Limits pass through sums, products and quotients

0:00You know how to read a limit: drive the input toward a place, watch what the outputs settle on, and never ask about the place itself. That does not scale: there are infinitely many ways to build a new function out of two old ones. What we want is a rule saying that if you know the two limits, you already know the limit of the combination. Almost everything computable here rests on that sentence, and it is not obvious: a limit is a statement about infinitely many nearby inputs, and nothing says two of them have to recombine.

0:39So we test it, which needs a reading that cannot cheat. Every reading here is taken the same way: eight inputs closing on the place from one side, five rounds of exact elimination, and three values that still have to agree at the end. Of the sixteen inputs two such runs offer, the number that ARE the place they are asked about is zero. Nothing here ever substitutes. Here is the rule, in four parts, for two functions at a place.

1:13The limit of their sum is the sum of the two limits. The limit of their difference is the difference of the two limits. The limit of their product is the product of the two limits. And the limit of their quotient is the quotient of the two limits, provided the bottom one is not zero. Four claims, all hanging off one bar across the top: both limits are assumed to exist first.

1:41That bar is the part nobody reads, and it does all the work. The fourth claim also carries a condition the other three do not. Neither piece of small print is decoration, and the rest of this is those two taken seriously. Let us score the thing rather than believe it. Nine rules go into a sweep: flat ones, straight ones, curved ones, two that flip sign at the place, and one that runs away.

2:12Of the nine, six have a reading at zero, and the readings that come up are nothing, one and three. Nine rules make eighty-one ordered pairs, and each gets two answers from code with nothing in common: one approaches the combination, the other reads the two limits separately and combines them. Neither is ever shown the other's answer. Each pair lands in whichever of five boxes those answers put it: both agreed, both disagreed, only the combination answered, only the two limits did, or neither did.

2:51For the sum: thirty-six, zero, two, zero, forty-three. For the difference: thirty-six, zero, three, zero, forty-two. For the product: thirty-six, zero, twenty-four, zero, twenty-one. Across all two hundred and forty-three scores, the number of times the two routes both answered and disagreed is zero. That is the rule, measured. Now the box that should surprise you: for the sum, two pairs had a limit and neither piece did. One of them: a rule reading minus one from below and one from above, added to its exact opposite.

3:33Neither has a limit; their sum is nothing everywhere except at the place, so its limit is nothing. The rule never told you that, because its hypothesis was never met: the implication runs one way only. Now the sharper question: can exactly ONE piece fail while the combination still has a limit? Take every pair whose sum has a limit and count how many have neither piece readable, exactly one, and both: two, zero, thirty-six.

4:06Zero in the middle. For the product: four, twenty, thirty-six. Same question, two operations, two different answers, and the reason is one line: if the sum has a limit and so does the first piece, the second is the difference of those two, and the difference claim hands it a limit whether you wanted one or not. A product has no such escape: x times a sign-flipping rule is the size of x, which reads nothing quite happily.

4:41Now the check you can do by hand: take x squared plus x, and drive x toward one. Three readings on their own: x squared reads one, x reads one, and x plus one reads two. Read x squared plus x as a sum: one plus one is two. Or notice the very same function is x times x plus one, and read it as a product: one times two is two.

5:09And straight off the whole function's own run, using no rule at all: two. Three routes, one answer: two decompositions of one function have to land in the same place, or the rule is not a rule. That they really are the same function is checked too: over every place both runs offer, the number where they differ is zero. Now something that is not a polynomial: the cosine, at nothing.

5:40A cosine is not a fraction, so it is never evaluated here - it is trapped. Its series alternates in sign and its terms shrink, so two consecutive partial sums bracket the true value between them. A tenth of the way out the low end is one hundred and ninety-nine over two hundred, and the whole bracket is one over two hundred and forty thousand wide. Both ends are exact fractions, so read both by approaching: the low end settles on one, and so does the high end.

6:17Squeezed between two readings that agree, the cosine's reading is one, and it was never evaluated. Now take x plus the cosine at nothing and read it whole through both ends: one, and one. The two pieces separately, added: nothing plus one is one. The sum claim does not care what kind of function you fed it. It cares only that both limits exist. Here is a rule you were probably taught separately: a fixed multiplier passes straight through a limit.

6:53It is not a separate rule; it is the product claim with one side held flat. So: what does a flat function read? Five constants are offered, and the flat function built from each reads its own value back: five, minus three, a half, nothing, seven quarters. Measured, not assumed. Substitute that into the product claim: one of the two limits is just the constant, so the constant comes out and the other limit stays.

7:25Score it. Multiply two x plus three by each constant and read at one: twenty-five, minus fifteen, five halves, nothing, thirty-five quarters. Read two x plus three on its own and multiply each constant in afterwards: the same five numbers. All five land in the box where both routes agreed - and it is still the product claim, hypothesis and all, so multiply a constant by a rule with no limit and neither route answers.

8:02The claims are stated for two functions, and you use them on ten. That step is an induction, performed silently: apply the two-term claim, then apply it again to the result. Ten terms are built here, none carrying a power above the third. Read the running sum of the first two, then the first three, and so on: five, nine, fourteen, twenty, twenty-seven, thirty-five, forty-four, fifty-four, sixty-five. Legitimate, because each step is one application of a claim already scored - but only while every term has a limit.

8:42So spoil it: replace the last two terms with a sign-flipping rule and its exact opposite. Of the ten, the number readable on their own is eight. Read the running sums whole: five, nine, fourteen, twenty, twenty-seven, thirty-five, forty-four, then nothing at nine terms, then forty-four again at ten. One term at a time and added: the same seven, then nothing, then nothing. At ten terms the whole sum reads perfectly and the term-by-term route cannot start.

9:18The extension is not free; it is the hypothesis, applied every time. Now the fourth claim's condition: provided the bottom limit is not zero. Almost everyone reads that as: provided the denominator is not zero near the place - a different sentence, and here is a function that separates them. Take x cubed minus four x squared plus four x, and go toward two. Of the sixteen places the two runs offer, the number where that denominator is zero is zero - a perfectly good non-zero number at every place you would put in a table.

10:02And its reading at two is nothing. The condition is on the reading, not on the values. For contrast, a denominator that really does vanish inside its own run: ten x minus nineteen is zero at one point nine, the first place the left-hand run offers. Its reading at two is one, so the condition is satisfied, and yet a ratio carrying it cannot be read at all - the run hits a place with nothing to divide by.

10:35A third kind of failure, and no algebra removes it. So what happens when the condition fails? Take x squared minus four, over x cubed minus four x squared plus four x, and go toward two. The top reads nothing; the bottom reads nothing. The quotient claim is unavailable and neither route answers - but they stop for different reasons. The two-limits route stops because there is nothing to divide by; the whole-combination route stops because its outputs do not settle on any one value.

11:14A tenth to the left it is minus three hundred and ninety over nineteen; a tenth to the right, four hundred and ten over twenty-one; further in, both grow without bound. Of thirteen quotient problems offered, the number the fourth claim settles outright, because the bottom reading is not zero, is eight. Of the other five, four put nothing over nothing and one puts something over nothing. Scored into the five boxes: eight agreed, none disagreed, three were answered only by the combination, none only by the two limits, and two by neither.

11:55Those three in the middle have limits, and the rule cannot reach them. Nothing over nothing is not an answer; it is a rule declining to apply. Watch what happens if you treat it as one. Here is the misconception written down as a rule: divide, and where there is nothing to divide by, say nothing. Put it to the same thirteen problems: eight agreed, three disagreed, none was answered only by the combination, two only by this rule, and none by neither.

12:31The three it gets wrong are the interesting ones: where the real answer is two it says nothing; where twelve, nothing; where twenty-seven, nothing. And twice more it answers confidently where there is no answer at all. So what decides those cases? Not this rule and not the fourth claim; something else has to be found, and that is the work that comes next. One last thing, about honesty rather than mathematics: nothing here is a proof of that rule.

13:05It has been scored, hard, on hundreds of cases, and scoring is not proving. What you have instead is more useful for now: you know what it claims and where it goes quiet. It demands that both limits already exist, and where they do, all four parts hold, whatever you fed in. Where they do not, the rule says nothing, and the combination may still have a limit you must find another way.

13:34The fourth part demands one thing more, on a limit and not on a set of values, and every hard limit you meet from here is a case where that demand fails. So the small print is not the exception to the subject. It is the door into it.