Combining continuous functions, and why a composite survives too

0:00Continuity is decided entirely by a limit. And limits already have an algebra: the limit of a sum is the sum of the limits, and the same for differences, products and quotients. So matching results for continuity should be expected, and they are. That is an argument for why the results are unsurprising. It is not a proof of any of them, and the difference matters, because what gets proved here is one line of work repeated, and what gets skipped is everything else.

0:35By the end of this video you will be able to say exactly which of these results you could reproduce from scratch, and which ones you are quoting. The first theorem comes in four parts. If two functions are continuous at an input, then so is their sum, so is their difference, so is their product, and so is their quotient. Three of those four carry no condition at all.

1:02The fourth carries exactly one: the divisor must not be nought at the input in question. Notice the shape of the assumption. Both functions are asked to be continuous at one particular input -- not on an interval, not everywhere. That is why the same theorem applies at a single join as easily as it applies across the whole line. The proof of the first part is four lines, and only one of them does any work.

1:35Line one: the limit of the sum, written out as the limit of the two values added together. That is the definition of what a sum of functions means, and nothing more. Line two: split it into the two limits. That is the algebra of limits from before, quoted. Line three: replace each limit by the value of its function at the input. This is the only line that uses continuity, and it uses it once.

2:06Line four: add the two values back up. That is the definition of a sum again. One line of mathematics, and three lines of bookkeeping around it. The other three parts are handed to the reader as similar, and they genuinely are. The difference is character for character the same proof with one sign changed. The product swaps the limit theorem for a sum with the limit theorem for a product, and nothing else moves.

2:38The quotient does the same again, and needs the divisor's limit at the input to be something other than nought -- which is exactly where the fourth part's condition comes from. Do that fourth one on paper rather than the first. It is the only one where the condition has to be produced rather than quoted, and producing it is what makes you remember it. Now a question almost nobody asks.

3:07How would we know if any of this had content? Take ten functions and one input, and ask each of them whether it is continuous there. Five of them are. The other five are broken there in five different ways: a jump, a value that refuses its own limit, a rule that runs away, a second jump, and the first jump read backwards. Now combine every ordered pair, all one hundred of them, under each of the four operations, and decide the result by the definition rather than by the theorem.

3:45Twenty-five of those pairs are pairs the theorem actually talks about: both functions continuous at the input. The other seventy-five are pairs it says nothing about. Keeping those two groups apart is the whole experiment. On the theorem's own ground the sum holds twenty-five times out of twenty-five, and fails nought times. The difference: twenty-five and nought. The product: twenty-five and nought. The quotient: fifteen and ten. Ten failures on its own ground -- which is not a hole in the theorem, it is the theorem's condition doing its job.

4:29Every single one of those ten has a divisor that vanishes at the input, and there are no others. Put the condition back in and the quotient row reads exactly like the other three: fifteen and nought. And no quotient at all survives a vanishing divisor here, so that condition is not merely sufficient -- in this population it is precisely the line. Now the other column, the seventy-five pairs the theorem never mentions.

5:02If those were all continuous too, the theorem would be telling us nothing. They are not. Sums: two survive out of seventy-five. Differences: five. Products: sixteen. Quotients: six. Mixed, every time, which is exactly what a real theorem looks like from outside its own hypothesis. And the survivors are worth knowing. The only two broken pairs whose sum survives are the two jumps that undo one another, and every broken function minus itself survives, which is all five of them.

5:44The sixteen surviving products are the interesting ones. Take a steady function that goes to nought at the input, and multiply it by something broken. There are twenty such pairs here, and sixteen of the products are continuous. A factor heading for nought can heal a broken partner. And the four it cannot heal are exactly the pairs whose broken factor runs away instead of staying bounded. Hold on to that, because it is the whole of the last item in this video.

6:20Two remarks follow, and each is one substitution. Put a constant in one slot of the product part and you have: a constant times a continuous function is continuous. Checked on every steady function here with four different constants, and it holds every time. Put minus one there and you have the negative of a continuous function. Put a constant in the numerator of the quotient part and you have the reciprocal of a continuous function, continuous wherever that function is not nought.

6:56Three of the five steady functions qualify, and the two that vanish at the input are refused -- which is the same condition again, not a new one. Those two moves plus the four parts assemble almost everything. A rational function is one polynomial divided by another. Polynomials are continuous everywhere. The fourth part finishes it, provided the divisor is not nought -- and the inputs where the divisor is nought are precisely the inputs the function does not have.

7:33So the exclusion is not an extra hypothesis you have to remember. It is the domain. On the domain, the condition holds at every input by construction. Now the sine, and here is where the standard route starts leaning on facts it does not prove. The argument writes the input as a fixed number plus a small increment, expands with the addition formula, and lets the increment go to nought.

8:04That leaves two terms. The first is the sine of the fixed number times the cosine of the increment. The second is the cosine of the fixed number times the sine of the increment. For the chain to land on the sine of the fixed number, you need two separate facts: the sine of the increment must go to nought, and the cosine of the increment must go to one.

8:31One of those two is usually announced up front, among the facts the argument says it will use. The other is used and never mentioned. That is easy to check, and here is how. Run the chain with a rival cosine that arrives at a half instead of one, and everything else untouched. The chain no longer closes on the sine of the fixed number. Run it with a rival sine that arrives at one instead of nought.

9:02It does not close there either. Both were tried at three different fixed numbers and the chain broke in the same way each time. So the argument consumes two facts, and the usual telling names one. This is not an error, and the conclusion is perfectly true. It tells you which line of your own version needs a sentence that the usual one does not have. The cosine goes the same way, by the same route and the same two facts.

9:35The tangent is then a quotient of two functions now known continuous, so the fourth part applies wherever the cosine is not nought. At seven ordinary inputs the sine, the cosine and the tangent are all unbroken, and the cosine vanishes at none of them. The cosine does have zeros, of course. Here it is changing sign between two inputs a hundred-millionth apart, which pins one of them down without ever naming the number.

10:07The sine changes sign the same way, at an input twice as far along. Those two sets of zeros are different sets: the sine vanishes at nought and the cosine does not. So the tangent and the secant, which have the cosine underneath, are perfectly happy at nought. The cosecant and the cotangent, which have the sine underneath, are the ones that lose it. Two of the six ratios are unbroken across the whole line, and the other four each lose infinitely many inputs.

10:44There is one way of combining functions the algebra cannot touch, and it is the most common one of all. Feeding the output of one function into another. Call the one applied first the inner function and the one applied second the outer function, and keep those words, because the theorem that handles this asks different things of the two. It says the composite is continuous at an input provided the inner function is continuous at that input, and the outer function is continuous at the value the inner function produces there.

11:23Two demands, landing on two different inputs. That asymmetry is the entire content of the theorem, and it is the thing most often lost. So let us score the misreading directly. Six outer rules, six inner rules, four inputs: a hundred and forty-four composites, of which a hundred and nineteen really are continuous, decided on the composite itself and not on its parts. The real theorem declares a hundred and fourteen of them and is wrong about none.

12:00The misreading -- both functions continuous at the same input -- declares a hundred and eighteen and is wrong about six. It also misses seven that the real one catches. So it is wrong in both directions, which anyone testing it only on well-behaved examples will never discover. The two readings part company on ten of the hundred and forty-four. On fifty-four of them the inner function moves the point, so the two questions genuinely land in different places.

12:37One more number from that table, and it is the important one. The real theorem is never wrong, and it misses five. Five composites that are genuinely continuous and that the theorem simply does not reach. Because it is a sufficient condition, not a test. If the theorem does not apply, you have learned nothing about the function -- you have learned something about the theorem. And there is a standard exercise built exactly on that gap.

13:09Take a rule about which you know only one thing: that it stays between minus one and one. A sine of a reciprocal is such a rule. Near nought nothing pins it down at all -- run the definition and not one candidate survives, so it has no limit there and no closure theorem can be applied to it. Now multiply it by a square. The product is continuous at nought, and that follows from the bound alone: no value of the wild rule is ever needed.

13:43Its limit there is nought, found the same way as every other limit in this video. And notice what the bound does not do. Away from nought, knowing only that the wild factor stays between minus one and one gives no verdict at any input at all. So the labour divides the other way round from usual: away from nought the two theorems settle it on the actual rules, and at nought the actual rules are no help and only the bound is.

14:16So here is the shape of the whole block. Four closure results, of which one substitution is proved and three are handed to you -- and they really are the same substitution, so write out the quotient and you have all four. One condition, on the divisor, at the one input under discussion and nowhere else. Two remarks that are the product and quotient parts with a constant dropped in.

14:45A composite theorem stated without proof, whose two demands land on two different inputs -- say both places out loud when you apply it. An argument for the sine that consumes two limits and names one. And a boundary: when neither theorem reaches an input, that is a fact about the theorems, and something else may still be true there.