E-StoreUNDERSTANDING LIMITS
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 xA is where Achilles is, xT is where the tortoise is, and t is seconds past noon, we have the following equation
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| Since Achilles and the tortoise must be in exactly the same spot when one passes the other, we solve for xA = xT. This being the case, we can subtract the two equations above to get: |
0 = 1000 meters - | 1 meters4 | eq. 2.0-2 |
t = 222 seconds
9
Zeno probably also had the algebraic wherewithal 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, t1, t2, t3, etc. At each tj Achilles is where the tortoise was at tj-1. That defines all the tj's except t1, which we will say is the moment of the starting signal. So the various tj's are:
t1 = 0 seconds eq. 2.0-3
t2 = 200 secondst3 = 220 secondst4 = 222 secondst5 = 222.2 secondst6 = 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 enormous role very soon. Your ability to solve rate problems will have to become second nature. If you need remediation on rate problems,
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, Δt, to run some distance, how long does it take Achilles to run that same distance ? Get that by dividing Δt by r. Why? (refer to the rate formula given abovE
ε represents how close we are to the limit.
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
ε =
2000000000 1999999998 − 9000000000 9000000000
2= eq. 2.0-89000000000
In fact, when you multiplied those nine 2's by 9, it became pretty clear that there is a pattern there. And if you multiply n 2's by 9, what you will always get is a 1 followed by n-1 9's, followed by an 8. And when you subtract that from 2 × 10n, you will always get 2. So that will always be the numerator of the difference. And the denominator? Clearly it will always be 9 × 10n. So for a decimal point followed by n 2's, we have:
2ε = eq. 2.0-99×10n
So, can you make ε as close to zero as you like by choosing the right n? and does ε stay at least that close for all subsequent n? The answer is clearly yes, because the numerator doesn't change as n gets big, but the denominator keeps getting more and more huge.
Well we have diverged a long way from the foot race. So what have we proved? We have proved that the sequence of times, tj, 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.
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