Differential Calculus: From Practice to Theory

Of course, the notion of “approaching infinity” is troublesome as well. That we speak in this fashion has more to do with the construction of the English language than with the mathematics we want to represent. If \(y\) is proceeding toward zero (an actual number) it makes sense to say “\(y\) approaches zero.” The notation “\(y\rightarrow0\)” captures this idea nicely. And since we usually pronounce the symbol \(\infty\) as “infinity” it feels natural to write \(x\rightarrow\infty\) and to speak this aloud as “\(x\) approaches infinity” as if \(\infty\) is a place we could get to. But as we’ve observed, this is absurd. Properly speaking the symbols “\(x\rightarrow\infty\)” should be spoken aloud as “\(x\) increases without bound,” but nearly everyone says “\(x\) approaches infinity” instead because it is a less cumbersome phrase. Like so much of mathematics this is perfectly clear once the underlying ideas have been internalized, but in the beginning it is can be quite confusing. Be patient with yourself.

The expression \(\tlimit{x}{\infty}{\frac{1}{1+x^2}}=0\) is making a very specific statement, one that is difficult to capture in a natural language like English. It says that the expression \(\frac{1}{1+x^2}\) is approaching a particular value (zero) as \(x\) approaches \(\infty\text{.}\) The equals sign doesn’t indicate that the function, \(\frac{1}{1+x^2}\) actually equals that value. Instead it is the limit of \(\frac{1}{1+x^2}\) which is equal to zero. Whether the function itself is ever equal to the limit, or not, is irrelevant. We’re interested in what it gets arbitrarily close to.

We are treading at the edge of some very deep waters here so we will not attempt to address all of the nuances yet. Instead we will simply interpret the formula \(\tlimit{x}{\infty}{\frac{1}{1+x^2}}=0\) to mean that as \(x\) approaches infinity (\(x\rightarrow\infty\)) the expression \(\frac{1}{1+x^2}\) approaches zero \(\left(\frac{1}{1+x^2}\rightarrow0\right)\text{.}\)

In Chapter 17 we will replace “arbitrarily close” with something much more precise. Indeed, finding a better way to say “arbitrarily close” will be the culmination of our efforts. This is what will allow us to finally replace differentials with a solid theoretical foundation.

But for now, we have enough to give our first definition of the limit concept.

We used the Witch for our first example because we are quite familiar with its properties from our earlier investigations. However we needn’t have started with so complex a function.

The limit concept is deeply abstract. We are introducing it here because, (1) the limit idea gives us a handy way to define asymptotes, and (2) when we return to it in Chapter 17 it will be helpful if you are comfortable with the notation and the essential underlying idea.

We have the following definition.

Yes, of course it does. In fact the following theorem is true.

This can be expressed in words with the phrase, “The limit of a sum is the sum of the limits.”

The following two theorems are not quite as intuitive as Theorem 12.1.6, but they are also true.

This can be expressed in words with the phrase “The limit of a product is the product of the limits.”

This can be expressed in words with the phrase “The limit of a quotient is the quotient of the limits.”

Like the differentiation rules, when these limit theorems are used together they allow us to break a large problem into smaller, more tractable, pieces. Here are some examples.

Of course, not every curve has a horizontal asymptote so not every function has a limit as \(x\) approaches \(\pm\infty\text{.}\) For example, consider the curves \(y=x^2\) and \(y=x^3.\)

Clearly these curves have no horizontal asymptotes because as \(x\rightarrow\pm\infty\) they don’t get close to a real number. In fact they either continue to rise or continue to drop as \(x\rightarrow\pm\infty\text{.}\)

It is very clear that the limit: \(\limit{x}{\infty}{(x-x)}\) is zero. After all, \(x-x\) is constantly equal to zero so we have \(\limit{x}{\infty}{(x-x)} = \limit{x}{\infty}{0} = 0\text{.}\) So it seems to be clear that \(\limit{x}{\infty}{(x-x)} = \infty-\infty = 0\) because anything minus itself has to be zero doesn’t it? Even grandmother - grandmother must be zero, right?

The source of the difficulty here is that only numbers can be subtracted. But neither \(\infty\) nor grandmother is a number so the expressions \(\infty-\infty\) and grandmother - grandmother are utter nonsense. Which is to say, they are undefined.

Limits can be deceptive so be careful. It can be helpful to have built in notational cues to remind ourselves that these are subtle problems. When we are working with limits we will use the notation \(\approach{\infty}\) to remind ourselves that \(\infty\) is not a number.

This says that the limit consists of something which is approaching \(\infty\) (increasing without bound) minus something else which is approaching \(\infty\) (again, increasing without bound). When we say it like this — more importantly when we think about it like this — we can see that the value of the limit will depend upon precisely how the two parts increase without bound.

Limits which appear to resolve to an expression like \(\approach{\infty} - \approach{\infty} \) are called indeterminate forms. Some indeterminate forms can be evaluated using what is known as L’Hôpital’s Rule. We will learn how to use L’Hôpital’s Rule in Section 12.4. But unfortunately L’Hôpital’s Rule only applies to some indeterminate forms so we still need all of our other tools.

In particular to evaluate \(\limit{x}{\infty}{(x^3-21x^2)}\) we’ll need Algebra.

We now know that this limit is not necessarily equal to zero, but how can we discover what it actually is equal to? As in much of Calculus we need to re-express our formulas algebraically to see what is going on.

Next we consider the limit \(\limit{x}{-\infty}{(x^3+21x^2)}\)

We have established that, because \(\infty\) is not a number, the expression \(\infty-\infty\) does not make sense. What about \(\frac{\infty}{\infty}\text{?}\) Is it true that \(\frac{\infty}{\infty}=1\text{?}\)

No, of course not. The expression \(\frac{\infty}{\infty}\) is meaningless for the same reason that \(\infty\) - \(\infty\) is meaningless.

We can now see that \(y(x)=\frac{2+2x^2+2x^4}{1+x^4}\) has a horizontal asymptote at \(y=2\text{.}\) If you haven’t already done so consider graphing this function to confirm.