Differential Calculus: From Practice to Theory

We want to ask, and answer, the question

as precisely as we can without invoking the notion of the infinitely large. Does this rephrasing work for you?

This seems to be better. We have side-stepped the issue of the infinitely large by saying “larger and larger.” But unfortunately this rephrasing of the question changes its meaning. To see what we mean, think of \(x\) as the radius of a balloon which we want to inflate as much as we possibly can. Suppose we know that when the radius of the balloon is \(8\) it will pop. To avoid popping our balloon, we blow in enough air during the first second to expand the radius to \(x(1)=4\text{.}\) In the next second we expand the radius to fill half of the remaining distance to \(x(2)=6\text{.}\) In the third we repeat the process, expanding it to \(x(3)=7\text{.}\) And we continue in this fashion. At each second the radius of our balloon expands to half of the remaining distance to \(8\text{.}\)

Clearly the balloon’s radius grows larger each second, so \(x\) grows “larger and larger” as required. But the intent of our original question was for \(x\) to become “infinitely large,” not for \(x\) to remain smaller than \(8\text{.}\)

And there is nothing special about \(8\text{.}\) If \(x\) moves half of the distance to \(16\) each second, or halfway to \(32\text{,}\) or even if it moves halfway to \(1,048,576\) at each step we have the same problem. In each case \(x\) is growing “larger and larger” but it is not growing in a manner that reflects what we think we mean when we say that \(x\) is growing “infinitely large.”

In this situation \(x\) is said to be bounded above because there is an upper bound on how large \(x\) can be, despite the fact that \(x\) is growing “larger and larger.”

Recall that in Section 12.1 we mentioned that the notation \(x\rightarrow\infty\) should be read aloud as “\(x\) increases without bound,” so we rephrase our question as follows:

This rephrasing poses our question unambiguously and without reference to the “infinitely large.” And the answer is very clear: \(y\) gets closer and closer to zero. But of course, the phrase “closer and closer” is also too vague, and for essentially the same reason that “larger and larger” is too vague to be useful.

Again, it is not our conception of the problem that is the difficulty. It is the language we’re using. As before, we must choose our words more carefully.

To capture the idea that \(y=\frac1x\) gets “closer and closer” to zero, without ever getting to zero let’s think this through, being careful to say exactly what we mean, no more, no less. To begin we ask, “Is there a value of \(x\) which forces \(y\) to be less than, say \(\frac{1}{2}\text{?}\)”

Recall that if \(0\lt a\lt b\) then \(\frac1a\gt\frac1b\text{.}\) Thus if \(\frac{1}{x}\lt \frac{1}{2}\text{,}\) then \(\frac{1}{\frac{1}{x}}\gt \frac{1}{\frac{1}{2}}\text{,}\) or \(x\gt 2\text{.}\)

So apparently any value of \(x\) strictly greater than \(2\) will guarantee that \(y=\frac1x\) is less than \(\frac{1}{2}\text{.}\)

Stop and think about that last sentence. Do you see that we’ve actually discovered more than the original question asked for? Our question was, “Is there a value of \(x\) which forces \(y\) to be less than \(1/2\text{?}\)” But we’ve actually found all of them. We’ve found that if \(x\) is any number greater than \(2\) then \(y=\frac1x\lt \frac12,\) regardless of which number we use.

Can we make \(y\lt 1/3?\) Sure. Exactly the same analysis will show that if \(x\gt 3\text{,}\) then \(y=\frac1x\lt \frac13\text{,}\) or if \(x\gt 4\text{,}\) then \(y\lt 1/4\text{,}\) and so on.

It should be clear that we needn’t have stopped at one millionth \(\left(10^{-6}\right)\text{.}\) The same argument will show that if we want \(y=\frac1x\lt 10^{-10},\) we need \(x\gt 10^{10}.\) And that if we want \(y=\frac1x\lt 10^{-1000},\) we need \(x\gt 10^{1000}.\)

So, by an appropriate choice of \(x\) we can make \(y=\frac1x\) as close to \(0\) as we choose. To be a little more precise we say that we can make \(y\) arbitrarily close to zero, even if it never actually is zero. And this is what we really intended when we said that \(y\) gets “closer and closer” (or “goes to”) zero.

Our discussion, in Subsection 17.2.1, of the meaning of \(\limit{x}{\infty}{\frac1x}=0\) is the key to redefining limits in a precise manner that would satisfy a skeptic such as Bishop Berkeley. For this particular limit, every challenge (How can we assure that \(y=\frac1x \) is within \(D\) of zero?) had it’s corresponding response (Take \(x\gt \frac1D \text{.}\)). But rather than doing this case-by-case, let’s do it for all cases.

We need answer all possible challenges at once. This seems like a lot to ask until we think about it a bit. All we really have to do is suppose that we have some small, positive, unspecified number and show that we can find out how large \(x\) has to be to make \(y=\frac1x\) less than that number. For the sake of being definite we’ll give our number a name. It is traditional to call it \(\eps\text{.}\)

Suppose that \(\eps\gt 0\) and we want to figure out how large to make \(x\) to guarantee that \(y=\frac1x\lt \eps.\) If we want \(y\lt \eps\text{,}\) first substitute \(\frac1x\) for \(y\) giving \(\frac1x\lt \eps\text{.}\) Solve this for \(x\) (remember that if \(a\lt b\) then \(\frac1a\gt \frac1b\)) so that \(x\gt \frac1\eps\text{.}\)

Do you see the significance of this? Because we left \(\eps\) unspecified (other than requiring it to be positive), we’ve met all possible challenges. If the challenge is to make \(y\) less than \(10^{-1000000000}\) our response is, “We’ve already done that. Just take \(\eps= 10^{-1000000000} \text{.}\)” Repeating the computation above gives \(x\gt10^{1000000000} \text{.}\)

Now for some function \(f(x)\gt0\text{,}\) the statement \(\tlimit{x}{\infty}{f(x)}=0\) has the following precise meaning:

To be sure, when we allow \(\eps\gt0\) to be arbitrary, but unspecified we skirt the edge of the infinitely small. But this is the point. If \(\eps\) is arbitrary then it can be as small as we need for it to be without ever being infinitely small. This is akin to Euclid allowing lines to be extended to any, unspecified, length without allowing them to be infinite in length. This is the idea underlying limits and limit notation.

Warning! Be aware that the way we tend to speak about limits and the meaning of the limit notation are inconsistent. This can present a lot of problems for the beginner. If we are speaking loosely, among friends, we would read this statement, \(\tlimit{x}{\infty}{\frac1x}=0\text{,}\) as: “The limit of one over \(x\) as \(x\) goes to infinity is equal to zero.”

This is a very poor way to express the idea we are trying to capture. To say that \(x\) “goes to” infinity completely undercuts everything we’ve said about infinity because it treats infinity as if it is an actual number we can get close to. It is not

When reading \(\tlimit{x}{\infty}{\frac1x}=0\) we should say “The limit of \(\frac1x \) as \(x\) increases without bound is zero”, but almost no one does. Generally, when talking about limits what we actually say is not what the notation means. That incongruity can be very confusing at first. In this text we will be very careful not to speak so casually. At least not until we have more experience with limits.

We have only begun, but this is enough for us to offer a first definition of the limit concept. We generalize slightly.

The parameter \(B\) is the lower bound that \(x\) has to exceed for \(f(x)\) to be less than \(\eps\text{.}\) In our first example we had \(B=2\text{,}\) in our last we had \(B=1/\eps\text{.}\) Naming the lower bound like this gives us a concrete way to specify how large \(x\) has to be. To great extent finding \(B\) is the whole problem. This is easier to see in an example.

Definition 17.2.3 works as long as \(f(x)\gt 0\text{,}\) but without this restriction it fails utterly as our next example shows:

Based on our experience in Example 17.2.6 with \(f(x) = -2-\frac1x\text{,}\) what would you say needs to be changed in Definition 17.2.3 to allow \(f(x)\le0\) as well?

The problem of course, is that if \(\eps\gt 0\) is given the statement \(f(x) \lt \eps\) doesn’t really capture the idea that \(f(x)\) is near the number zero, only that it is less than the number \(\eps\text{.}\) For example, \(-1000\) is less than \(\eps\) but it is nowhere near zero. What we need is a way to measure how far \(f(x)\) is from zero, regardless of whether in the positive or the negative direction.

That is exactly what the absolute value function measures. For example, both \(3\) and \(-3\) are the same distance from zero, the first one in the positive and the second one in the negative direction. That is \(\abs{3}=\abs{-3}=3\text{.}\)

We need to modify our definition so that the distance from \(y\) to zero is less than \(\eps\text{.}\) We want \(\abs{y}\lt \eps,\) not just \(y\lt \eps\text{.}\)

Notice that in Example 17.2.8 and Example 17.2.10 that the scrapwork was an essential part of the solution, but in the formal proof the results of the scrapwork were so abbreviated as to almost not be present. This is part of the formalism of mathematical writing. We try to make sure that everything that needs to be said is said — and absolutely nothing more. In this instance in particular, we are not obligated to explain where the bound \(B\) came from, only that it works. It can take time to become comfortable with this presentation style.

Even with practice very few people can read a formal proof without doing the computations necessary to show that all of the claims made are actually true. Keep paper and pencil handy at all times to help you follow the argument.

Definition 17.2.7 only tells us what it means when the limit of some function as \(x\rightarrow\infty\) is zero. But as we observed in Example 17.2.6 as \(x\) increases without bound \(f(x)=-2-\frac1x\) approaches \(-2\text{,}\) not \(0\text{.}\) We’ll need something more general, but all of the important ideas have been introduced. We generalize Definition 17.2.7 as follows.

Notice that if \(L=0\) this reduces to Definition 17.2.7.

It should be clear how to define a limit at \(-\infty\text{.}\) All of the same issues of clarity and precision that we encountered before come up here as well. The only difference is that we have to change the sense of our inequalities to reflect that \(x\) is decreasing without bound.

In the last section, we were able to transform our intuitive understanding of horizontal asymptotes into a rigorous definition of \(\limit{x}{\pm\infty}{f(x)}\) which did not make reference to infinity. Here we will do the same for vertical asymptotes \(\left(\limit{x}{a}{f(x)}=\pm \infty\right)\text{.}\) Along the way, we will develop a more rigorous way of addressing one sided limits as well.

We next want to rigorously define the meanings of the symbols in formula (17.4) without referring to the nebulous word “infinity.”

For example, if we want \(\frac{1}{x}>100\text{,}\) then this will happen for \(0\lt x\lt \frac{1}{100}\text{.}\) If we want \(\frac{1}{x}>1000\text{,}\) then this will happen when \(0\lt x\lt \frac{1}{1000}\text{.}\) It seems reasonable that given a positive number \(B\text{,}\) we should be able to get \(\frac{1}{x}>B\text{,}\) provided \(0\lt x\lt\frac{1}{B}\text{.}\)

Using the example \(\rlimit{x}{0}{\frac1x} =\infty \) as a guide, we make the following definition.

Using Definition 17.2.22 as a guide we make the following definition.

The following problem shows that Definition 17.2.28 isn’t strictly necessary because we could define such limits viaDefinition 17.2.22.

Well, yes. Sort of. It tells us that the expression \(\limit{x}{0}{\frac{1}{x}}\) is without meaning; that it is undefined.

What we have here is Sherlock Holmes’ Maxim 9.2.17 taken to its extreme. Instead of eliminating all but one possibility we have eliminated all possibilities. Therefore we forced to conclude that \(\limit{x}{0}{\frac{1}{x}}\) has no meaning, — that it does not exist.

Thus we have the following definition.

It is common to take the statement in Problem 17.2.37 as the definition of \(\limit{x}{a}{f(x)}=\infty \) and then to prove Definition 17.2.33 as a theorem. You did this in Problem 17.2.37. Logically the two statements are equivalent so it doesn’t matter which is taken as the definition. Each statement can be used to prove the other as a theorem.

Finally, notice that in Problem 17.2.37 , we had the conditions \(a-\delta \lt x\lt a+\delta \) and \(x\neq a\text{.}\) This can be more succinctly written as \(0\lt \left|x-a\right|\lt \delta \text{.}\) Keep this in mind as you read the next section on limits that don’t involve infinity.