Section 2 Limits

(Think back to how you solved rate problems in algebra. Remember that x = v * t where x is distance, v is speed, and t is time. Even if you hated rate problems in algebra, you have to get good at them now, because rates will play an enormous role very soon. Your ability to solve rate problems will have to become second nature.

By what factor does Achilles' speed exceed the tortoise's? Get that by dividing the tortoise's speed into Achille's speed. Call that number r for ratio. Now, if it takes the tortoise an amount of time, delta_t, to run some distance, how long does it take Achilles to run that same distance ? Get that by dividing delta_t by r. Why? (refer to the rate formula given above).

Now observe that during each interval, Achilles runs the same distance that the tortoise ran in the previous interval. Here is the pattern of the intervals, which we call delta_t's:

   delta_t1  =  t2 - t1  = 200 seconds -   0 seconds = 200 seconds
   delta_t2  =  t3 - t2  = 220 seconds - 200 seconds =  20 seconds
   delta_t3  =  t4 - t3  = 222 seconds - 220 seconds =   2 seconds
   .
   .
   .                                                              eq. 2.0-4

We already know that Achilles passes the tortoise at

           2
   t = 222 - seconds                                              eq. 2.0-5
           9

You instructor will word it in a different way. He or she will say, "given any epsilon greater than zero, you need to be able to find an n big enough so that |t - t_j| < epsilon for all j > n no matter how close to zero epsilon is."

Don't let the Greek symbol or the mathematical formalism intimidate you. The statement of the problem I gave means precisely the same thing as the one your instructor gives. A limit means simply, I tell you how close I need to be, and you can tell me what I have to do to be that close. In the formal version of this, epsilon is the "how close I gotta be" and n is the "what I gotta do to be that close."

If I say, for example, I've got to be within a thousandth of a second of t (i.e. epsilon = 0.001 seconds), you can tell me, go past t₇ (i.e. n = 7) and you will find all values of t_j that are within a thousandth of a second of t. If I said I've got to be within a millionth of a second, you could tell me what? (go ahead and extend the table of t_j's to find out).

But so far we haven't proved the assertion. To prove it, we must show that we can always find how far to go no matter how close we have to be. Let's do that now.

We can see that the jth term of the sequence given in 2.0-3 is a bunch of 2's, beginning in the hundreds column and extending rightward. There are always j-1 2's in each decimal. So we can easily write the decimal for any term in the series. The 13th, for example, is: t₁₃ = 222.222222222 (count 'em -- twelve 2's).

It's pretty clear that after t₃ all the t_j's are within 1 of 222. So all we will prove is that the fractional part of the sequence (that is the part to the right of the decimal point) converges to a limit of 2/9.

So let's find the difference between a bunch of 2's to the right of the decimal point and 2/9. Recall that if we have n 2's to the right of the decimal point, what we really have is the counting number made of n digits of 2's divided by 10ⁿ. So, for example, nine 2's to the right of the decimal point means:

                    222222222
   0.222222222  =   ---------                                     eq. 2.0-6
                       109

               | 2     222222222 |
    epsilon  = | -  -  --------- |                                eq. 2.0-7
               | 9        109    |

We have to put these fractions over a common denominator in order to subract them. So to the right-hand one we have to multiply top and bottom by 9. To the left-hand one we have to multiply top and bottom by 10⁹. We get:

                | 2000000000     1999999998 |          2
   epsilon   =  | ----------  -  ---------- |  =   ----------     eq. 2.0-8
                | 9000000000     9000000000 |      9000000000

               2
   epsilon = -----                                                eq. 2.0-9
             9*10n

Well we have diverged a long way from the foot race. So what have we proved? We have proved that the sequence of times, t_j, does indeed converge to a limit, and that limit is the same time that we predicted earlier, using algebra, that Achilles should be passing the tortoise. We can tell Mr. Zeno that even if he can never see how Achilles gets to 222 2/9 seconds, we can show that he certainly gets as close as anybody could possibly ask.

And since we know that Achilles does indeed pass the tortoise, our notion of a limit bears a true connection to reality. Zeno's dilemma was that the sequence of times leading up to the passing represented a completed infinite, yet the infinite is defined as that which never ends. Zeno's problem was that he had a poor definition of "infinite." Infinites do end all the time. And they end in a limit.

2.1 How Close to the Edge Dare You Go?

When I was a kid, my aunt went and bought this tiny little car. Undoubtably the real reason she did it was because she was going through some financial difficulties at the time, and it was all she could afford. But what she told her kids (my cousins) was that small cars were good. Big cars hog the road, don't drive well, destroy the earth, and so on. Well one day my aunt and cousins showed up packed into this thing. The eldest of my cousins got out and proceded to repeat the whole "small car good, big car bad," line to me. We then walked down the driveway, and there was our big old station wagon, which my dad had assured me was a great car. So I asked my cousin, "Is this car too big?" Realizing that a little tact was called for, he replied, "No, but if it were any bigger, it would be too big."

Well, I found that response perfectly maddening. You mean, if you added just one inch, it would be too big? What if you just added a coat of paint to the bumper -- that would make it a little bigger. Would it be too big then? It was clear to me that there was no place where too-biggness started, only a place where small-enough ended. What was maddening had nothing to do with my cousin. It had everything to do with discovering something new about how the world works. What was the biggest a car could be and not be too big? Our station wagon filled that bill. But what was the smallest a car could be and still be too big? In my cousin's world, no such thing could possibly exist.

It took days for me to resolve this dilemma in my own head. And what resolved it was thinking about all the little-boy macho games I had played with other boys -- like the time a bunch of us found our way to the roof of a condemned building and dared each other how close to the edge of the roof we could stand. Clearly there was a place where you could not stand, and if you tried to stand there you would find yourself plummeting toward the ground. But there was no place that was the absolute limit of where you could stand without falling, and that, of course, was the whole point of the game. No matter how close you stood to the edge, some other guy could stoke up his courage and skooch just a tad closer to disaster than you were.

And this brings us to another kind of a limit. How close can you stand to mathematical disaster without falling off the edge? But what is mathematical disaster? Well, division by zero, for one. Thou shalt divide by any of the fruits of this garden, but by the tree of zero, thou shalt not divide. You mean I can divide by a tiny little number like 0.00000001, but not by zero? Yep. Even by 0.0000000000001? Yep -- even tinier than that. Just don't go dividing by zero.

So there is that spot we can't stand on without falling. But if I dare you to stand closer to it than I am, you can always do it.

So suppose I had the function:

          x2  - 3*x + 2
   y  =   -------------                                           eq. 2.1-1
             x - 1

         (x - 1) * (x - 2)
   y  =  -----------------                                        eq. 2.1-2
              x - 1

   y = x - 2                                                      eq. 2.1-3

Let's go back to the macho boys standing on the roof of the condemned building, but imagine that it's a sloped roof. So the altitude of each boy's feet varies according to where on the roof he is standing. Based upon how far right or left a boy is standing, we can determine how high his feet are. Imagine that red trace of our graph represents the roof. The hole in the graph is a hole in the roof. If one of the boys tries to stand there, he will fall all the way to the basement and his mother will be angry with him. But the boys persist in a game of who-can-stand-nearest-the-hole. So how high would a boy's feet be if he could stand on the hole?

In a way, that sounds like an absurd question -- rather like the proverbial tree falling in the woods with nobody around. You could even be a smartass about it and say his feet would be at the same altitude as the rest of him, all twisted up in the basement. Yet we have a sense that there is a serious answer to this based upon the behavior of the rest of the roof.

Let's say we measure altitude from the basement, which on our graph we'll say is at y = -10 meters (that is the basement is 10 meters below the x-axis), and that each square on the graph is 1 meter. We sense that the answer to how high a boy's feet would be if he could stand on the hole ought to be 9 meters. But why?

We can learn much from the game they are playing. We see that the closer a boy stands to the hole, the closer his feet are to being exactly 9 meters above the basement. How close to 9 meters above the basement would you like a boy's feet to be? No matter how close that is, you can dare him to stand close enough to the hole that his feet will be within that degree of closeness to 9 meters above the basement.

And that's the point. If you can tell me how close the boy's feet have to be to 9 meters, I can tell you how close he has to stand to the hole. That is what makes 9 meters a limit. So we can say, the limit of altitude as the boy's position goes toward the position of the hole is 9 meters. Or to express it in the standard math shorthand:

                  lim                  altitude of his feet  =  9 meters
   boy's position --> position of hole

      lim   af  =  9 meters
    x --> xh

Now let's analyse equations 2.1-1 and 2.1-2 again. As we already observed, everywhere except at x = 1 either of these functions is identical to y = x - 2. At precisely x = 1, the functions given in equations 2.1-1 and 2.1-2 have no meaning. But as x gets very close to 1, y = x - 2 gets very close to -1. Not only that, you can get y = x - 2 as close to -1 as you'd like simply by skooching x close enough to 1. We can see this trend without ever having to evaluate y at the forbidden value of x = 1. And so we say, the limit of y as x goes toward 1 is -1. Or using the mathematical shorthand:

           (x - 1) * (x - 2)
     lim   -----------------   =  (x - 2)  =  -1
   x --> 1      (x - 1)

It is with near certainty that your teacher will call upon you to prove this or that limit is equal to such and such. And he or she will want you to use the traditional method of delta and epsilon to do it. Before I demonstrate that for you, I want you to make the connection between this and what we have been talking about. In the case of the boy on the roof, I could tell you how close I wanted the altitude of his feet to be to 9 meters, and you could respond with how close to the hole he had to be. That made 9 meters the limit. In the case of equations 2.1-1 and 2.1-2, I could tell you how close I wanted y to be to -1 and you could tell me how close x had to be to 1 in order to make that happen. That made -1 the limit for y.

In the delta and epsilon proofs, the epsilon is the how close the dependent value has be, and the delta is the how close you have to make the independent variable to the "goes toward" value in order to make that happen.

In the case of the boy on the roof, the epsilon is how close his feet are to being exactly 9 meters above the basement, and the delta is how close to the hole he has to stand in order for that to happen.

In equations 2.1-1 and 2.1-2, the epsilon is how close y is to -1 and the delta is how close x is to 1.

The typical way you see it phrased in a math text is:

We say that:

              lim    f(x)   =   y
            x --> a

if and only if for each epsilon > 0 there exists a delta such that |f(x) - y| < epsilon whenever |x - a| < delta.

The test is, I name how close to y I want f(x) to be. That is I say I want f(x) to be within epsilon of y. If, no matter how small an epsilon I name, you can name how close x has to be to a in order to guarantee what I asked for, then you have proved the limit.

So in the example of equations 2.1-1 and 2.1-2, we have:

             (x - 1) * (x - 1)
    f(x)  =  -----------------                                    eq. 2.1-3
                  (x - 1)

In general, if you can show a formula that, given an epsilon, turns out a delta that always meets the test, then you have met your teacher's test, and he or she will mark your work correct.

The Main Point to Remember from 2.0 and 2.1

We have discussed two kinds of limits so far, a limit that a sequence of numbers approaches, and a limit that a function approaches. They may seem very different, but they aren't. In both cases, a limit is like a contract. It says, if you need to be this close to something (and it doesn't matter how close this close is), then I can tell you what you need to do to be this close. That is what the delta-epsilon proofs are all about. That is what limits are always about.

The first type of limit is similar to a race-car mechanic telling the machinist how accurately the engine parts must be cut, and the machinist telling him how many dollars it will cost. However accurate the mechanic wants it, the machinist can deliver it with that accuracy. But the tighter the tolerance, the more the cost.

The second is similar to aiming a rifle. You tell the marksman how close to the bullseye he has to hit, and he can tell you how still he has to hold the rifle in order to make that happen.

Here is a graphic that illustrates the delta-epsilon method of proving the second kind of limit. The two green horizontal lines are no more than epsilon above the limit as x goes toward 2 and no more than epsilon below the limit. The two blue vertical lines are spaced delta on either side of x = 2. A delta-epsilon proof requires that you show that you can always make the two horizontal lines as close as you like to the limit provided you make the two vertical lines sufficiently close to whatever x is going toward (in this case 2). As you bring the two vertical lines closer and closer to what x is going toward, the two horizontal lines must close in on the limit.

If you remember these underlying ideas about limits, learning the details of how to do limits proofs will be a lot easier for you.