. The easiest way to do this will be to define a $\delta$ in terms of $\epsilon$ so that our conditions will always be met. In order to do so, we might start by evaluating our $\epsilon$ sentence first:

$|(6x+1)-31|=|6x-30|=6|x-5|<\epsilon$

Now, let’s solve that inequality for $|x-5|$, since that phrase appears in our $\delta$ sentence: $|x-5|<\frac{\epsilon}{6}$. Now, we can see that whatever arbitrary $\epsilon$ we choose, we can set $\delta =\frac{\epsilon}{6}$ to ensure that our conditions are always met. Thus we have proved that $lim_{x\rightarrow 5} 6x+1=31$.

# Non-Standard Analysis

Now, let’s contrast this with a definition for limits using non-standard analysis (from Keisler):

Let $f(x)$ be a function defined over an interval containing $x=c$, except perhaps at $x=c$. Then we say that,

$lim_{x\rightarrow c} f(x)=L$

if whenever $x$ is infinitely close but not equal to $c$, the function $f(x)$ is infinitely close to $L$.

This single definition does the same amount of work as all 9 of the classical definitions for limits. The key to its powerful nature is the use of Hyperreal numbers— infinitesimals and infinite numbers, in particular. Let’s look at our earlier example again to see how this works.

To evaluate $lim_{x\rightarrow 5} 6x+1=31$, all we need to do is substitute a number which is infinitely close, but not equal, to 5 into the function. So, let’s say we have some infinitesimal number, $\alpha>0$. Now, we just evaluate:

$f(5+\alpha)=6(5+\alpha)+1=31+6\alpha$

Given the rules of infinitesimals, we know that $6\times \alpha$ is also infinitesimal, so our answer is infinitely close to 31 and we have proved our hypothesis. So much easier and more straightforward than the classical method!

# To Infinity and Beyond

But what about limits involving the lemniscate, $\infty$? These prove to be much easier with the non-standard method, as well. Let’s take another simple example: $lim_{x\rightarrow \infty} \frac{1}{x^2+2}=0$

Using the classical definition, we need to find some $m>0$ for any $\epsilon>0$ such that it is true that $|\frac{1}{x^2+2} -0|<\epsilon$ whenever $x>m$. So, solving our first inequality for $x$ gives us:

$|\frac{1}{x^2+2}-0|=\frac{1}{|x^2+2|}<\epsilon$

$|x^2+2|>\frac{1}{\epsilon}$

$x^2>\frac{1}{\epsilon}-2$

$x>\sqrt{\frac{1}{\epsilon}-2}$

For the sake of ease, we have assumed that $x$ will not be a complex number and that $x^2$ will therefore always be positive. Taking $m=\sqrt{\frac{1}{\epsilon}-2}$ will give us a value which always makes our set of inequalities true, thus proving that $lim_{x\rightarrow \infty} \frac{1}{x^2+2}=0$

Now, we can contrast the non-standard approach. In this case, since the lemniscate, $\infty$, is still not a number, we can simply substitute in any positive, infinite Hyperreal number for $c$ in our non-standard definition. Let’s use $C$ to denote that infinite Hyperreal and $C+\alpha$ for a number infinitely close, but not equal, to $C$.

This gives us $\frac{1}{(C+\alpha )^2+2}$. Since $C$ is infinite, we can know that $C+\alpha$ is infinite. The square of any infinite number is also infinite, and adding any finite number to an infinite number still yields an infinite number. Thus, the denominator of our fraction must be an infinite number. The reciprocal of an infinite is an infinitesimal, and the only Real number infinitely close to any infinitesimal number is zero. Thus, we have proved that $lim_{x\rightarrow \infty} \frac{1}{x^2+2}=0$.

So much more wonderfully simple! We didn’t even need to manipulate the expression algebraically. All we needed to do was think about the properties of the numbers.

# Rigor and Intuition need not compete!

Modern calculus classes rarely spend much time– if any– discussing the formal definition of a limit. When I took Calculus I in high school, the lesson was omitted entirely. When I again took Calculus I in college (my high school class had not been an Advanced Placement course, so my university required I take it again), the textbook we used contained a small blurb mentioning the formal definition– and only the first classical definition, at that– but none of our lessons required us to actually use it. Because of that, the notion of limits always bothered me as a student. The way I had been taught to do them, limits seemed like a pseudo-math guessing game instead of the rigorous mathematical tools which I had been used to utilizing. Many of my friends and tutoring pupils also have difficulty understanding exactly what limits are, for this very reason.

Using non-standard analysis would eliminate these issues. We would be able to teach our students precisely what limits are and how to rigorously prove them in an intuitive way. Limits which are tricky and complicated to prove using the classical $\epsilon - \delta$ definitions are often quite simple– even trivial– to prove using the non-standard definition. No more skipping over the topic because it is too complicated for many students. No more asking them to trust us and just use shortcut algorithms without understanding the thing which they are cutting short. No more need to sacrifice rigor for the sake of pedagogy.

Perhaps it is time to really give a non-standard approach to teaching calculus the shot it deserves.