Around 500 BC, there was a Greek philosopher named Zeno. You've probably heard of him, or at least of the dilemma he raised. He stated it in a number of different ways, but I shall concentrate on just one.

The hero, Achilles, and the tortoise agree to a foot race. The tortoise complains that Achilles has the gift of speed, but alas he, the tortoise, does not. So he petitions Achilles for a handicap. And Achilles agrees to allow the tortoise a 1000 meter head start. Surely, Achilles thinks to himself, I can still win this race.

Now Achilles certainly does have the gift of speed and can do a steady 5 meters per second without even breaking a sweat. So that is how fast Achilles decides to run. But the poor tortoise redlines at just half a meter per second. And knowing that he'll have to do his very best even to have a chance against the speedster, Achilles, he must go just as fast has he can go.

The starting signal sounds precisely at noon. The problem is to figure out exactly what time Achilles passes the tortoise.

Well, a little elementary algebra quickly gives us an answer. If
`x _{A}` is where Achilles is,

^{ }æ 1 meters ö x_{T}= 1000 meters + ç~~t ÷ equation of the tortoise~~_{ }è 2 second ø eq. 2.0-1^{ }æ meters ö x_{A}= 0 meters + ç 5~~t ÷ equation of Achilles~~_{ }è second ø

Since Achilles and the tortoise must be in exactly the same spot when one
passes the other, we solve for `x _{A} = x_{T}`. This being the case, we
can subtract the two
equations above to get:

æ 1 meters ö 0 = 1000 meters - ç 4~~t ÷ eq 2.0-2 è 2 second ø 2 t = 222~~~~seconds 9~~

Zeno probably also had the algebraic wherewithall to solve the problem in
a similar way. But he observed that after 200 seconds, Achilles runs the
1000 meter distance of the tortoise's head start. At that time he is where
the tortoise started. But in that amount of time, the tortoise has run
100 meters. So 20 seconds later (that is at `t = 220` seconds) Achilles
catches up to that spot. But in those 20 seconds, the tortoise has run
another 10 meters. It takes Achilles only 2 seconds to catch up to that
spot. But by then, the tortoise has run another meter. Zeno realized
that this process goes on forever, with Achilles always catching up to
where the tortoise was, and the tortoise always running a little bit
farther. So how can Achilles ever pass the tortoise?

We actually have several examples of things that tend to a limit in
this parable. We have the sequence of times, `t _{1}, t_{2}, t_{3},`
etc.
At each

t_{1}= 0 seconds eq. 2.0-3 t_{2}= 200 seconds t_{3}= 220 seconds t_{4}= 222 seconds t_{5}= 222.2 seconds t_{6}= 222.22 seconds . . .

I'm sure you can see the pattern here, but can you prove it? Yes, *you*
can, and you're going to do so right now.

(Think back to how you solved rate
problems in algebra. Remember that ` x = vt`
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

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, `Dt`, to run some distance, how long does
it take
Achilles to run that same distance ? Get that by dividing `Dt`
by `r`.
Why? (refer to the rate formula given above).

Don't let theDt(that is pronounced, "Delta-Tee") throw you. TheDjust means "difference." SoDtjust means the difference between tot's.

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:

DtNotice that the_{1}= t_{2}- t_{1}= 200 seconds - 0 seconds = 200 seconds Dt_{2}= t_{3}- t_{2}= 220 seconds - 200 seconds = 20 seconds Dt_{3}= t_{4}- t_{3}= 222 seconds - 220 seconds = 2 seconds . . . eq. 2.0-4

We already know that Achilles passes the tortoise at

2 t = 222from the algebra we did before. Can we show that the sequence of~~seconds eq. 2.0-5 9~~

You instructor will word it in a different way. He or she will say,
"given any `e` greater than zero, you need to be able to
find an `n`
big enough so that ` |t - t` for all

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, `e`
(that's Greek letter, 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. ` e = 0.001 seconds`), you can tell me,
go past

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 `j`th 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: ` t`
(count 'em -- twelve 2's).

It's pretty clear that after `t _{3}` all the

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
`10 ^{n}`. So, for example, nine

222222222 0.222222222 =So, in the example where~~eq. 2.0-6 10~~^{9}

^{ }ç 2 222222222^{ }÷ e = ç~~-~~^{ }÷ eq. 2.0-7_{ }ç 9 10^{9}÷

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 `10 ^{9}`.
We get:

In fact, when you multiplied those nine^{ }ç 2000000000 1999999998 ÷ 2 e = ç~~-~~~~÷ =~~~~eq. 2.0-8~~_{ }ç 9000000000 9000000000 ÷ 9000000000

2 e =So, can you make~~eq. 2.0-9 9×10~~^{n}

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

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.

Section 2.1: How Close to the Edge Dare You Go?