Two standard limits, the first squeezed out of an inequality and the second reduced to it

0:00Here is a quotient. On top, e raised to the power x, minus one. Underneath, x. Ask it what it does at nothing, and it has no answer for you. The top reads exactly nothing there. The bottom reads exactly nothing there. Nothing over nothing is not a number that is hard to find. It is not a number at all. So the value at the place is not what anybody is after. What is wanted is what the quotient closes in on as x is driven toward nothing without ever arriving.

0:37That is a limit, and it is the whole of this video. There is a second quotient with the same trouble, and by the end the two of them will turn out to be one piece of work rather than two. The usual move, when a quotient reads nothing over nothing, is to factorise the top and cancel. That works when the top is a polynomial, because a polynomial that reads nothing at a place carries that place inside it as a factor, waiting to be taken out.

1:11This top is not a polynomial, and that can be measured rather than asserted. Take nine readings of it. Fit the one polynomial of degree one through the first two, and ask it to predict a tenth reading. Then degree two through the first three. And so on, up to degree eight. Of those eight fits, the number that predict the further reading is nothing at all. Not one of them.

1:39Do the identical thing to a genuine cubic and six of the eight come back right, which is what says the fitting works when there is something there to find. So there is no factor to take out. Cancelling is not available here, and something else has to be handed over instead. What gets handed over is an inequality. Not a rearrangement, not an identity - a trap. The quotient is caught between two expressions.

2:11Below it sits one over, one plus the size of x. Above it sits one, plus the size of x multiplied by e minus two. Read those two slowly, because three separate things about them matter and each one is easy to skate over. The size of x, not x. The comparisons at both ends inclusive, not strict. And the whole statement made only for x between minus one and one, with nothing itself taken out.

2:43Every one of those three is doing work, and the next three parts of this video are the three of them, one at a time. Start with the size of x, because this is where a careless reading does the most damage. Drop the size bars from that lower bound and you get one over, one plus x. It looks like the same thing. It is a different statement. Put both versions to forty places spread across the interval.

3:15The bound as it stands holds at all forty. It fails at none of them, and there is not one place where the two sides are too close to separate. The version without the bars holds at twenty of the forty, fails at nineteen, and at one place it cannot even be asked, because its denominator is nothing there. The nineteen failures are every negative place in the run, and the one place it cannot be asked is minus one - the left end of the very interval the statement is made on.

3:50So the bars are not decoration. Without them the lower bound is false on half its own interval and undefined at one end of it. Now the second thing: both comparisons are inclusive. At or below, at or above. That is not a matter of taste, and it is not a copy of the trigonometric inequality this argument is modelled on, where the comparisons are strict. It is forced, and forced at one place.

4:23Take x equal to one. The quotient becomes e minus one. The upper bound becomes one, plus one times e minus two - which is also e minus one. Those are not two numbers that happen to agree to a lot of decimal places. They are the same combination of powers of e, coefficient by coefficient, and no run of numbers could ever settle that. Carry the two sides as exact bookkeeping in e and subtract them, and what is left is nothing at all.

4:58Do the same subtraction at the other end of the interval and what is left is not nothing, so the bookkeeping is capable of saying no. A strict comparison would therefore be false at the right-hand end. Inclusive is the only thing it can be. Third: the interval. Minus one to one, with nothing punched out of the middle. The punched-out point is obvious enough - the quotient is not there to be compared with anything.

5:29The two ends are less obvious, and they are not there for tidiness. Take eight places outside the interval, four on each side, and put the same two comparisons to them. The upper comparison holds at four of the eight and fails at four. Split by side, it fails at all four of the places above the interval and holds at all four below it. The lower comparison holds at all eight.

5:59So the interval is carrying exactly one of the two bounds, on exactly one side of nothing. The upper bound is a local statement and the lower one, on this evidence, is not. None of which matters for the limit, by the way. A limit at nothing only ever consults places near nothing, and an interval of radius one is enormously more room than that needs. One more thing has to be said about this inequality before it is used, and it is the most important sentence in the video.

6:37It is not proved. It is stated, and then used twice, and that is all. The trigonometric inequality that this argument copies does get a proof - a geometric one, with a picture. This one gets none. That is not a mistake, and it is not a criticism. It is a fact about what is being leaned on, and a viewer is entitled to know which of the bricks in an argument have been made and which have been handed over.

7:08Testing is not proving, and nothing in this video proves it either. What testing did do was catch a misreading - that is exactly what the missing size bars turned out to be. A test cannot establish the thing. It can tell you that you have written down the thing that was meant. Now the argument. If a quantity is trapped between two others, and both of those close in on the same value, the trapped one has nowhere left to go.

7:41So the work is to show that both ends of this trap close in on one. Take six bounds, a tenth down to a millionth, and ask how many of them each end eventually gets and stays inside of one. The lower end gets inside all six. The upper end gets inside all six. A bound that simply sits where it is gets inside none of them, which is what says those two sixes are readings and not formalities.

8:11But the two ends get there by different routes, and it is worth naming which. The lower end is a quotient, and a quotient of limits is only available when the bottom does not close in on nothing. Its bottom is one plus the size of x, which closes in on one. Watch what the condition is protecting. Something that does close in on nothing does so on all six bounds - and a quotient built on that bottom instead settles inside none of the eight spans it is offered. It runs away.

8:48The upper end is a product: a fixed number multiplied by something closing in on nothing. Each of the six bounds is beaten a decade further in than the last. And a product with one factor closing in on nothing does not have to close in on nothing. One built with a second factor that has no limit at all never gets under a single one of the six. Now the squeeze itself, and here is the thing to watch: the quotient is not consulted.

9:23For each of the six bounds, ask how far in the whole trap has to be brought before it stays inside that much of one, all the way in from there. The answers are a tenth, a hundredth, a thousandth, and on down - one decade per bound. Two words in that question are load-bearing. All the way in. A trap that is tight at one offset and wide again further in is not a trap, it is a coincidence, and the machinery is given one to prove it: it accepts that one at none of the six bounds.

9:59A trap only one end of which closes is not a trap either, and that one is refused at all six too. And a trap that never closes at all - two fixed numbers either side of one - is refused at every bound. So the trap is real, it closes, and it closes on one. The last thing to check is that the quotient is actually inside it. Over eighteen places closing in from both sides, the number of times it sits outside the trap is nothing.

10:34That settles the first standard limit. As x is driven toward nothing, e to the x minus one, over x, closes in on one. Read straight off the run, the four closest readings sit together inside seven of eight spans, and they close on one within all eight of them. They close on two within none of them, which is the check that the reading is a reading. And the same quotient with a size put on its denominator instead - a small change, and one nothing traps - settles inside none of the eight spans, closes on one within none, and closes on nothing within none.

11:17It has no limit at all. The trap is what made the difference. Here is the second quotient. The logarithm of one plus x, over x. Nothing over nothing again at the same place, and the obvious expectation is that it needs an inequality of its own. It does not. It needs a change of variable. Call the top u. Then u is the logarithm of one plus x, which says exactly that e to the u is one plus x, which says that x is e to the u minus one.

11:55Put that back. The logarithm quotient is u over, e to the u minus one. That is the first quotient upside down. So one inequality has bought two theorems, and the second one arrives without a single new comparison being asserted. A change of variable is only legitimate if the new variable is doing what the old one did: closing in on nothing. So that gets measured too. Over sixteen places, ask how far in each of six bounds on the size of u is beaten.

12:33The first bound is beaten from a hundredth in, and each one after that a decade further, which is the pattern of something closing in on nothing. Notice it takes a hundredth rather than a tenth. The logarithm is not symmetric about this place, and it is the left-hand side that decides. A new variable that does not close in on nothing gets under none of the six. Two more checks, because the substitution is the whole argument here.

13:06First, the search that finds u has to be right: over those sixteen places, the number of times e raised to the power it found fails to bracket what it was asked for is nothing. Ask it for the power of a different number and all sixteen fail. Second, the swap itself: the number of places where the logarithm quotient and the first quotient upside down disagree is nothing. Make the swap with the wrong new variable and all sixteen disagree.

13:40So the second limit is one, and its own run says so: closing on one within seven of the eight spans and on two within none. Two standard limits are now available, and from here on the work is rewriting rather than analysis. First trap: the exponent and the denominator have to be the same thing. Take e to the three x, minus one, over x. It is tempting to call that one, because the standard limit is one.

14:13The run says three. It closes on three within seven of the eight spans, and it closes on one within none of them. The repair is arithmetic. Multiply and divide by three. Now the quotient has three x underneath as well as inside, which is the standard form, and the three that was pulled out is sitting in front. Three times one is three. With four in the exponent the run closes on four, and again on one within none of the spans.

14:47The number in the exponent comes out in front. It does not go away. Second move: an exponent that is a sum. Take e to the two plus x, minus e squared, over x, as x closes in on nothing. The standard limit starts from one, and this starts from e squared, so it looks out of reach. It is not, because e raised to a sum splits into a product. e to the two plus x is e squared times e to the x.

15:20Pull e squared out of both terms on top and what is left inside the bracket is exactly the standard quotient. The answer is e squared, and the run agrees without being told: it closes on the value the series gives for e squared, seven point three eight nine zero, and it closes on one within none of the spans. The same move works at a place that is not nothing. e to the x minus e cubed, over x minus three, as x closes in on three, comes out at e cubed - twenty point zero eight five five.

16:00And with five in place of three it comes out at e to the fifth, one four eight point four one three. Three different runs, three different values, each of them the number that was pulled out. Third move: a difference on top gets split. Take e to the x, minus the sine of x, minus one, over x. Split the top into two pieces over the same bottom: e to the x minus one over x, and the sine of x over x.

16:34The first closes on one, by everything this video has just done. The second closes on one as well, and that is the trigonometric standard limit, which arrives here from earlier work rather than from this inequality. One take away one is nothing, and the run agrees: it closes on nothing within all eight spans, and on one within none. This is worth pausing on. A standard limit is not just a result. It is an input. Once you have two of them, a problem stops being analysis and becomes a matter of arranging the pieces so that the known shapes appear.

17:19Fourth move: shifting the place. Take the logarithm of x, over x minus one, as x closes in on one. The place is not nothing, and both standard limits are statements about nothing. So move the question. Write x as one plus h. Then x closing in on one is h closing in on nothing, and the expression becomes the logarithm of one plus h, over h - which is the second standard limit exactly.

17:53The answer is one. The measurement of the shift is worth seeing, because the two variables are genuinely different things. Over the run, x itself closes on one within all six bounds and on nothing within none of them, while h closes on nothing within all six and on one within none. That is the entire content of shifting a point. Not a trick - a change in what is being driven where.

18:24One problem in this set needs more than the exponential limit, and it is worth doing in full. x times e to the x minus one, over one minus the cosine of x, as x closes in on nothing. Split it as e to the x minus one over x, multiplied by x squared over one minus the cosine of x. The first factor closes on one. The second needs an identity: one minus the cosine of x is twice the square of the sine of half of x.

19:00That identity is tested here rather than assumed. Over eighteen places the number of times the two sides disagree is nothing, and with the angle not halved they disagree at all eighteen. With the identity in, the second factor becomes two, divided by the square of the sine of half x over half x. That bracket closes on one, so the factor closes on two. One times two is two, and the run says two: closing on two within seven of the eight spans and on one within none of them.

19:38Three separate results in one problem, which is what makes it the hardest thing here. And step back at the end. One inequality, asserted and never proved, carried through a squeeze, gave the first limit. A change of variable turned the second one into the first upside down. Everything after that was rewriting. That is the shape worth remembering. Not eight answers - two inputs, and four ways of arranging a problem until they fit.