Section 2 Limits (continued)

2.2 Combining Limits: Two Tall Tales

Suppose you are the skipper of a boat that is lost in the fog. Well, you're not completely lost, because you knew where you were when the fog descended over you. You know exactly how far and in what direction you have to go in order to get back to port. You have an accurate compass and speed indicator. But you have a problem. Your watch is broken. Your only option is to begin making for the mouth of the harbor at a fixed speed, and count the time in your head. Your time count has got to be accurate. If you go too far, you'll end up on the rocks. If you don't go far enough, you'll never find the mouth of the harbor mouth in the fog.

Let's say it will take exactly two hours to get to the mouth of the harbor. The accuracy of your time count, and therefore the accuracy of where you end up, is dependent upon how hard you can concentrate on counting seconds for two hours. The harder you concentrate, the more accurate you will be. In fact, you can be as accurate as you like just by concentrating hard enough.

So we can say that the mouth of the harbor is the limit of where you'll end up as your concentration goes toward perfection. But that is not the end of the problem. When you reach what you believe to be the harbor mouth, the fog is so thick that you can see nothing. You still have to get to your mooring in the harbor. Since you know exactly how far it is from the harbor mouth to your mooring, you set course for the mooring counting seconds again. And again, your accuracy depends upon your degree of concentration.

Where you end up after both legs of the journey is the sum of the distances you went on each leg. And with what accuracy do you reach your destination? Is the mooring the limit of where you end up as your concentration goes toward perfection? In other words, if you executed everything perfectly, the distance to the mooring would be the sum of the distance to the mouth of the harbor and the distance from the mouth of the harbor to the mooring, right? But you are unable to navigate with complete perfection because you are counting the seconds without a watch.

So is the sum of the limits the same as the limit of the sum? Keep in mind, the spot from which you began the second leg is not entirely known. You only know that it is as close to the mouth of the harbor as you care to make it. From there, you assume you are at the harbor mouth, and execute the second leg, as best you can, based upon that assumption. So does the limit hold for arriving at the mooring?

The answer is, yes it does. Suppose that you must arrive within 10 meters of the mooring in order to see it in the fog. Where you end up after the first leg is the harbor mouth plus some error, which we shall call epsilon₁. Where you end up after the second leg is where you got to on the first leg plus the distance to the mooring plus some other error, which we shall call epsilon₂. So the total is distance to the harbor mouth plus distance from the harbor mouth to the mooring plus epsilon₁ plus epsilon₂.

If you had gone exactly the sum of the two distances, with no error at all, you would end up right at the mooring. But you didn't. You have to add in the errors. So you end up within epsilon₁ + epsilon₂ of the mooring. But remember that you can make both epsilon₁ and epsilon₂ as small as you like by concentrating hard enough. So, is it clear that you can also make their sum as small as you like? If you have to end up within 10 meters of the mooring, you can simply concentrate enough to make epsilon₁ less than 5 meters and epsilon₂ less than 5 meters (or any other combination that is guaranteed to add up to less than 10 meters).

And what if the fog were so thick that you had to get within 5 meters of the mooring? could you do that as well? Yes, if you concentrate hard enough to make each leg accurate to within 2.5 meters. No matter how close you have to get to the mooring, you can concentrate hard enough on each of the two legs to get that close. And by our definition of limits, that means that the limit of your final destination of the two legs as your concentration goes toward perfection is the mooring.

The point of this illustration is that the sum of the limits is always equal to the limit of the sum.

Here is another illustration. On a particular socialist collective farm, each peasant family is allocated a one hundred kilo sack of rice per harvest. Once the harvest is in (and before the allocations are made), the commisar is required to inform the politbureau how many kilos of rice the collective will have available to supply to the national distillery. The commisar knows to the last gram how much has been harvested. So he must take that amount, subtract from it one hundred kilos per family times the number of families, and report the result to the politbureau. But there are two factors that he can never be quite sure of. One is the the exact census of his collective -- it seems the census takers take from the distillery as well. The other is the exact capacity of the one-hundred kilo sacks he has been supplied with from the textile mill. The mill, it seems, is just across the road from the distillery.

The politbureau is aware of these problems, and as a result, it limits the production of the distillery right around harvest time. Of course, it can't curtail it altogether -- there would be a revolution if that ever happened. But clearly, the less spirits distilled around harvest time, the better will be the census numbers, and the closer to one hundred kilos will be the capacity of the grain sacks. And for the sake of this illustration, we will say that the limit of the census count as available spirits goes toward zero is, in fact, the exact population count. Likewise, the limit of the capacity of the grain sacks as available spirits goes toward zero is exactly one hundred kilos.

Suppose the harvest is 500 thousand kilos and the census indicates that there are 3000 families on the collective. How much rice will the collective actually send to the distillery? Or a better question is, what is the limit as available spirits goes toward zero of the amount of grain the collective will send to the distillery? Your intuition tells you that it should be 200 thousand kilos, right? You got that by using the commisar's formula. And your intuition turns out to be correct.

But in applying the commisar's formula, you took the product of two limits -- the limit of the census numbers as available spirits goes toward zero, and the limit of the grain-sack capacity as available spirits goes toward zero. What your intuition told you is that the product of the limits is equal to the limit of the products. But can you prove it?

If the limit of the census numbers as available spirits goes to zero is the actual number, then we have:

   n_f  =  3000 families + epsilon_c

where n_f is the actual number of families, and epsilon_c is the error. By the definition of a limit, if I tell you how accurate the census has to be -- that is how close to zero epsilon_c has to be, then you can tell me how scarse to make distilled spirits so that the census takers will return numbers at least that accurate.

Likewise if the limit of the grain-sack capacity as available spirits goes to zero is one hundred kilos, then we have:

   k_s  =  100 kilos + epsilon_s

where k_s is the actual capacity of the sacks and epsilon_s is their error. Once again, if I tell you how accurate the sacks have to be -- that is how close to zero epsilon_s has to be, then you can tell me how scarse to make distilled spirits so that the sacks will be so.

The commisar's formula is:

   k_p  =  500,000 kilos  -  (3000 families * 100 kilos per family)

where k_p is the predicted number of kilos to be sent to the distillery. But the formula we are more interested in is the one for the actual number of kilos to be sent to the distillery:

   k_a  =  500,000 kilos  -  (n_f * k_s)

       =  500,000 kilos  -

         ( (3000 families + epsilon_c) * (100 kilos per family + epsilon_s) )

where k_a is the actual number of kilos that will be sent to the distillery. Multiplying this out using the distributive law, we have:

   k_a  =  500,000 kilos  -  300,000 kilos  -

          (3000 families * epsilon_s)  -

          (100 kilos per family * epsilon_c)  -

          (epsilon_c * epsilon_s)

So what is the overall error for k_p? that is, what is the difference between k_p and k_a? That would be:

    epsilon_p  =  (3000 families * epsilon_s)  +

                 (100 kilos per family * epsilon_c)  +

                 (epsilon_c * epsilon_s)

where epsilon_p is the overall error in the commisar's prediction.

Now remember that you can always find an amount of spirits to produce sufficiently small to make either epsilon_c or epsilon_s as close to zero as you'd like either of them to be. Can you explain how that means, according to the formula given above, that you can also find an amount of spirits sufficiently small to make epsilon_p as close to zero as you'd like as well? Hint: remember in the last illustration we had two epsilon components to make a third one -- and what we did was find a way to ensure that each of the components would be no more than half of the maximum we wanted their sum to be. In this example we have the sum of three terms. So instead of half, you could work things so that each term was no more than how much of the maximum allowable total (which in this case is epsilon_p)?

The point, again, of this illustration is that the product of two limits is always equal to the limit of the product.

Exercises

1) Suppose you know that for some function, f(x), it has a limit, L as x goes toward a, and suppose you know that L is not zero. So we have:

   lim   f(x)  =  L        (L <> 0)                              eq 2.2-1
 x --> a

Show that if x is close enough to a, then you are guaranteed that f(x) will not be zero. In other words, there is an interval around a such that if x is in that interval, f(x) cannot possibly be zero. Another way of putting it is that there exists a delta > 0 such that f(x) is never zero whenever |x - a| <= delta.

First, write the contract for 2.2-1. You should be able to do that yourself by now. If not, it is defined for you back in 2.1.

Now, since we know that L is not zero, then we know that it is some distance from zero. See what happens to your contract if you stipulate that epsilon < |L| (or in other words, if epsilon is less than that distance). Explain how that sets a range around L in which f(x) is guaranteed not to be zero. Then explain how the contract guarantees that there is a delta that makes f(x) always fall inside that range whenever |x - a| <= delta.

2) I don't have a cute parable for the quotient of limits. Clearly, if you can show that whenever

   lim   f(x)  =  L                                              eq 2.2-2
 x --> a

and L is not zero, it guarantees that

           1         1
   lim    ----   =   -                                           eq.2.2-3
 x --> a  f(x)       L

then you can apply the product rule for limits to show that the quotient of the limits is the same as the limit of the quotient. Of course, the limit of the denominator must not be zero.

Proving that 2.2-2 implies 2.2-3 is a coached exercise:

First, what is the contract in 2.2-2? That should be easy. And what is the contract in 2.2-3? It is not much harder, but it is what you have to show is the inescapable consequence of the contract of 2.2-2. For the second contract, use the symbols epsilon_r and delta_r. The r here stands for reciprocal.

Before you procede, make a mental note that it is given in the problem that L is not zero.

Your contract for 2.2-3 should contain the difference of two fractions. Go ahead and put them over a common denominator. Look carefully at the common denominator. Is there any way it could be zero? Can you use what you learned in the first exercise to show that you can always find a delta_r > 0 small enough that this denominator is guaranteed not to be zero whenever |x - a| <= delta_r?

So you've now guaranteed that you can make it so the denominator is never zero. For some x that satisfies |x - a| <= delta_r, the denominator will be as close to zero as it can possibly be. Call that denominator d_worst, and substitute that in for the denominator you have. If the inequality with d_worst holds, then the one with the old denominator must hold as well. Can you explain why?

Notice that d_worst is a constant.

We haven't even looked at the numerator yet. It should contain a familiar difference. In fact, you've seen something like it in the contract you made for 2.2-2. Doesn't that contract guarantee that you can make this difference as close to zero as you like by choosing x sufficiently close to a? And if you can make the numerator as close to zero as you like, and the denominator is constant, can you argue that you can make the whole quotient as close to zero as you like? If so, you've proved the theorem and your done.