Limits

Historically, the calculus of infinitesimals as created by Newton and Leibniz was reinterpreted in the nineteenth century by Cauchy, Bolzano, and Weierstrass in terms of limits. All mathematicians learned both languages, and switched back and forth between them effortlessly, like the lady I overheard in a Southern California supermarket telling her mother, “Let’s get that one, con los nuts.” Those who had been trained in infinitesimals might hear a statement using the language of limits, but translate it mentally into infinitesimals; to them, every statement about limits was really a statement about infinitesimals. To their younger colleagues, trained using limits, every statement about infinitesimals was really to be under- stood as shorthand for a limiting process. When Robinson laid the rigorous foundations for the hyper- real number system in the 1960’s, a common objection was that it was really nothing new, because every statement about infinitesimals was really just a different way of expressing a corresponding statement about limits; of course the same could have been said about Weierstrass’s work of the preceding century! In reality, all practitioners of calculus had realized all along that different approaches worked better for different problems; Problem 4.13 is an example of a result that is much easier to prove with infinitesimals than with limits.

The Weierstrass definition of a limit is this:

Definition of the limit

We say that is the limit of the function f(x) as approaches , written

$\lim_{x\rightarrow a}f(x)=l$

if the following is true: for any real number $\epsilon$ , there exists another real number $\delta$ such that for all in the interval $a-\delta \leq x\leq a+\delta$ , the value of lies within the range from $l-\epsilon$ to $l+\epsilon$ .

Intuitively, the idea is that if I want you to make f(x) close to ,just have to tell you how close, and you can tell me that it will be that close as long as x is within a certain distance of a.

In terms of infinitesimals, we have:

Definition of the limit

We say that is the limit of the function f(x) as approaches , written

$\lim_{x\rightarrow a}f(x)=l$

if the following is true: for any infinitesimal number dx, the value of f(a+dx) is finite, and the standard part of f(a+dx) equals .

The two definitions are equivalent. As remarked previously, the derivative dx/dt can be defined as the limit $\textrm{lim}\Delta _{t\rightarrow 0}(\Delta x/\Delta t)$ , and if we use the Weierstrass definition of the limit, this means that the derivative can be defined entirely in terms of the real number sys- tem, without the user of hyperreal numbers.

Sometimes a limit can be evaluated simply by plugging in numbers:

Example

Evaluate

$\lim_{x\rightarrow 0}\frac{1}{1+x}$

Plugging in x=0 , we find that the limit is 1.

In some examples, plugging in fails if we try to do it directly, but can be made to work if we massage the expression into a different form:

Example

Evaluate

$\lim_{x\rightarrow 0}\frac{\frac{2}{x}+7}{\frac{1}{x}+8686}$

Plugging in x=0 fails because division by zero is undefined.

Intuitively, however, we expect that the limit will be well defined, and will equal 2, because for very small values of , the numerator is dominated by the 2/x term, and the denominator by the 1/x term, so the 7 and 8686 terms will matter less and less as gets smaller and smaller.

To demonstrate this more rigorously, a trick that works is to multiply both the top and the bottom by , giving

$\frac{2+7x}{1+8686x}$

which equals 2 when we plug in x=0 , so we find that the limit is zero.

This example is a little subtle, because when equals zero, the function is not defined, and moreover it would not be valid to multiply both the top and the bottom by . In general, it’s not valid algebra to multiply both the top and the bottom of a fraction by 0, because the result is 0/0, which is undefined. But we didn’t actually multiply both the top and the bottom by zero, because we never let equal zero. Both the Weierstrass definition and the definition in terms of infinitesimals only refer to the properties of the function in a region very close to the limiting point, not at the limiting point itself.

This is an example in which the function was not well defined at a certain point, and yet the limit of the function was well defined as we approached that point. In a case like this, where there is only one point missing from the domain of the function, it is natural to extend the definition of the function by filling in the “gap tooth.” Example below shows that this kind of filling- in procedure is not always possible.

Example

Investigate the limiting behavior of 1/x^2 as approaches 0, and 1.

At x=1 , plugging in works, and we find that the limit is 1.

Figure 3.5 Example 40 the function

1/x^2
At x=0 , plugging in doesn’t work, because division by zero is undefined. Applying the definition in terms of infinitesimals to the limit as approaches 0, we need to find out whether 1/(0+dx)^2 is finite for infinitesimal dx , and if so, whether it al- ways has the same standard part. But clearly $1/(0+dx)^2=dx^{-2}$ is always infinite, and we conclude that this limit is undefined.

Example

Figure 3.6

$\textrm{tan}^{-1}(1/x)$
Investigate the limiting behavior of $f(x)=\textrm{tan}^{-1}(1/x)$ as approaches 0.

Plugging in doesn’t work, because division by zero is undefined.

In the definition of the limit in terms of infinitesimals, the first requirement is that f(0+dx) be finite for infinitesimal values of . The graph makes this look plausible, and indeed we can prove that it is true by the transfer principle. For any real we have $-\pi/2\leq f(x)\leq \pi / 2$ , and by the transfer principle this holds for the hyperreals as well, and therefore f(0+dx) is finite.

The second requirement is that the standard part of f(0+dx) have a uniquely defined value. The graph shows that we really have two cases to consider, one on the right side of the graph, and one on the left. Intuitively, we expect that the standard part of f(0+dx) will equal $\pi /2$ for positive , and $-\pi /2$ for negative, and thus the second part of the definition will not be satisfied. For a more formal proof, we can use the transfer principle. For real with 0< x< 1 , for example, is always positive and greater than 1, so we conclude based on the transfer principle that f(0+dx)>1 for positive infinitesimal . But on similar grounds we can be sure that f(0+dx)<-1 when is negative and infinitesimal. Thus the standard part of can have different values for different infinitesimal values of , and we conclude that the limit is undefined.