Limit: Difference between revisions

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Motivation

Quick summary

The term "limit" in mathematics is closely related to one of the many senses in which the term "limit" is used in day-to-day English. In day-to-day English, there are two uses of the term "limit":

Limit as something that one approaches, or is headed toward
Limit as a boundary or cap that cannot be crossed or exceeded

The mathematical term "limit" refers to the first of these two meanings. In other words, the mathematical concept of limit is a formalization of the intuitive concept of limit as something that one approaches or is headed toward.

For a function $f$ , the notation:

$lim_{x \to c} f (x)$

is meant to say "the limit, as $x$ approaches $c$ , of the function value $f (x)$ " and thus, the mathematical equality:

$lim_{x \to c} f (x) = L$

is meant to say "the limit, as $x$ approaches $c$ , of the function value $f (x)$ , is $L$ ." In a rough sense, what this means is that as $x$ gets closer and closer to $c$ , $f (x)$ eventually comes, and stays, close enough to $L$ .

Graphical interpretation

The graphical interpretation of " $lim_{x \to c} f (x) = L$ " is that, if we move along the graph $y = f (x)$ of the function $f$ in the plane, then the graph approaches the point $(c, L)$ whether we make $x$ approach $c$ from the left or the right. However, this interpretation works well only if $f$ is continuous on the immediate left and immediate right of $c$ .

This interpretation is sometimes termed the "two finger test" where one finger is used to follow the graph for $x$ slightly less than $c$ and the other finger is used to follow the graph for $x$ slightly greater than $c$ .

Two key ideas

The concept of limit involves two key ideas, both of which help explain why the definition is structured the way it is:

Arbitrarily close: The limit depends on how things behave arbitrarily close to the point involved. The notion of "arbitrarily close" is difficult to quantify non-mathematically, but what it means is that any fixed distance is too much. For instance, if doing $lim_{x \to 2} f (x)$ , we can take points close to 2 such as 2.1, 2.01, 2.001, 2.0001, 2.0000001, 2.000000000000001. Any of these points, viewed in and of itself, is too far from 2 to offer any meaningful information. It is only the behavior in the limit, as we get arbitrarily close, that matters.
Trapping of the function close by: For a function to have a certain limit at a point, it is not sufficient to have the function value come close to that point. Rather, for $lim_{x \to c} f (x) = L$ to hold, it is necessary that for $x$ very close to $c$ , the function value $f (x)$ is trapped close to $L$ . It is not enough that it keeps oscillating between being close to $L$ and being far from $L$ .

Full timed transcript: [SHOW MORE]

0:00:15.549,0:00:19.259

Vipul: Okay, so in this talk, I'm going to go over the basic

0:00:19.259,0:00:24.619 motivation behind the definition of limit, and not so much the

0:00:24.619,0:00:28.099 epsilon-delta definition. That was just an intuitive idea, and a few somewhat

0:00:28.099,0:00:29.680 non-intuitive aspects of that.

0:00:29.680,0:00:36.680 Here I have the notation: "limit as x approaches c of f(x) is L" is

0:00:37.540,0:00:42.079 written like this. Limit ... Under the limit, we write where the

0:00:42.079,0:00:46.180 domain point goes, so x is approaching a value, c, and c could be an

0:00:46.180,0:00:51.059 actual number. x, however, will always be a variable letter. This x

0:00:51.059,0:00:54.519 will not be a number. c could be a number like zero, one, two, three,

0:00:54.519,0:00:55.329 or something.

0:00:55.329,0:01:02.050 f(x). f is the function. We are saying that as x approaches some

0:01:02.050,0:01:06.640 number c, f(x) approaches some number L, and thatâs what this is:

0:01:06.640,0:01:09.030 Limit as x approaches c of f(x) is L.

0:01:09.030,0:01:15.259 Now what does this mean? Roughly what it means is that as x is coming

0:01:15.259,0:01:22.259 closer and closer to c, f(x) is sort of hanging around L. Itâs coming

0:01:22.410,0:01:28.720 closer and closer to L. By the way, there are two senses in which the

0:01:28.720,0:01:32.429 word limit is used in the English language: One meaning its limit in

0:01:32.429,0:01:36.310 this approach sense, which is the mathematical meaning of limit.

0:01:36.310,0:01:41.319 There is another sense in which the word limit is used in the English

0:01:41.319,0:01:46.220 language, which is limit as a boundary or a as a gap or as a bound.

0:01:46.220,0:01:53.160 We may say, there is a limit to how many apples you can eat from the

0:01:53.160,0:01:58.640 food vault or something, and that sense of limit is not used ... for

0:01:58.640,0:02:02.110 that sense of limit you do not use the word "limit" in mathematics. For

0:02:02.110,0:02:05.899 that sense of limit, you use the word bound. In mathematics, we

0:02:05.899,0:02:11.800 reserve the use of the word limit only for this approach sense. Just

0:02:11.800,0:02:18.800 so we donât get confused in mathematics. As I said, the idea is that

0:02:21.120,0:02:25.760 as x approaches c, f(x) approaches L, so as x is coming closer and

0:02:25.760,0:02:29.480 closer to c, the distance between x and c is becoming smaller and

0:02:29.480,0:02:32.740 smaller, the distance between f(x) and L is also roughly becoming

0:02:32.740,0:02:37.980 smaller and smaller. This doesnât quite work unless your function is

0:02:37.980,0:02:41.250 increasing or decreasing near c, so you could have various

0:02:41.250,0:02:46.750 complications with oscillatory functions, so the point is this notion

0:02:46.750,0:02:52.170 doesnât really â¦ it's not very clear what we mean here without further

0:02:52.170,0:02:55.470 elaboration and without a clear definition.

0:02:55.470,0:03:02.470 I'm going to sort of move up toward the definition, and before we go

0:03:02.970,0:03:09.180 there, I want to say, that there is a graphical concept of limit,

0:03:09.180,0:03:13.430 which you may have seen in school. (well, if youâve seen limits in

0:03:13.430,0:03:17.110 school, which hopefully you have. This video is sort of more of a

0:03:17.110,0:03:21.500 review type than learning it for the first time). Let's try to

0:03:21.500,0:03:24.630 understand this from that point of view.

0:03:24.630,0:03:31.630 Let's say, you have a function whose graph looks something like this.

0:03:35.990,0:03:42.990 This is x of c, so this is the value x of c, and this is a graph of

0:03:44.069,0:03:48.310 the function, these curves are the graph of the function, so where x

0:03:48.310,0:03:53.900 is less than c, the graph is along this curve. For x greater than c,

0:03:53.900,0:03:58.120 the graph is this curve. So x less than c, the graph is this curve; x

0:03:58.120,0:04:01.740 greater than c, the graph is this curve. At x equal to c, the value

0:04:01.740,0:04:06.330 is that filled dot.

0:04:06.330,0:04:13.330 You can see from here that as x is approaching c from the left, so if

0:04:13.880,0:04:17.819 you take values of x, which are slightly less than c, the function

0:04:17.819,0:04:23.259 values â¦ so the function, the graph of it, the function values are

0:04:23.259,0:04:27.449 their prospective Y coordinates, so this is x, this is Y, this is the

0:04:27.449,0:04:34.449 graph. Y is f(x). When x is to the initial left of c, the value, Y

0:04:35.749,0:04:42.749 value, the Y approach f(x) value is â¦ are these values, so this or

0:04:44.610,0:04:51.610 this. As x approaches c from the left, the Y values are approaching

0:04:53.699,0:04:57.240 the Y coordinate of this open circle.

0:04:57.240,0:05:04.240 In a sense, if you just were looking at the limit from the left for x

0:05:05.680,0:05:10.830 approaching c from the left, then the limit would be the Y coordinate

0:05:10.830,0:05:16.279 of this open circle. You can also see an x approaches c from the

0:05:16.279,0:05:22.749 right, so approaches from here â¦ the Y coordinate is approaching the Y

0:05:22.749,0:05:29.749 coordinate of this thing, this open circle on top. There are actually

0:05:31.009,0:05:38.009 two concepts here, the left-hand limit is this value. We will call this L1. The right-hand limit is this value,

0:05:45.599,0:05:49.349 L2, so the left-hand limit, which is the notation as limit as x

0:05:49.349,0:05:56.349 approaches c from the left of f(x) is L1, the right-hand limit from the

0:05:58.089,0:06:05.089 right, thatâs plus of f(x), is L2, and the value f of c is some third

0:06:08.059,0:06:15.059 number. We donât know what it is, but f of c, L1, L2, are in this case

0:06:16.770,0:06:18.360 all different.

0:06:18.360,0:06:25.360 What does this mean as far as the limit is concerned? Well, the

0:06:25.900,0:06:28.259 concept of limit is usually a concept of two sides of limit, which

0:06:28.259,0:06:33.419 means that in this case the limit as x approaches c of f(x) does not

0:06:33.419,0:06:36.289 exist because you have a left-hand limit, and you have a right-hand

0:06:36.289,0:06:39.860 limit, and they are not equal to each other. The value, as such,

0:06:39.860,0:06:43.279 doesnât matter, so whether the value exists, what it is, does not

0:06:43.279,0:06:46.379 affect this concept of limit, but the real problem here is that the

0:06:46.379,0:06:48.490 left-hand limit and right-hand limit are not equal. The left-hand

0:06:48.490,0:06:55.490 limit is here; the right-hand limit is up here.

0:06:59.050,0:07:03.499 This graphical interpretation, you see the graphical interpretation is

0:07:03.499,0:07:07.749 sort of that. For the left-hand limit, you basically sort of follow

0:07:07.749,0:07:11.499 the graph on the immediate left and see where it's headed to and you

0:07:11.499,0:07:15.789 get the Y coordinate of that. For the right-hand limit, you follow

0:07:15.789,0:07:21.129 the graph on the right and see where they're headed to, and add the Y

0:07:21.129,0:07:22.240 coordinate of that.

0:07:22.240,0:07:29.240 Let me make an example, where the limit does exist. Let's say you

0:07:42.899,0:07:48.449 have a picture, something like this. In this case, the left-hand limit

0:07:48.449,0:07:52.610 and right-hand limit are the same thing, so this number, but the

0:07:52.610,0:07:55.889 values are different. You could also have a situation where the value

0:07:55.889,0:08:00.460 doesnât exist at all. The function isn't defined at the point, but

0:08:00.460,0:08:03.139 the limits still exist because the left-hand limit and right-hand

0:08:03.139,0:08:04.719 limit are the same.

0:08:04.719,0:08:09.979 Now, all these examples, they're sort of a crude way of putting this

0:08:09.979,0:08:13.710 idea, which is called the two-finger test. You may have heard it in

0:08:13.710,0:08:18.399 some slightly different names. The two-finger test idea is that you

0:08:18.399,0:08:23.929 use one finger to trace the curve on the immediate left and see where

0:08:23.929,0:08:28.259 thatâs headed to, and use another finger to trace the curve on the

0:08:28.259,0:08:33.640 immediate right and see where thatâs headed to, and if your two

0:08:33.640,0:08:38.270 fingers can meet each other, then the place where they meet, the Y

0:08:38.270,0:08:41.870 coordinate of that, is the limit. If, however, they do not come to

0:08:41.870,0:08:46.940 meet each other, which happens in this case, one of them is here, one

0:08:46.940,0:08:51.120 is here, and then the limit doesnât exist because the left-hand limit

0:08:51.120,0:08:53.509 and right-hand limit are not equal.

0:08:53.509,0:08:59.819 This, hopefully, you have seen in great detail where youâve done

0:08:59.819,0:09:05.779 limits in detail in school. However, what I want to say here is that

0:09:05.779,0:09:11.850 this two-finger test is not really a good definition of limit. Whatâs

0:09:11.850,0:09:13.600 the problem? The problem is that you could have really crazy

0:09:13.600,0:09:18.790 function, and it's really hard to move your finger along the graph of

0:09:18.790,0:09:25.220 the function. If the function sort of jumps around a lot, it's really

0:09:25.220,0:09:29.440 hard, and it doesnât really solve any problem. It's not really a

0:09:29.440,0:09:35.100 mathematically pure thing. It's like trying to answer the

0:09:35.100,0:09:39.540 mathematical question using a physical description, which is sort of

0:09:39.540,0:09:41.579 the wrong type of answer.

0:09:41.579,0:09:45.610 While this is very good for a basic intuition for very simple types of

0:09:45.610,0:09:50.040 functions, it's not actually the correct idea of limit. What kind of

0:09:50.040,0:09:56.990 things could give us trouble? Why do we need to define our

0:09:56.990,0:10:03.209 understanding of limit? The main thing is functions which have a lot

0:10:03.209,0:10:07.980 of oscillation. Let me do an example.

0:10:07.980,0:10:14.980 I'm now going to write down a type of function where, in fact, you

0:10:18.220,0:10:21.899 have to develop a pure cut concept of limit to be able to answer this

0:10:21.899,0:10:28.899 question precisely. This is a graph of a function, sine 1 over x.

0:10:28.959,0:10:32.920 Now this looks a little weird. It's not 1 over sine x; that would

0:10:32.920,0:10:39.920 just equal secant x. It's not that. It's sine of 1 over x, and this

0:10:44.879,0:10:50.220 function itself is not defined at x equals zero, but just the fact

0:10:50.220,0:10:52.660 that thatâs not defined, isn't good enough for us to say the limit

0:10:52.660,0:10:55.139 doesn't [inaudible 00:10:36] we actually have to try to make a picture

0:10:55.139,0:10:57.660 of this and try to understand what the limit is here.

0:10:57.660,0:11:04.660 Let's first make the picture of sine x. Sine-x looks like that. How

0:11:12.560,0:11:19.560 will sine 1 over x look? Let's start of where x is nearly infinity.

0:11:20.100,0:11:25.759 When x is very large positive, 1 over x is near zero, slightly

0:11:25.759,0:11:30.660 positive, just slightly bigger than zero, and sine 1 over x is

0:11:30.660,0:11:36.879 therefore slightly positive. It's like here. It's going to start up

0:11:36.879,0:11:42.810 with an S [inaudible 00:11:21] at zero. Then it's going to sort of go

0:11:42.810,0:11:49.420 this path, but much more slowly, each one, then it's going to go this

0:11:49.420,0:11:56.420 path, but in reverse, so like that. Then it's going to go this path,

0:11:57.149,0:12:00.740 but now it does all these oscillations, all of these oscillations. It

0:12:00.740,0:12:03.569 has to go faster and faster.

0:12:03.569,0:12:10.569 For instance, this is pi, this 1 over pi, then this is 2 pi, this

0:12:12.329,0:12:16.990 number is 1 over 2 pi, then the then next time it reaches zero will be

0:12:16.990,0:12:21.160 1 over 3 pi, and so on. Whatâs going to happen is that near zero it's

0:12:21.160,0:12:24.579 going to be crazily oscillating between minus 1, and 1. The frequency

0:12:24.579,0:12:29.170 of the oscillation keeps getting faster and faster as you come closer

0:12:29.170,0:12:34.050 and closer to zero. The same type of picture on the left side as

0:12:34.050,0:12:40.360 well; it's just that it's an odd function. It's this kind of picture.

0:12:40.360,0:12:47.360 I'll make a bigger picture here ... I'll make a bigger picture on another

0:12:53.649,0:13:00.649 one. all of these oscillation should be between minus 1 and 1, and we

0:13:22.439,0:13:29.399 get faster so we get faster and faster, and now my pen is too thick.

0:13:29.399,0:13:31.600 It's the same, even if you used your finger instead of the pen to

0:13:31.600,0:13:38.600 place it, it would be too thick, it's called the thick finger problem.

0:13:38.850,0:13:45.060 Iâm not being very accurate here, but just the idea. The pen or

0:13:45.060,0:13:49.199 finger is too thick, but actually, there's a very thin line, and it's

0:13:49.199,0:13:52.519 an infinitely thin line of the graph, which goes like that.

0:13:52.519,0:13:59.519 Let's get back to our question: What is limit as x approaches zero,

0:14:02.699,0:14:09.699 sine 1 over x. I want you to think about this a bit. Think about like

0:14:13.439,0:14:18.050 the finger test. You move your finger around, move it like this,

0:14:18.050,0:14:21.579 this, this â¦ you're sort of getting close to here but still not quite

0:14:21.579,0:14:28.579 reaching it. It's â¦ where are you headed? It's kind of a little

0:14:31.610,0:14:36.879 unclear. Notice, it's not that just because we plug in zero doesnât

0:14:36.879,0:14:39.170 make sense, the limit doesn't... Thatâs not the issue. The issue is

0:14:39.170,0:14:43.249 that after you make the graph, it's unclear whatâs happening.

0:14:43.249,0:14:49.329 One kind of logic is that the other limit is zero? Why? Well, it's

0:14:49.329,0:14:52.949 kind of balance around here. It's a bit above and below, and it keeps

0:14:52.949,0:14:59.949 coming close to zero. That any number of the form x is 1 over N pi,

0:15:00.329,0:15:07.329 sine 1 over x is zero. It keeps coming close to zero. As x

0:15:07.990,0:15:12.459 approaches zero, this number keeps coming close to zero.

0:15:12.459,0:15:17.449 If you think of limit as something thatâs approaching, then as x

0:15:17.449,0:15:24.449 approaches zero, sine 1 over x is sort of coming close to zero, is it?

0:15:31.230,0:15:36.550 It's definitely coming near zero, right? Anything you make around

0:15:36.550,0:15:41.920 zero, any small â¦ this you make around zero, the graph is going to

0:15:41.920,0:15:42.399 enter that.

0:15:42.399,0:15:47.269 On the other hand, it's not really staying close to zero. It's kind of

0:15:47.269,0:15:50.300 oscillating with the minus 1 and 1. However, smaller interval you

0:15:50.300,0:15:54.540 take around zero on the x thing, the function is oscillating between

0:15:54.540,0:15:57.600 minus 1 and 1. It's not staying faithful to zero.

0:15:57.600,0:16:02.249 Now you have kind of this question: What should be the correct

0:16:02.249,0:16:09.249 definition of this limit? Should it mean that it approaches the

0:16:10.029,0:16:15.100 point, but maybe goes in and out, close and far? Or should it mean it

0:16:15.100,0:16:18.879 approaches and stays close to the point? That is like a judgment you

0:16:18.879,0:16:22.629 have to make in the definition, and it so happens that people who

0:16:22.629,0:16:28.639 tried defining this chose the latter idea; that is, it should come

0:16:28.639,0:16:33.089 close and stay close. So thatâs actually key idea number two we have

0:16:33.089,0:16:38.290 here the function â¦ for the function to have a limit at the point, the

0:16:38.290,0:16:43.639 function needs to be trapped near the limit, close to the point in the

0:16:43.639,0:16:45.079 domain.

0:16:45.079,0:16:49.459 This is, therefore, it doesnât have a limit at zero because the

0:16:49.459,0:16:54.420 function is oscillating too widely. You cannot trap it. You cannot

0:16:54.420,0:17:01.059 trap the function values. You cannot say thatâ¦ you cannot trap the

0:17:01.059,0:17:08.059 function value, say, in this small horizontal strip near zero. You

0:17:08.319,0:17:11.650 cannot trap in the area, so that means the limit cannot be zero, but

0:17:11.650,0:17:15.400 the same logic works anywhere else. The limit cannot be half, because

0:17:15.400,0:17:20.440 you cannot trap the function in a small horizontal strip about half

0:17:20.440,0:17:22.130 whereas x approaches zero.

0:17:22.130,0:17:26.440 We will actually talk about this example in great detail in our future

0:17:26.440,0:17:30.330 with you after we've seen the formal definition, but the key idea you

0:17:30.330,0:17:33.890 need to remember is that the function doesnât just need to come close

0:17:33.890,0:17:37.340 to the point of its limit. It actually needs to stay close. It needs

0:17:37.340,0:17:41.050 to be trapped near the point.

0:17:41.050,0:17:44.810 The other important idea regarding limits is that the limit depends

0:17:44.810,0:17:50.370 only on the behavior very, very close to the point. What do I mean by

0:17:50.370,0:17:56.580 very, very close? If you were working it like, the real goal, you may

0:17:56.580,0:18:02.300 say, it's like, think of some really small number and you say that

0:18:02.300,0:18:07.050 much distance from it. Let's say I want to get the limit as x

0:18:07.050,0:18:14.050 approaches 2...I'll just write it here. I want to get, let's say,

0:18:23.520,0:18:30.520 limit has x approaches 2 of some function, we may say, well, we sort

0:18:30.550,0:18:37.550 of â¦ whatâs close enough? Is 2.1 close enough? No, thatâs too far.

0:18:38.750,0:18:43.380 What about 2.0000001? Is that close enough?

0:18:43.380,0:18:47.420 Now, if you werenât a mathematician, you would probably say, "Yes,

0:18:47.420,0:18:54.420 this is close enough." The difference is like ... so it's

0:18:57.040,0:19:04.040 10^{-7}. It's really only close to 2 compared to our usual sense of

0:19:12.990,0:19:16.670 numbers, but as far as mathematics is concerned, both of these numbers

0:19:16.670,0:19:21.110 are really far from 2. Any individual number that is not 2 is very

0:19:21.110,0:19:22.130 far from 2.

0:19:22.130,0:19:29.130 What do I mean by that, well, think back to one of our

0:19:29.670,0:19:36.670 pictures. Here's a picture. Supposed I take some points. Let's say

0:19:41.970,0:19:47.640 this is 2, and suppose I take one point here, which is really close to

0:19:47.640,0:19:50.970 2, and I just change the value of the function at that point. I

0:19:50.970,0:19:55.200 change the value of the function at that point, or I just change the

0:19:55.200,0:19:59.990 entire picture of the graph from that point rightward. I just take

0:19:59.990,0:20:05.940 this picture, and I change it to, let's say â¦ so I replace this

0:20:05.940,0:20:11.410 picture by that picture, or I replace the picture by some totally new

0:20:11.410,0:20:15.250 picture like that picture. I just change the part of the graph to the

0:20:15.250,0:20:21.440 right of some point, like 2.00001, whatever. Will that effect the

0:20:21.440,0:20:25.770 limit at 2? No, because the limit at 2 really depends only on the

0:20:25.770,0:20:27.520 behavior if you're really, really close.

0:20:27.520,0:20:32.040 If you take any fixed point, which is not 2, and you change the

0:20:32.040,0:20:35.000 behavior sort of at this time that point or farther away than that

0:20:35.000,0:20:42.000 point, then the behavior close to 2 doesnât get affected. Thatâs the

0:20:42.820,0:20:46.660 other key idea here. Actually I did these in [inaudible 00:20:30].

0:20:46.660,0:20:52.060 Thatâs how it is coming, actually, but I'll just say it again.

0:20:52.060,0:20:56.570 The limit depends on the behavior arbitrarily close to the point. It

0:20:56.570,0:21:00.210 doesnât depend on the behavior at any single specific other point. It

0:21:00.210,0:21:06.910 just depends on the behavior as you approach the point and any other

0:21:06.910,0:21:11.330 point is far away. It's only sort of together that all the other

0:21:11.330,0:21:16.230 points matter, and it's only them getting really close that

0:21:16.230,0:21:19.790 matters. The other thing is that the function actually needs to be

0:21:19.790,0:21:26.790 tracked near the point for the limit notion to be true. This type of

0:21:26.860,0:21:29.650 picture where it's oscillating between minus 1 and 1, however close

0:21:29.650,0:21:35.150 you get to zero, keeps oscillating, and so you cannot trap it around

0:21:35.150,0:21:40.590 any point. You cannot trap the function value in any small enough

0:21:40.590,0:21:47.590 strip. In that case, the limit doesnât exist. In subsequent videos,

0:21:48.550,0:21:54.630 we'll see Epsilon definition, we'll do a bit of formalism to that, and

0:21:54.630,0:22:00.640 then we'll come back to some of these issues later with the formal

0:22:00.640,0:22:01.870

understanding.

Definition for finite limit for function of one variable

Two-sided limit

Suppose $f$ is a function of one variable and $c \in R$ is a point such that $f$ is defined to the immediate left and immediate right of $c$ (note that $f$ may or may not be defined at $c$ ). In other words, there exists some value $t > 0$ such that $f$ is defined on $(c - t, c + t) ∖ {c} = (c - t, c) \cup (c, c + t)$ .

For a given value $L \in R$ , we say that:

$lim_{x \to c} f (x) = L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

For every $ε > 0$ (the symbol $ε$ is a Greek lowercase letter pronounced "epsilon")
there exists $δ > 0$ such that (the symbol $δ$ is a Greek lowercase letter pronounced "delta")
for all $x \in R$ satisfying $0 < | x - c | < δ$ (explicitly, $x \in (c - δ, c) \cup (c, c + δ) = (c - δ, c + δ) ∖ {c}$ ),
we have $| f (x) - L | < ε$ (explicitly, $f (x) \in (L - ε, L + ε)$ ).

The limit (also called the two-sided limit) $lim_{x \to c} f (x)$ is defined as a value $L \in R$ such that $lim_{x \to c} f (x) = L$ . By the uniqueness theorem for limits, there is at most one value of $L \in R$ for which $lim_{x \to c} f (x) = L$ . Hence, it makes sense to talk of the limit when it exists.

Full timed transcript: [SHOW MORE]

0:00:15.809,0:00:20.490 Vipul: In this talk, I'm going to introduce the definition, the formal epsilon delta definition

0:00:20.490,0:00:24.669 of a two-sided limit for a function of a one variable, that's called f.

0:00:24.669,0:00:31.349 I'm going to assume there is a point c and c doesn't actually have to be in the domain of f.

0:00:31.349,0:00:38.030 Thus f doesn't have to be defined at c for this notion to make sense rather f is defined around c.

0:00:38.030,0:00:44.909 What that means is f is defined on some open set containing c.

0:00:51.009,0:01:03.009 Let's make a picture here so you have c, c + t, c -- t.

0:01:03.040,0:01:11.040 What this is saying is there is some t probably small enough so that the function is defined

0:01:12.549,0:01:18.590 in here and may be it's not defined at the point c.

0:01:18.590,0:01:31.590 This set for some t>0. The function is defined on the immediate left of c and it is defined

0:01:31.999,0:01:34.770 on the immediate right of c.

0:01:34.770,0:01:38.890 We need that in order to make sense of what I'm going to say.

0:01:38.890,0:01:44.590 We say that limit as x approaches c of f(x) is L where L is some other real number or

0:01:44.590,0:01:49.679 maybe it's the same real number [as c], so we say this limit equals L, now I'll write the definition

0:01:49.679,0:01:56.679 in multiple lines just to be clear about the parts of the definition.

0:01:56.770,0:02:39.770 For every epsilon > 0. This is epsilon. There exists delta > 0 such that for all x in R satisfying...what?

0:02:41.070,0:02:45.070 Rui: Satisfying |x -- c| ...

0:02:45.659,0:02:53.659 Vipul: [|x-c|] should be not equal to zero so zero less than, exclude the point c itself,

0:02:54.810,0:02:56.930 less than delta. What do we have?

0:02:56.930,0:02:59.459 Rui: We have y is within.

0:02:59.459,0:03:04.260 Vipul: Well y is just f(x).

0:03:04.260,0:03:10.290 Rui: f(x_0)

0:03:14.290,0:03:16.819 Vipul: Well f(x) minus the claimed limit is?

0:03:17.219,0:03:18.040 Rui: L.

0:03:18.640,0:03:22.890 Vipul: You're thinking of continuity which is a little different but here we have this less than?

0:03:22.890,0:03:24.569 Rui: Epsilon.

0:03:24.569,0:03:37.569 Vipul: Epsilon. Let me now just re-write these conditions in interval notation.

0:03:37.830,0:03:40.031 What is this saying x in what interval? [ANSWER!]

0:03:40.040,0:03:43.519 Rui: c +- ...

0:03:43.519,0:03:49.840 Vipul: c- delta to c + delta excluding the point c itself, that is what 0 < [|x -- c|] is telling us.

0:03:49.840,0:03:56.530 It is telling us x is within delta distance of c, but it is not including c.

0:03:56.530,0:04:10.530 Another way of writing this is (c -- delta,c) union (c, c + delta)

0:04:12.810,0:04:19.340 x is either on immediate delta left of c or it's on the immediate delta right of c.

0:04:21.040,0:04:31.040 You do something similar on the f(x) side so what interval is this saying, f(x) is in what? [ANSWER!]

0:04:31.720,0:04:35.930 Rui: L -- epsilon, L + epsilon.

0:04:35.930,0:04:42.930 Vipul: Awesome. Instead of writing the conditions in this inequality form you could have written

0:04:43.919,0:04:47.590 them in this form, so instead of writing this you could have written this or this, instead

0:04:47.590,0:04:49.580 of writing this you could have written this.

0:04:50.080,0:04:59.500 If this statement is true, the way you read this is you say limit as x approaches c of f(x) equals L.

0:04:59.500,0:05:07.500 Okay. Now how do we define the limit?

0:05:11.169,0:05:18.169 It's the number L for which the above holds. This should be in quotes.

0:05:22.009,0:05:29.009 If a number L exists for which.

0:05:34.220,0:05:41.220 Now what would you need in order to show that this definition makes sense?

0:05:47.919,0:05:52.919 Rui: I don't think I understand your question.

0:06:03.090,0:06:09.090 Vipul: What I mean is, what I wanted to ask was what would you need to prove in order

0:06:09.990,0:06:14.889 to say the notion of the limit makes sense? Well, you need to show that there is uniqueness here.

0:06:14.740,0:06:19.080 It cannot happen that the limit is some number L and the limit is another number M so you

0:06:19.080,0:06:20.539 need to show uniqueness.

0:06:20.539,0:06:27.330 You need to show that if this holds for one number L it cannot also hold for another number.

0:06:27.330,0:06:32.050 Once you have shown that then it you could define it like this.

0:06:32.050,0:06:38.440 Now I should say "if it exists."

0:06:38.440,0:06:42.120 What I'm saying is that there is a uniqueness theorem which we will prove some other time.

0:06:42.120,0:06:49.120 Which says that if this is true for one number it cannot be true for any other number so

0:06:49.440,0:06:54.740 this statement is true for at the most one value of L and if there is such a value of

0:06:54.740,0:06:55.050

L that's called the limit.

Note: Although the definition customarily uses the letters $ε$ and $δ$ , any other letters can be used, as long as these letters are different from each other and from the letters already in use. The reason for sticking to a standard letter choice is that it reduces cognitive overload.

Left hand limit

Suppose $f$ is a function of one variable and $c \in R$ is a point such that $f$ is defined on the immediate left of $c$ (note that $f$ may or may not be defined at $c$ ). In other words, there exists some value $t > 0$ such that $f$ is defined on $(c - t, c)$ .

For a given value $L \in R$ , we say that:

$lim_{x \to c^{-}} f (x) = L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

For every $ε > 0$
there exists $δ > 0$ such that
for all $x \in R$ satisfying $0 < c - x < δ$ (explicitly, $x \in (c - δ, c)$ ),
we have $| f (x) - L | < ε$ (explicitly, $f (x) \in (L - ε, L + ε)$ .

The left hand limit (acronym LHL) $lim_{x \to c^{-}} f (x)$ is defined as a value $L \in R$ such that $lim_{x \to c^{-}} f (x) = L$ . By the uniqueness theorem for limits (one-sided version), there is at most one value of $L \in R$ for which $lim_{x \to c^{-}} f (x) = L$ . Hence, it makes sense to talk of the left hand limit when it exists.

Right hand limit

Suppose $f$ is a function of one variable and $c \in R$ is a point such that $f$ is defined on the immediate right of $c$ (note that $f$ may or may not be defined at $c$ ). In other words, there exists some value $t > 0$ such that $f$ is defined on $(c, c + t)$ .

For a given value $L \in R$ , we say that:

$lim_{x \to c^{+}} f (x) = L$

if the following holds (the single sentence is broken down into multiple points to make it clearer):

For every $ε > 0$
there exists $δ > 0$ such that
for all $x \in R$ satisfying $0 < x - c < δ$ (explicitly, $x \in (c, c + δ)$ ),
we have $| f (x) - L | < ε$ (explicitly, $f (x) \in (L - ε, L + ε)$ .

The right hand limit (acronym RHL) $lim_{x \to c^{+}} f (x)$ is defined as a value $L \in R$ such that $lim_{x \to c^{+}} f (x) = L$ . By the uniqueness theorem for limits (one-sided version), there is at most one value of $L \in R$ for which $lim_{x \to c^{+}} f (x) = L$ . Hence, it makes sense to talk of the right hand limit when it exists.

Full timed transcript: [SHOW MORE]

0:00:15.940,0:00:20.740 Vipul: In this talk, I'm going to give definitions of one-sided limits.

0:00:20.740,0:00:25.650 So it is going to be the left hand limit and the right hand limit, and I'm going to basically

0:00:25.650,0:00:42.650 compare it with the definition of two-sided limit which was in a previous video. Let's just write this down--left-hand limit.

0:00:44.110,0:00:48.679 Let me first remind you what the definition of two-sided limit says.

0:00:48.679,0:00:57.679 So here's what it says. It says limit as x approaches c, f(x) = L

0:00:58.469,0:01:03.140 so f has to be defined on the immediate left and the immediate right of c.

0:01:03.140,0:01:07.960 It says that this is true if the following holds so for every epsilon greater than zero

0:01:07.960,0:01:13.960 there exists a delta > 0 such that for all x which are within delta of c

0:01:14.000,0:01:22.771 either delta on the left of c or within a delta on the right of c we have that f(x) is within an epsilon

0:01:23.650,0:01:30.530 distance of L. Okay. Now with the left and right hand limit

0:01:30.530,0:01:37.460 what we are trying to do we are trying to consider only one-sided approaches on the, on the x

0:01:39.000,0:01:41.510 What will change when we do the left-hand limit,

0:01:42.001,0:01:44.641 what will be different from this definition? [ANSWER!]

0:01:45.710,0:01:48.330 Rui: We approach c from the left.

0:01:48.330,0:01:52.790 Vipul: We'll approach c from the left so what part of this definition will change? [ANSWER!]

0:01:52.790,0:01:54.880 Rui: From the fourth line?

0:01:54.880,0:01:56.890 Vipul: You mean this line?

0:01:56.890,0:02:06.810 Rui: Oh for all x within c distance, within delta distance of c

0:02:06.810,0:02:08.700 Vipul: So what will change?

0:02:08.700,0:02:14.020 Rui: We will not have (c, c + delta).

0:02:14.020,0:02:18.390 Vipul: This part won’t be there. We will just be concerned about whether when x is

0:02:18.390,0:02:23.000 delta close on the left side of c, f(x) is here...

0:02:23.000,0:02:28.000 Will we change this one also to only include the left? [ANSWER!]

0:02:28.000,0:02:30.000 Or this one will remain as it is?

0:02:30.300,0:02:31.500 Rui: I think it will remain.

0:02:31.500,0:02:33.460 Vipul: It will remain as it is because we

0:02:33.460,0:02:35.340 are just saying as x approaches c from the left

0:02:35.340,0:02:36.340 f(x) approaches L.

0:02:36.340,0:02:43.340 We are not claiming that f(x) approaches L from the left, okay? Let me make a number line picture.

0:02:51.750,0:02:56.130 We will do a full geometric understanding of the thing later. Right now it's just very [formal].

0:02:56.130,0:03:00.850 So the function is defined on the immediate left of c, maybe not defined at c. It is defined

0:03:00.850,0:03:01.920 on the immediate left of c.

0:03:01.920,0:03:06.410 We don’t even know if the function is defined on the right of c and what we are

0:03:06.410,0:03:13.410 saying is that for any epsilon, so any epsilon around L you can find a delta such that if you restrict

0:03:13.800,0:03:20.800 attention to the interval from c minus delta to c [i.e., (c- delta, c) in math notation]

0:03:21.450,0:03:23.130 then the f value there is within the epsilon distance of L.

0:03:24.130,0:03:28.959 Now the f value could be epsilon to the left or the right so we take left hand limit on

0:03:28.959,0:03:33.840 the domain side it doesn’t have to approach from the left on the other side.

0:03:33.840,0:03:40.690 Let me just write down the full definition. We want to keep this on the side.

0:03:40.690,0:04:03.690 What it says that for every epsilon > 0 there exists

0:04:05.180,0:04:16.680 by the way, the understanding of the what this definition really means will come in another video you may have seen before this or after this

0:04:16.680,0:04:21.209 ... for all x ... [continuing definition]

0:04:21.209,0:04:26.500 Now we should also change it if we are writing in this form so how will it read now?

0:04:26.500,0:04:28.030 Rui: For all x ...

0:04:35.000,0:04:38.000 Vipul: So will you put x – c or c – x? [ANSWER!]

0:04:38.330,0:04:40.990 Rui: It will be x – c, oh c – x.

0:04:41.000,0:04:46.760 Vipul: c – x. Because you want c to be bigger than x. You want x to be on the left of c.

0:04:46.850,0:05:01.850 What would this read, i.e. x is in (c – delta,c). Okay.

0:05:05.000,0:05:11.460 What do we have? We have the same thing. This part doesn’t change.

0:05:13.000,0:05:19.000 Rui: f(x) is within epsilon distance of L.

0:05:34.400,0:05:40.400 Vipul: Why do I keep saying this thing about the L approach doesn’t have to be from the left?

0:05:41.000,0:05:44.350 What’s the significance of that? Why is that important? [ANSWER!]

0:05:45.000,0:05:51.000 Rui: It’s important because we don’t know whether the function is decreasing or increasing

0:05:51.620,0:05:52.370 at that point.

0:05:52.370,0:05:55.750 Vipul: Yeah, so if your function is actually increasing than L will also be approached

0:05:55.750,0:06:01.590 from the left, and if it’s decreasing it will be approached from the right, but sometimes

0:06:01.590,0:06:07.590 it’s neither increasing nor decreasing, but it's still true it approaches from one side, so that’s a little complicated but the way

0:06:07.590,0:06:12.150 this comes up is that when you are dealing with composition of functions, so when you

0:06:12.150,0:06:16.710 are doing one function and then applying another function to that and you have some results

0:06:16.710,0:06:18.440 with one-sided limits.

0:06:18.440,0:06:30.440 Let me just write this down. If you have one-sided limits and you have composition,

0:06:31.610,0:06:39.550 so you are doing one function and then doing another you have to be very careful.

0:06:45.050,0:06:48.350 You need to be very careful when you are doing one-sided limits and composition.

0:06:48.360,0:06:57.360 Why? Because if you have g of f(x) and x approaches to c from the left, f(x) approaches L but

0:06:57.850,0:06:59.280 not necessarily from the left.

0:06:59.280,0:07:03.560 You then you have another thing which is as f(x) approaches L from the left, g of that

0:07:03.560,0:07:09.280 approaches something you just need to be careful that when you compose things the sidedness

0:07:09.280,0:07:10.930 could change each time you compose.

0:07:10.930,0:07:14.590 Rui: Can you write a composition of the function out?

0:07:14.590,0:07:17.870 Vipul: Not in this video. We will do that in another video.

0:07:17.870,0:07:23.800 That’s something we will see in a subsequent video but this is just something to keep in

0:07:23.800,0:07:27.770 mind so when you see that it will ring a bell.

0:07:30.770,0:07:31.880 Let us do... what other side is left? [pun unintended!] Rui: Right? Vipul: Right!

0:07:31.880,0:07:36.690 Vipul: By the way, you probably already know this if you have seen limits intuitively so

0:07:36.690,0:07:42.300 I'm not stressing this too much but left hand limit is really the limit as you approach

0:07:42.300,0:07:49.300 from the left. You are not moving toward the left you are moving from the left to the point.

0:07:50.160,0:07:55.940 Right hand limit will be approach from the right to the point so it is right, moving from

0:07:55.940,0:07:59.330 the right, so the words left and right are describing where the limit is coming *from*,

0:07:59.330,0:08:06.330 not the direction which it is going to.

0:08:12.569,0:08:17.650 Now you can just tell me what will be the corresponding thing. To make sense of this

0:08:17.650,0:08:19.819 notion we need f to be defined where? [ANSWER!]

0:08:19.819,0:08:21.699 Rui: On its right.

0:08:21.699,0:08:29.199 Vipul: On the immediate right of c. If it is not defined on the immediate right it doesn’t

0:08:29.389,0:08:36.389 even make sense to ask this question what the right hand limit is.

0:08:37.280,0:08:38.550 How will that be defined?

0:08:38.550,0:08:44.240 Rui: For every epsilon greater than zero

0:08:44.240,0:08:51.240 Vipul: The epsilon is the interval on which you are trying to trap the function value.

0:08:51.500,0:08:54.279 Rui: There exists epsilon

0:08:54.279,0:08:55.890 Vipul: No, delta

0:08:55.890,0:09:14.890 Rui: delta> 0 such that for all x with x – c > 0

0:09:15.040,0:09:22.040 Vipul: The general one is for all x with 0<|x-c|<delta because you want to capture both the intervals.

0:09:23.170,0:09:29.270 In this one, the left hand limit one, we just captured the left side interval.

0:09:29.270,0:09:39.270 Now in the right one we just want to capture the right side interval, so as you said 0< x- c < delta.

0:09:44.180,0:09:51.480 In the picture, the function is defined, say c to c + t and you are really saying you can

0:09:52.290,0:10:00.290 find delta if x is in here [between c and c + delta] which actually... this is not including c, it is all the points

0:10:00.390,0:10:05.390 in the immediate right of c. We have? [ANSWER!]

0:10:06.000,0:10:13.000 Rui: The absolute value of f(x) – L is less than epsilon.

0:10:20.010,0:10:22.010 Vipul: So f(x) is? Are we here? We have everything?

0:10:23.010,0:10:23.260 Rui: Yes.

0:10:26.190,0:10:30.890 Vipul: We have both of these here? So do you see what’s the main difference between these

0:10:30.890,0:10:37.430 two and the actual [two-sided limit] definition?

0:10:37.430,0:10:42.930 For every epsilon there exists delta... the first second and fourth line remain the same.

0:10:42.930,0:10:47.440 It is this line where you are specifying where the x are that’s different.

0:10:47.440,0:10:53.000 In the two-sided thing the x could be either place.

0:10:53.300,0:10:55.200 For the left hand limit the x,

0:10:55.720,0:10:59.000 you just want x here [in (c - delta, c)] and

0:10:59.000,0:11:07.000 for the right hand limit you just want x in (c,c + delta).

0:11:07.000,0:11:09.000

Okay? [END!]

Relation between the limit notions

The two-sided limit exists if and only if (both the left hand limit and right hand limit exist and they are equal to each other).

Definition of finite limit for function of one variable in terms of a game

The formal definitions of limit, as well as of one-sided limit, can be reframed in terms of a game. This is a special instance of an approach that turns any statement with existential and universal quantifiers into a game.

Two-sided limit

Consider the limit statement, with specified numerical values of $c$ and $L$ and a specified function $f$ :

$lim_{x \to c} f (x) = L$

Note that there is one trivial sense in which the above statement can be false, or rather, meaningless, namely, that $f$ is not defined on the immediate left or immediate right of $c$ . In that case, the limit statement above is false, but moreover, it is meaningless to even consider the notion of limit.

The game is between two players, a Prover whose goal is to prove that the limit statement is true, and a Skeptic (also called a Verifier or sometimes a Disprover) whose goal is to show that the statement is false. The game has three moves:

First, the skeptic chooses $ε > 0$ , or equivalently, chooses the target interval $(L - ε, L + ε)$ .
Then, the prover chooses $δ > 0$ , or equivalently, chooses the interval $(c - δ, c + δ) ∖ {c}$ .
Then, the skeptic chooses a value $x$ satisfying $0 < | x - c | < δ$ , or equivalently, $x \in (c - δ, c + δ) ∖ {c}$ , which is the same as $(c - δ, c) \cup (c, c + δ)$ .

Now, if $| f (x) - L | < ε$ (i.e., $f (x) \in (L - ε, L + ε)$ ), the prover wins. Otherwise, the skeptic wins (see the subtlety about the domain of definition issue below the picture).

We say that the limit statement

$lim_{x \to c} f (x) = L$

is true if the prover has a winning strategy for this game. The winning strategy for the prover basically constitutes a strategy to choose an appropriate $δ$ in terms of the $ε$ chosen by the skeptic. Thus, it is an expression of $δ$ as a function of $ε$ .

We say that the limit statement

$lim_{x \to c} f (x) = L$

is false if the skeptic has a winning strategy for this game. The winning strategy for the skeptic involves a choice of $ε$ , and a strategy that chooses a value of $x$ (constrained in the specified interval) based on the prover's choice of $δ$ .

Slight subtlety regarding domain of definition: The domain of definition issue leads to a couple of minor subtleties:

A priori, it is possible that the $x$ chosen by the skeptic is outside the domain of $f$ , so it does not make sense to evaluate $f (x)$ . In the definition given above, this would lead to the game being won by the skeptic. In particular, if $f$ is not defined on the immediate left or right of $c$ , the skeptic can always win by picking $x$ outside the domain.
It may make sense to restrict discussion to the cases where f is defined on the immediate left or right of c. Explicitly, we assume that f is defined on the immediate left and immediate right, i.e., there exists t>0 such that f is defined on the interval (c−t,c+t)∖{c}. In this case, it does not matter what rule we set regarding the case that the skeptic picks x outside the domain. To simplify matters, we could alter the rules in any one of the following ways, and the meaning of limit would remain the same as in the original definition:
- We could require (as part of the game rules) that the prover pick $δ$ such that $(c - δ, c + δ) ∖ {c} \subseteq d o m f$ . This pre-empts the problem of picking $x$ -values outside the domain.
- We could require (as part of the game rules) that the skeptic pick $x$ in the domain, i.e., pick $x$ with $0 < | x - c | < δ$ and $x \in d o m f$ .
- We could alter the rule so that if the skeptic picks $x$ outside the domain, the prover wins (instead of the skeptic winning).

Full timed transcript: [SHOW MORE]

0:00:15.589,0:00:21.160 Vipul: In this video, I'm going to go over the usual definition of limit and think of

0:00:21.160,0:00:24.930 it in terms of a game.

0:00:24.930,0:00:26.390 The game is as follows.

0:00:26.390,0:00:27.340 Consider this statement.

0:00:27.340,0:00:31.509 You are saying limit as x approaches c of f(x) is L.

0:00:31.509,0:00:32.029 Okay.

0:00:32.029,0:00:35.160 There are two players to this game.

0:00:35.160,0:00:38.600 One is the prover and one is the skeptic.

0:00:38.600,0:00:44.550 The prover's goal is to show that this claim is true so the prover is trying to convince

0:00:44.550,0:00:48.730 the skeptic that this limit as x approaches c of f(x) is L,

0:00:48.730,0:01:01.160 the skeptic will try to ask tough questions and see if the prover can still manage to show this.

0:01:01.160,0:01:04.059 The way the game is structured is as follows.

0:01:04.059,0:01:08.899 Let me just go over the individual components of the statement for the limit and I will

0:01:08.899,0:01:10.610 translate each one.

0:01:10.610,0:01:17.610 I will explain the game and then explain how it corresponds to the definition you've seen.

0:01:20.219,0:01:27.219 We begin with the skeptic chooses epsilon > 0.

0:01:35.840,0:01:42.840 This is the part of the definition which reads for every epsilon > 0.

0:01:47.099,0:01:53.289 That's the first clause of the definition and that's basically the skeptic is choosing

0:01:53.289,0:01:54.579 epsilon > 0.

0:01:54.579,0:01:59.299 What is the skeptic trying to do when choosing epsilon > 0?

0:01:59.299,0:02:06.299 What the skeptic is effectively doing is choosing this interval L -- epsilon to L + epsilon.

0:02:14.400,0:02:18.220 The skeptic is effectively trying to choose this interval L -- epsilon to L + epsilon.

0:02:18.220,0:02:26.110 What is the skeptic trying the challenge the prover into doing when picking this interval? [ANSWER!]

0:02:26.110,0:02:29.890 Rui: Whether the prover can trap.

0:02:29.890,0:02:35.180 Vipul: The skeptic is trying to challenge (and this will become a clearer a little later).

0:02:35.180,0:02:41.790 The idea is, the skeptic is trying to challenge the prover into trapping the function when

0:02:41.790,0:02:47.620 the input x is close to c, trapping the function output within this interval and that's

0:02:47.620,0:02:52.459 not clear which is why we need to continue its definition.

0:02:52.459,0:02:58.609 The prover chooses. What does the prover choose? [ANSWER!]

0:02:58.609,0:03:00.260 Rui: delta.

0:03:00.260,0:03:07.260 Vipul: delta > 0 and this corresponds to the next part of the definition which says

0:03:08.480,0:03:15.480 there exists delta > 0.

0:03:19.749,0:03:26.749 In this picture, which I have up here, this is the value c.

0:03:28.840,0:03:31.989 This is c + delta and this is c -- delta.

0:03:31.989,0:03:41.349 This is c and L, so c is the x coordinate, L is the function value or limited the function value.

0:03:41.349,0:03:48.349 The skeptic chooses this strip like this from L -- epsilon to L + epsilon by choosing epsilon

0:03:51.450,0:03:56.109 so the skeptic just chooses the number absent what it is effectively doing is to choose

0:03:56.109,0:04:01.790 this strip, L -- epsilon to L + epsilon. The prover then chooses a delta.

0:04:01.790,0:04:03.829 What's the prover effectively choosing?

0:04:03.829,0:04:07.290 The prover is effectively choosing this interval.

0:04:07.290,0:04:14.230 Okay so that's this interval.

0:04:14.230,0:04:20.209 It is c -- delta to c + delta except you don't really care about the point c itself,

0:04:20.209,0:04:26.490 (but that's a little subtlety we don't have to bother about), so the skeptic is choosing

0:04:26.490,0:04:29.780 the interval like this. The prover is choosing the interval like this.

0:04:29.780,0:04:33.340 How is the skeptic choosing the interval? By just specifying the value of epsilon.

0:04:33.340,0:04:34.880 How is the prover choosing [the interval around c]?

0:04:34.880,0:04:45.880 By just specifying a value of delta. Okay. Now what does the skeptic now do? [ANSWER!]

0:04:46.500,0:04:52.979 Rui: Skeptic will check.

0:04:53.079,0:05:00.079 Vipul: There is something more to choose (right?) before checking.

0:05:02.710,0:05:06.599 What does the definition say? For every epsilon > 0 there exists a delta greater than zero

0:05:06.599,0:05:07.259 such that ... [COMPLETE!]

0:05:07.259,0:05:08.580 Rui: For every.

0:05:08.580,0:05:13.220 Vipul: For every x such that something. The skeptic can now pick x.

0:05:13.220,0:05:17.000 Rui: That's what I meant by checking.

0:05:17.000,0:05:21.940 Vipul: The skeptic could still, like, pick a value to challenge the prover.

0:05:21.940,0:05:28.940 The skeptic chooses x but what x can the skeptic choose?

0:05:29.169,0:05:31.810 Rui: Within the...

0:05:31.810,0:05:36.590 Vipul: This interval which the prover has specified.

0:05:36.590,0:05:43.590 The skeptic is constrained to choose x within the interval.

0:05:44.250,0:05:49.639 That's the same as c -- delta ... Is this all coming?

0:05:49.639,0:05:50.330 Rui: Yes.

0:05:50.330,0:05:57.330 Vipul: c -- delta, c union c to c + delta.

0:05:59.110,0:06:15.110 The way it's written is for every x in this interval.

0:06:16.849,0:06:21.349 Lot of people write this in a slightly different way.

0:06:21.349,0:06:28.349 They write it as ...

0:06:28.400,0:06:31.720 (You should see the definition video before this.)

0:06:31.720,0:06:37.729 (I'm sort of assuming that you have seen the definition -- this part [of the screen] so you can map it)

0:06:37.729,0:06:40.000 so a lot of people write it like this.

0:06:40.000,0:06:45.190 It is just saying x is within delta distance of c but it's not equal to c itself.

0:06:45.190,0:06:50.949 Now it's time for the judge to come in and decide who has won.

0:06:50.949,0:06:55.930 How does the judge decide? [ANSWER!]

0:06:55.930,0:07:01.360 Rui: For the x that the skeptic chooses and see the corresponding y.

0:07:01.360,0:07:03.289 Vipul: The f(x) value.

0:07:03.289,0:07:10.289 Rui: If the f(x) value is within the horizontal strip then the prover wins.

0:07:12.509,0:07:30.000 Vipul: If |f(x) -- L| < epsilon which is the same as saying f(x) is in what interval? [ANSWER!]

0:07:30.000,0:07:41.620 L- epsilon to L + epsilon then the prover wins. Otherwise? [ANSWER!]

0:07:42.120,0:07:46.120 Rui: The skeptic wins.

0:07:46.120,0:07:53.120 [But] the skeptic can choose a really dumb [stupid] x.

0:07:54.039,0:07:57.610 Vipul: That's actually the next question I want to ask you.

0:07:57.610,0:08:01.240 What does it actually mean to say that this statement is true?

0:08:01.240,0:08:04.770 Is it just enough that the prover wins? That's not enough.

0:08:04.770,0:08:07.909 What do you want to say to say that this statement is true?

0:08:07.909,0:08:11.210 Rui: For every x in the interval.

0:08:11.210,0:08:16.289 Vipul: For every x but not only for every x you should also say for every epsilon.

0:08:16.289,0:08:22.139 All the moves that the skeptic makes, the prover should have a strategy, which works for all of them.

0:08:22.139,0:08:25.710 So, this statement is true [if] ...

0:08:25.710,0:08:29.800 This is true if the prover has what for the game? [ANSWER!]

0:08:30.539,0:08:35.050 Rui: Winning strategy. Vipul: Winning what? Rui: Strategy.

0:08:35.050,0:08:38.669 Vipul: Yeah. True if the prover has a winning strategy.

0:08:38.669,0:08:44.910 It is not just enough to say that the prover won the game some day but the prover should

0:08:44.910,0:08:50.220 be able to win the game regardless of how smart the skeptic is or regardless of how

0:08:50.220,0:08:53.960 experienced the skeptic is or regardless of how the skeptic plays.

0:08:53.960,0:09:00.960 That's why all the moves of the skeptic are prefaced with a "for every." Right?

0:09:02.230,0:09:07.560 Whereas all the moves of the prover are prefaced, (well there is only one move really of the

0:09:07.560,0:09:11.180 prover) are prefaced with "there exists" because the prover controls his own choices.

0:09:11.180,0:09:15.360 When it is the prover's turn it's enough to say "there exists" but since the prover doesn't

0:09:15.360,0:09:21.590 control what the skeptic does all the skeptic moves are prefaced with "for every."

0:09:21.590,0:09:26.150 By the way, there is a mathematical notation for these things.

0:09:26.150,0:09:31.730 There are mathematical symbols for these, which I'm not introducing in this video,

0:09:31.730,0:09:37.920 but if you have seen them and got confused then you can look at the future video where

0:09:37.920,0:09:40.500

I explain the mathematical symbols.

Full timed transcript: [SHOW MORE]

0:01:26.720,0:01:33.720 Ok, so in this talk, we are going to give the definition of what it means to say that this statement,

0:01:34.250,0:01:37.940 the one up here, is false.

0:01:37.940,0:01:41.300 So far we've looked at what it means for this statement to be true.

0:01:41.300,0:01:44.960 Now we are going to look at what it means for the statement to be false.

0:01:44.960,0:01:48.340 Basically, you just use the same definition, but you would change a little bit of what

0:01:48.340,0:01:49.490 it looks like.

0:01:49.490,0:01:54.130 Let me first remind you of the limit game because that is a very nice way of thinking

0:01:54.130,0:01:57.380 about what it means to be true and false.

0:01:57.380,0:01:58.860 What does the limit game say?

0:01:58.860,0:02:01.680 It is a game between two players, a prover and a skeptic.

0:02:01.680,0:02:04.680 What is the goal of the prover? [ANSWER!]

0:02:04.680,0:02:06.310 Rui: To show he is right.

0:02:06.310,0:02:07.930 Vipul: To show that this is true.

0:02:07.930,0:02:08.489 Rui: True.

0:02:08.489,0:02:12.830 Vipul: The skeptic is trying to show that this is false, or at least trying to come

0:02:12.830,0:02:16.730 up with the strongest evidence to suggest that this is false.

0:02:16.730,0:02:18.090 How does the game proceed?

0:02:18.090,0:02:23.349 The skeptic begins by choosing an epsilon greater than zero.

0:02:23.349,0:02:25.200 What is the skeptic effectively trying to pick?

0:02:25.200,0:02:30.769 The skeptic is effectively trying to pick this neighborhood of L and trying to challenge

0:02:30.769,0:02:36.579 the prover to trap the function value for x within that neighborhood.

0:02:36.579,0:02:40.719 What's that neighborhood the skeptic is secretly picking? [ANSWER!]

0:02:40.719,0:02:43.909 Rui: L -- epsilon [to L + epsilon]

0:02:43.909,0:02:50.909 Vipul: Ok, the prover chooses a delta greater than zero so the prover is now basically trying

0:02:53.040,0:03:00.040 to pick a neighborhood of c, the point near the domain points, and

0:03:02.650,0:03:09.650 then the skeptic will then pick a value x, which is within the interval delta distance of c except the point c itself.

0:03:10.120,0:03:16.200 That's either delta interval on the left or delta interval on the right of c.

0:03:16.200,0:03:20.569 Then the judge comes along and computes this value, absolute value f(x) minus...Are we,

0:03:20.569,0:03:21.739 is this in the picture?

0:03:21.739,0:03:22.700 Rui: Yes.

0:03:22.700,0:03:27.329 Vipul: If it is less than epsilon then the prover would have won, but now we want to

0:03:27.329,0:03:34.329 see if the skeptic wins if it is greater or equal to epsilon, that means f(x) is not in

0:03:35.569,0:03:36.129 the epsilon...

0:03:36.129,0:03:37.249 Rui: Neighborhood.

0:03:37.249,0:03:42.459 Vipul: This video assumes you have already seen the previous videos where we give these

0:03:42.459,0:03:48.689 definitions and so I'm sort of reviewing it quickly, but not explaining it in full detail.

0:03:48.689,0:03:54.069 So, the skeptic wins if f(x) is outside this interval, that means the prover failed to

0:03:54.069,0:03:58.069 rise to the skeptic's challenge of trapping the function.

0:03:58.069,0:04:05.069 Let's now try to work out concretely what the definition would read.

0:04:06.590,0:04:10.439 The skeptic is the one in control because you want to figure out whether the skeptic

0:04:10.439,0:04:12.639 has a winning strategy.

0:04:12.639,0:04:17.690 Ok, so let me just say this clearly, this is just saying when does the skeptic win?

0:04:17.690,0:04:21.090 Now in order to say this limit statement is false, we need something stronger. What do

0:04:21.090,0:04:25.360 we need to say this is false? [ANSWER!]

0:04:25.360,0:04:26.450 The skeptic should have...

0:04:26.450,0:04:28.820 Rui: Should have a winning strategy.

0:04:28.820,0:04:30.410 Vipul: A winning strategy.

0:04:30.410,0:04:34.229 The skeptic should have a strategy so that whatever the prover does, the skeptic has

0:04:34.229,0:04:36.139 some way of winning.

0:04:36.139,0:04:41.229 What should this read...if you actually translate it to the definition?

0:04:41.229,0:04:44.169 Rui: There exists an...

0:04:44.169,0:04:46.000 Vipul: There exists epsilon

0:04:46.000,0:04:51.000 Rui: ...an epsilon greater than zero.

0:04:58.000,0:05:00.000 Vipul: Okay. Such that...

0:05:00.280,0:05:07.210 Rui: For every delta greater than zero.

0:05:07.210,0:05:10.870 Vipul: So the skeptic, when it's the skeptic's move the skeptic says "there exists."

0:05:10.870,0:05:14.310 If anything works, the skeptic can pick that, but when it's the provers move, the skeptic

0:05:14.310,0:05:15.699 has no control.

0:05:15.699,0:05:30.699 This should read, for every delta greater than zero...What will the next part read?

0:05:31.770,0:05:33.930 Rui: There exists an x.

0:05:33.930,0:05:40.930 Vipul: Exists x in this interval.

0:05:44.289,0:05:45.340 Rui: Yeah.

0:05:45.340,0:05:50.159 Vipul: Which you often see it written in a slightly different form.

0:05:50.159,0:05:57.159 Maybe, I don't have space here, so here it is also written as "0 ...", are we down here?

0:05:59.960,0:06:01.560 Rui: Yes.

0:06:01.560,0:06:04.470 Vipul: This is the form it's usually written in concise definitions.

0:06:04.470,0:06:20.710 We have this...So the definition, maybe it's not clear, but the definition would read like that.

0:06:20.710,0:06:25.419 So there exists Epsilon greater than zero such that for every delta greater than zero there

0:06:25.419,0:06:30.879 exists x, in here, which you could also write like this, such that, I guess I should put

0:06:30.879,0:06:35.310 the "such that." [writes it down]

0:06:35.310,0:06:39.849 Such that. absolute value of f(x) -- L is greater than or equal to epsilon

0:06:39.849,0:06:44.680 Let me just compare it with the usual definition for the limit to exist.

0:06:44.680,0:06:47.750 The colors are in a reverse chrome.

0:06:47.750,0:06:52.860 That's fine. For every epsilon greater than zero became there exists epsilon greater than

0:06:52.860,0:06:55.879 zero because the player who is in control has changed.

0:06:55.879,0:06:59.789 There exists delta greater than zero became for every delta greater than zero, for all

0:06:59.789,0:07:05.139 x with this became their exists x satisfying this condition.

0:07:05.139,0:07:07.629 What happened to the last clause?

0:07:07.629,0:07:12.099 The less than Epsilon begin greater than or equal to.

0:07:12.099,0:07:17.069 The last clause just got reversed in meaning, all the others, we just changed the quantifier

0:07:17.069,0:07:22.389 from "for all" to "there exists" and from "there exists" to "for all" and that is just because

0:07:22.389,0:07:25.770 we changed who is winning.

0:07:25.770,0:07:30.439 If you have seen some logic, if you ever see logic, then there are some general rules of

0:07:30.439,0:07:33.650 logic as to how to convert a statement to its opposite statement.

0:07:33.650,0:07:38.610 This is a general rule that "for all" becomes

"there exists" and "there exists" becomes "for all."

Non-existence of limit

The statement $lim_{x \to c} f (x)$ does not exist could mean one of two things:

$f$ is not defined around $c$ , i.e., there is no $t > 0$ for which $f$ is defined on $(c - t, c + t) ∖ {c}$ . In this case, it does not even make sense to try taking a limit.
$f$ is defined around $c$ , around $c$ , i.e., there is $t > 0$ for which $f$ is defined on $(c - t, c + t) ∖ {c}$ . So, it does make sense to try taking a limit. However, the limit still does not exist.

The formulation of the latter case is as follows:

For every
$L \in R$
, there exists
$ε > 0$
such that for every
$δ > 0$
, there exists
$x$
satisfying
$0 < | x - c | < δ$
and such that
$| f (x) - L | \geq ε$
.

We can think of this in terms of a slight modification of the limit game, where, in our modification, there is an extra initial move by the prover to propose a value $L$ for the limit. The limit does not exist if the skeptic has a winning strategy for this modified game.

An example of a function that does not have a limit at a specific point is the sine of reciprocal function. Explicitly, the limit:

$lim_{x \to 0} \sin (\frac{1}{x})$

does not exist. The skeptic's winning strategy is as follows: regardless of the $L$ chosen by the prover, pick a fixed $ε < 1$ (independent of $L$ , so $ε$ can be decided in advance of the game -- note that the skeptic could even pick $ε = 1$ and the strategy would still work). After the prover has chosen a value $δ$ , find a value $x \in (0 - δ, 0 + δ) ∖ {0}$ such that the $\sin (1 / x)$ function value lies outside $(L - ε, L + ε)$ . This is possible because the interval $(L - ε, L + ε)$ has width $2 ε$ , hence cannot cover the entire interval $[- 1, 1]$ , which has width 2. However, the range of the $\sin (1 / x)$ function on $(0 - δ, 0 + δ) ∖ {0}$ is all of $[- 1, 1]$ .

Full timed transcript: [SHOW MORE]

0:00:15.500,0:00:19.140 Vipul: Okay. This talk is going to be about certain misconceptions

0:00:19.140,0:00:22.440 that people have regarding limits and these are misconceptions that

0:00:22.440,0:00:25.840 people generally acquire after...

0:00:25.840,0:00:29.180 These are not the misconceptions that people have before studying limits,

0:00:29.180,0:00:32.730 these are misconceptions you might have after studying limits,

0:00:32.730,0:00:35.059 after studying the epsilon delta definition.

0:00:35.059,0:00:38.550 I'm going to describe these misconceptions in terms of the limit game,

0:00:38.550,0:00:41.900 the prover skeptic game of the limit. Though the misconceptions

0:00:41.900,0:00:45.850 themselves can be, sort of, don't depend on the understanding of the

0:00:45.850,0:00:49.059 game but to understand exactly what's happening, it's better to think

0:00:49.059,0:00:51.010 of it in terms of the game.

0:00:51.010,0:00:55.370 First recall the definition. So limit as x approaches c of f(x) is a

0:00:55.370,0:01:01.629 number L; so c and L are both numbers, real numbers. f is a function,

0:01:01.629,0:01:06.380 x is approaching c. And we said this is true if the following -- for

0:01:06.380,0:01:10.180 every epsilon greater than zero, there exists a delta greater than

0:01:10.180,0:01:14.800 zero such that for all x which are given delta distance of c, f(x) is

0:01:14.800,0:01:17.590 within epsilon distance of L. Okay?

0:01:17.590,0:01:24.590 Now, how do we describe this in terms for limit game?

0:01:26.530,0:01:33.530 KM: So, skeptic starts off with the first part of the definition.

0:01:34.990,0:01:38.189 Vipul: By picking the epsilon? Okay, that's the thing written in

0:01:38.189,0:01:42.939 black. What's the skeptic trying to do? What's the goal of the skeptic?

0:01:42.939,0:01:49.100 KM: To try and pick an epsilon that would not work.

0:01:49.100,0:01:53.450 Vipul: So the goal of the skeptic is to try to show that the statement is false.

0:01:53.450,0:01:54.100 KM: Yeah.

0:01:54.100,0:01:57.790 Vipul: Right? In this case the skeptic should try to start by choosing

0:01:57.790,0:02:02.220 an epsilon that is really -- the goal of the skeptic is to pick an

0:02:02.220,0:02:04.500 epsilon that's really small, what is the skeptic trying to challenge

0:02:04.500,0:02:07.920 the prover into doing by picking the epsilon? The skeptic is trying to

0:02:07.920,0:02:11.959 challenge the prover into trapping the function close to L when x is

0:02:11.959,0:02:17.040 close to c. And the skeptic specifies what is meant by "close to L" is

0:02:17.040,0:02:19.860 by the choice of epsilon. Okay?

0:02:19.860,0:02:24.900 When picking epsilon the skeptic is effectively picking this interval, L -

0:02:24.900,0:02:30.700 epsilon, L + epsilon). Okay? And basically that's what the skeptic is

0:02:30.700,0:02:33.680 doing. The prover is then picking a delta. What is the goal of the

0:02:33.680,0:02:36.239 prover in picking the delta? The prover is saying, "Here's how I can

0:02:36.239,0:02:40.099 trap the function within that interval. I'm going to pick a delta and

0:02:40.099,0:02:43.520 my claim is that if the x value within delta distance of c, except the

0:02:43.520,0:02:47.000 point c itself, so my claim is for any x value there the function is

0:02:47.000,0:02:48.260 trapped in here."

0:02:48.260,0:02:52.819 So, the prover picks the delta and then the skeptic tries to meet the

0:02:52.819,0:02:56.709 prover's claim or rather, test the prover's claim by picking an x

0:02:56.709,0:02:59.670 which is within the interval specified by the prover and then they

0:02:59.670,0:03:03.379 both check whether f(x) is within epsilon distance [of L]. If it is

0:03:03.379,0:03:07.940 then the prover wins and if it is not, if this [|f(x) - L|]is not less

0:03:07.940,0:03:09.989 than epsilon then the skeptic wins. Okay?

0:03:09.989,0:03:13.659 So, the skeptic is picking the neighborhood of the target point which

0:03:13.659,0:03:17.030 in this case is just the open interval of radius epsilon, the prover

0:03:17.030,0:03:21.940 is picking the delta which is effectively the neighborhood of the domain

0:03:21.940,0:03:25.760 point except the point c as I've said open interval (c - delta, c +

0:03:25.760,0:03:30.870 delta) excluding c and then the skeptic picks an x in the neighborhood

0:03:30.870,0:03:35.700 specified by prover and if the function value is within the interval

0:03:35.700,0:03:38.830 specified by the skeptic then the prover wins.

0:03:38.830,0:03:41.989 Now, what does it mean to say the statement is true in terms of the

0:03:41.989,0:03:43.080 game?

0:03:43.080,0:03:50.080 KM: So, it means that the prover is always going to win the game.

0:03:51.849,0:03:55.629 Vipul: Well, sort of. I mean the prover may play it stupidly. The

0:03:55.629,0:04:00.750 prover can win the game if the prover plays well. So, the prover has a

0:04:00.750,0:04:03.230 winning strategy for the game. Okay?

0:04:05.230,0:04:10.299 The statement is true if the prover has a winning strategy for [the

0:04:10.299,0:04:14.090 game] and that means the prover has a way of playing the game such that

0:04:14.090,0:04:17.320 whatever the skeptic does the prover is going to win the game. The

0:04:17.320,0:04:20.789 statement is considered false if the skeptic has a winning strategy

0:04:20.789,0:04:23.370 for the game which means the skeptic has a way of playing so that

0:04:23.370,0:04:25.729 whatever the prover does the skeptic can win the game.

0:04:25.729,0:04:27.599 Or if the game doesn't make sense at all ...

0:04:27.599,0:04:29.460 maybe the function is not defined on

0:04:29.460,0:04:31.050 the immediate left and right of c.

0:04:31.050,0:04:32.370 If the function isn't defined then we

0:04:32.370,0:04:34.160 cannot even make sense of the statement.

0:04:34.160,0:04:36.990 Either way -- the skeptic has a winning strategy

0:04:36.990,0:04:37.770 or the game doesn't make sense --

0:04:41.770,0:04:43.470 then the statement is false.

0:04:43.470,0:04:47.660 If the prover has a winning strategy the statement is true.

0:04:47.660,0:04:54.660 With this background in mind let's look at some common misconceptions.

0:04:56.540,0:05:03.540 Okay. Let's say we are trying to prove that the limit as x approaches

0:05:27.620,0:05:31.530 2 of x^2 is 4, so is that statement correct? The statement we're

0:05:31.530,0:05:32.060 trying to prove?

0:05:32.060,0:05:32.680 KM: Yes.

0:05:32.680,0:05:35.960 Vipul: That's correct. Because in fact x^2 is a continuous function

0:05:35.960,0:05:40.160 and the limit of a continuous function at the point is just the

0:05:40.160,0:05:43.030 value at the point and 2^2 is 4. But we're going to now try to prove

0:05:43.030,0:05:48.530 this formally using the epsilon-delta definition of limit, okay? Now

0:05:48.530,0:05:51.229 in terms of the epsilon-delta definition or rather in terms of this

0:05:51.229,0:05:55.160 game setup, what we need to do is we need to describe a winning

0:05:55.160,0:06:01.460 strategy for the prover. Okay? We need to describe delta in terms of

0:06:01.460,0:06:05.240 epsilon. The prover essentially ... the only move the prover makes is

0:06:05.240,0:06:09.130 this choice of delta. Right? The skeptic picked epsilon, the prover

0:06:09.130,0:06:12.810 picked delta then the skeptic picks x and then they judge who won. The

0:06:12.810,0:06:15.810 only choice the prover makes is the choice of delta, right?

0:06:15.810,0:06:16.979 KM: Exactly.

0:06:16.979,0:06:20.080 Vipul: The prover chooses the delta in terms of epsilon.

0:06:20.080,0:06:24.819 So, here is my strategy. My strategy is I'm going to choose delta as,

0:06:24.819,0:06:29.509 I as a prover is going to choose delta as epsilon over the absolute

0:06:29.509,0:06:33.690 value of x plus 2 [|x + 2|]. Okay?

0:06:33.690,0:06:36.880 Now, what I want to show that this strategy works. So, what I'm aiming

0:06:36.880,0:06:39.840 is that if ... so let me just finish this and then you can tell me where

0:06:39.840,0:06:43.419 I went wrong here, okay? I'm claiming that this strategy works which

0:06:43.419,0:06:47.130 means I'm claiming that if the skeptic now picks any x which is within

0:06:47.130,0:06:54.130 delta distance of 2; the target point,

0:06:56.710,0:07:01.490 then the function value is within epsilon distance of 4, the claimed

0:07:01.490,0:07:04.080 limit. That's what I want to show.

0:07:04.080,0:07:08.300 Now is that true? Well, here's how I do it. I think, I started by

0:07:08.300,0:07:13.539 picking this expression, I factored it as |x - 2||x + 2|. The absolute

0:07:13.539,0:07:16.810 value of product is the product of the absolute values so this can be

0:07:16.810,0:07:21.599 split like that. Now I see, while we know that |x - 2| is less than

0:07:21.599,0:07:24.979 delta and this is a positive thing. So we can either less than delta

0:07:24.979,0:07:31.979 times absolute value x plus 2. Right? And this delta is epsilon over

0:07:35.599,0:07:37.620 |x + 2| and we get epsilon.

0:07:37.620,0:07:40.460 So, this thing equals something, less than something, equals

0:07:40.460,0:07:43.580 something, equals something, you have a chain of things, there's one

0:07:43.580,0:07:47.720 step that you have less than. So overall we get that this expression,

0:07:47.720,0:07:53.740 this thing is less than epsilon. So, we have shown that whatever x the

0:07:53.740,0:08:00.370 skeptic would pick, the function value lies within the epsilon

0:08:00.370,0:08:05.030 distance of the claimed limit. Whatever the skeptic picks (x within the

0:08:05.030,0:08:09.240 delta distance of the target point).

0:08:09.240,0:08:16.240 Does this strategy work? Is this a proof? What's wrong with this?

0:08:24.270,0:08:31.270 Do you think there's anything wrong with the algebra down here?

0:08:33.510,0:08:40.510 KM: Well, we said that ...

0:08:40.910,0:08:47.910 Vipul: So, is there anything wrong in the algebra here? This is this,

0:08:50.160,0:08:51.740 this is less than delta, delta ... So, this part

0:08:51.740,0:08:52.089 seems fine, right?

0:08:52.089,0:08:52.339 KM: Yes.

0:08:52.330,0:08:55.640 Vipul: There's nothing wrong in the algebra here. So, what could be

0:08:55.640,0:09:00.310 wrong? Our setup seems fine. If the x value is within delta distance

0:09:00.310,0:09:03.350 of 2 then the function value is within epsilon this is 4. That's

0:09:03.350,0:09:05.360 exactly what we want to prove, correct?

0:09:05.360,0:09:11.120 So, there's nothing wrong this point onward. So, the error happened

0:09:11.120,0:09:14.440 somewhere here. Where do you think that part you think what is wrong

0:09:14.440,0:09:21.160 here? In the strategy choice step? What do you think went wrong in the

0:09:21.160,0:09:24.010 strategy choice step?

0:09:24.010,0:09:28.850 What? Okay, so let's go over the game. Skeptic will choose the epsilon,

0:09:28.850,0:09:29.760 then?

0:09:29.760,0:09:35.130 KM: Then the prover chooses delta.

0:09:35.130,0:09:36.080 Vipul: Prover chooses delta. Then?

0:09:36.080,0:09:39.529 KM: Then the skeptic has to choose the x value.

0:09:39.529,0:09:42.470 Vipul: x value. So, when the prover is deciding the strategy, when the

0:09:42.470,0:09:45.860 prover is choosing the delta, what information does the prover have?

0:09:45.860,0:09:48.410 KM: He just has the information epsilon.

0:09:48.410,0:09:50.500 Vipul: Just the information on epsilon. So?

0:09:50.500,0:09:57.060 KM: So, in this case the mistake was that because he didn't know the x value yet?

0:09:57.060,0:10:03.100 Vipul: The strategy cannot depend on x.

0:10:03.100,0:10:04.800 KM: Yeah.

0:10:04.800,0:10:09.790 Vipul: So, the prover is sort of picking the delta based on x but the

0:10:09.790,0:10:12.660 prover doesn't know x at this stage when picking the delta. The delta

0:10:12.660,0:10:15.910 that the prover chooses has to be completely a function of epsilon

0:10:15.910,0:10:19.680 alone, it cannot depend on the future moves of the skeptic because the

0:10:19.680,0:10:23.700 prover cannot read the skeptic's mind. Okay? And doesn't know what the

0:10:23.700,0:10:24.800 skeptic plans to do.

0:10:24.800,0:10:31.800 So that is the ... that's the ... I call this ... can you see what I

0:10:42.240,0:10:43.040 call this?

0:10:43.040,0:10:45.399 KM: The strongly telepathic prover.

0:10:45.399,0:10:51.470 Vipul: So, do you know what I meant by that? Well, I meant the prover

0:10:51.470,0:10:58.470 is sort of reading the skeptic's mind. All right? It's called

0:11:07.769,0:11:10.329 telepathy.

0:11:10.329,0:11:17.329 Okay, the next one.

0:11:25.589,0:11:30.230 This one says that the function defined this way. Okay? It's defined

0:11:30.230,0:11:34.829 as g(x) is x when x is rational and zero when x is irrational. So,

0:11:34.829,0:11:41.829 what would this look like? Well, it's like this. There's a line y

0:11:42.750,0:11:49.510 equals x and there's the x-axis and the graph is just the irrational x

0:11:49.510,0:11:52.750 coordinate parts of this line and the rational x coordinate parts of

0:11:52.750,0:11:56.350 this line. It's kind of like both these lines but only parts of

0:11:56.350,0:11:58.529 them. Right?

0:11:58.529,0:12:02.079 Now we want to show that limit as x approaches zero of g(x) is

0:12:02.079,0:12:06.899 zero. So just in here, do you think the statement is true? That x goes

0:12:06.899,0:12:09.910 to zero, does this function go to zero?

0:12:09.910,0:12:10.610 KM: Yes.

0:12:10.610,0:12:17.610 Vipul: Because both the pieces are going to zero. That's the inclusion. Okay?

0:12:20.610,0:12:24.089 This is the proof we have here. So the idea, we again think about it

0:12:24.089,0:12:27.790 in terms of the game. The skeptic first picks the epsilon, okay? Now

0:12:27.790,0:12:30.779 that we would have to choose the delta, but there are really two cases

0:12:30.779,0:12:35.200 on x, right? x rational and x irrational. So the prover chooses the

0:12:35.200,0:12:39.459 delta based on sort of whether the x is rational or irrational, so if

0:12:39.459,0:12:43.880 the x is rational then the prover just picks delta equals epsilon, and

0:12:43.880,0:12:48.339 that's good enough for rational x, right? Because for rational x the

0:12:48.339,0:12:51.410 slope of the line is one so picking delta as epsilon is good enough.

0:12:51.410,0:12:55.760 For irrational x, if the skeptic's planning to choose an irrational x

0:12:55.760,0:12:59.730 then the prover can just choose any delta actually. Like just pick

0:12:59.730,0:13:03.880 the delta in advance. Like delta is one or something. Because if x is

0:13:03.880,0:13:10.430 irrational then it's like a constant function and therefore, like, for

0:13:10.430,0:13:14.970 any delta the function is trapped within epsilon distance of the given

0:13:14.970,0:13:16.970 limit. Okay?

0:13:16.970,0:13:19.950 So the prover sort of makes two cases based on whether the skeptic

0:13:19.950,0:13:26.950 will pick a rational or an irrational x and sort of based on that if

0:13:27.040,0:13:30.730 it's rational this is the prover's strategy, if it's irrational then

0:13:30.730,0:13:34.050 the prover can just do any delta.

0:13:34.050,0:13:37.630 Can you tell me what's wrong with this proof?

0:13:37.630,0:13:44.630 KM: So, you're still kind of basing it on what the skeptic is going to

0:13:44.750,0:13:45.800 pick next.

0:13:45.800,0:13:49.100 Vipul: Okay. It's actually pretty much the same problem [as the

0:13:49.100,0:13:55.449 preceding one], in a somewhat minor form. The prover is sort of making

0:13:55.449,0:13:59.959 cases based on what the skeptic is going to do next, and choosing a

0:13:59.959,0:14:01.940 strategy according to that. But the prover doesn't actually know what

0:14:01.940,0:14:05.089 the skeptic is going to do next, so the prover should actually have a

0:14:05.089,0:14:08.970 single strategy that works in both cases. If cases will be made to

0:14:08.970,0:14:12.209 prove that the strategy works so the prover has to have a single

0:14:12.209,0:14:12.459 strategy.

0:14:12.449,0:14:15.370 Now in this case the strategy we can choose the prover just, the

0:14:15.370,0:14:18.779 prover can pick delta as epsilon because that will work in both cases.

0:14:18.779,0:14:20.019 KM: Exactly.

0:14:20.019,0:14:23.320 Vipul: Yeah. But in general if you have two different piece

0:14:23.320,0:14:26.579 definitions then the way you would do it so you would pick delta as

0:14:26.579,0:14:30.300 the min [minimum] of the delta that work in the two different pieces,

0:14:30.300,0:14:32.910 because you sort of want to make sure that both cases are covered. But

0:14:32.910,0:14:36.730 the point is you have to do that -- take the min use that rather than

0:14:36.730,0:14:39.730 just say, "I'm going to choose my delta based on what the skeptic is

0:14:39.730,0:14:42.589 going to move next." Okay?

0:14:42.589,0:14:49.120 This is a minor form of the same misconception that that was there in

0:14:49.120,0:14:56.120 the previous example we saw.

0:15:04.620,0:15:11.620 So, this is what I call the mildly telepathic prover, right? The

0:15:14.970,0:15:18.579 prover is still behaving telepathically predicting the skeptic's future

0:15:18.579,0:15:23.740 moves but it's not so bad. The prover is just making, like, doing a

0:15:23.740,0:15:25.470 coin toss type of telepathy. That isn't the only one the prover is

0:15:25.470,0:15:30.790 actually, deciding exactly what x skeptic would take. But it's still

0:15:30.790,0:15:32.790 the same problem and the reason why I think people will have this

0:15:32.790,0:15:36.329 misconception is because they don't think about it in terms of the

0:15:36.329,0:15:38.970 sequence in which the moves are made, and the information that each

0:15:38.970,0:15:45.970 body has at any given stage of the game.

0:15:50.889,0:15:57.889 Let's do this one.

0:16:10.930,0:16:15.259 So, this is a limit game, right? Let's say that limit as x approaches

0:16:15.259,0:16:22.259 1 of 2x is 2, okay? How do we go about showing this? Well, the idea is

0:16:23.699,0:16:27.990 let's play the game, right? Let's say the skeptic it picks epsilon as

0:16:27.990,0:16:34.990 0.1, okay? The prover picks delta as 0.05. The skeptic is then picking

0:16:35.139,0:16:38.790 epsilon as 0.1, the skeptic is saying, "Please trap the function

0:16:38.790,0:16:43.800 between 1.9 and 2.1. Okay? Find the delta small enough so that the

0:16:43.800,0:16:48.389 function value is dropped between 1.9 and 2.1. The prover picks delta

0:16:48.389,0:16:55.389 as 0.05 which means the prover is now getting the input value trap

0:16:57.850,0:17:04.850 between 0.95 and 1.05. That's 1 plus minus this thing. And now the

0:17:05.439,0:17:09.070 prover is claiming that if the x value is within this much distance of

0:17:09.070,0:17:13.959 1 except the value equal to 1, then the function value is within 0.1

0:17:13.959,0:17:17.630 distance of 2. So, the skeptic tries picking x within the interval

0:17:17.630,0:17:23.049 prescribed by the prover, so maybe the skeptic picks 0.97 which is

0:17:23.049,0:17:26.380 within 0.05 distance of 1.

0:17:26.380,0:17:31.570 And then they check that f(x) is 1.94, that is at the distance of 0.06

0:17:31.570,0:17:38.570 from 2. So, it's within 0.1 of the claimed limit. Who won the game?

0:17:38.780,0:17:42.650 If the thing is within the interval then who wins?

0:17:42.650,0:17:43.320 KM: The prover.

0:17:43.320,0:17:46.720 Vipul: The prover wins, right? So, the prover won again so therefore

0:17:46.720,0:17:52.100 this limit statement is true, right? So, what's wrong with this as a

0:17:52.100,0:17:57.370 proof that the limit statement is true? How is this not a proof that

0:17:57.370,0:18:03.870 the limit statement is true? This what I've written here, why is that

0:18:03.870,0:18:05.990 not a proof that the limit statement is true?

0:18:05.990,0:18:11.960 KM: Because it's only an example for the specific choice of epsilon and x.

0:18:11.960,0:18:16.200 Vipul: Yes, exactly. So, it's like a single play of the game, the

0:18:16.200,0:18:20.470 prover wins, but the limit statement doesn't just say that the prover

0:18:20.470,0:18:24.380 wins the game, it says the prover has a winning strategy. It says that

0:18:24.380,0:18:27.660 the prover can win the game regardless of how the skeptic plays;

0:18:27.660,0:18:31.070 there's a way for the prover to do that. This just gives one example

0:18:31.070,0:18:34.640 where the prover won the game, but it doesn't tell us that regardless

0:18:34.640,0:18:37.280 of the epsilon the skeptic takes the prover can pick a delta such that

0:18:37.280,0:18:41.090 regardless of the x the skeptic picks, the function is within the

0:18:41.090,0:18:45.530 thing. So that's what they should do. Okay?

0:18:45.530,0:18:51.160 Now you notice -- I'm sure you notice this but the way the game and the

0:18:51.160,0:18:58.160 limit definition. The way the limit definition goes, you see that all

0:18:59.870,0:19:04.260 the moves of the skeptic be right "for every" "for all." Right? And

0:19:04.260,0:19:07.390 for all the moves of the prover it's "there exists." Why do we do

0:19:07.390,0:19:11.140 that? Because we are trying to get a winning strategy for the prover,

0:19:11.140,0:19:14.309 so the prover controls his own moves. Okay?

0:19:14.309,0:19:15.250 KM: Exactly.

0:19:15.250,0:19:18.630 Vipul: So, therefore wherever it's a prover move it will be a there

0:19:18.630,0:19:22.240 exists. Where there is a skeptic's move the prover has to be prepared

0:19:22.240,0:19:29.240 for anything the skeptic does. All those moves are "for every."

0:19:30.559,0:19:33.850 One last one. By the way, this one was called, "You say you want a

0:19:33.850,0:19:36.870 replay?" Which is basically they're just saying that just one play is

0:19:36.870,0:19:40.890 not good enough. If the statement is actually true, the prover should

0:19:40.890,0:19:45.370 be willing to accept the skeptic ones, the reply and say they want to

0:19:45.370,0:19:47.679 play it again, the prover should say "sure" and "I'm going to win

0:19:47.679,0:19:53.320 again." That's what it would mean for the limit statement to be true.

0:19:53.320,0:20:00.320 One last one. Just kind of pretty similar to the one we just saw. Just

0:20:16.690,0:20:23.690 a little different.

0:20:39.020,0:20:46.020 Okay, this one, let's see. We are saying that the limit as x

0:20:50.450,0:20:56.900 approaches zero of sin(1/x) is zero, right? Let's see how we prove

0:20:56.900,0:21:01.409 this. If the statement true ... well, do you think the statement is

0:21:01.409,0:21:08.409 true? As x approach to zero, is sin 1 over x approaching zero? So

0:21:13.980,0:21:20.980 here's the picture of sin(1/x). y-axis. It's an oscillatory function

0:21:22.010,0:21:27.870 and it has this kind of picture. Does it doesn't go to zero as x

0:21:27.870,0:21:29.270 approaches zero?

0:21:29.270,0:21:30.669 KM: No.

0:21:30.669,0:21:35.539 Vipul: No. So, you said that this statement is false, but I'm going to

0:21:35.539,0:21:38.700 try to show it's true. Here's how I do that. Let's say the skeptic

0:21:38.700,0:21:44.510 picks epsilon as two, okay? And then the prover ... so, the epsilon is

0:21:44.510,0:21:48.520 two so that's the interval of width two about the game limit zero. The

0:21:48.520,0:21:55.150 prover picks delta as 1/pi. Whatever x the skeptic picks, okay?

0:21:55.150,0:22:02.150 Regardless of the x that the skeptic picks, the function is trapped within epsilon of the game limit. Is that

0:22:10.340,0:22:16.900 true? Yes, because sin (1/x) is between minus 1 and 1, right? Therefore

0:22:16.900,0:22:20.100 since the skeptic picked an epsilon of 2, the function value

0:22:20.100,0:22:24.030 is completely trapped in the interval from -1 to 1, so therefore the

0:22:24.030,0:22:27.919 prover managed to trap it within distance of 2 of the claimed limit zero.

0:22:27.919,0:22:30.970 Okay? Regardless of what the skeptic does, right? It's not just saying

0:22:30.970,0:22:34.370 that the prover won the game once, it's saying whatever x the skeptic

0:22:34.370,0:22:40.740 picks the prover can still win the game. Right? Regardless if the

0:22:40.740,0:22:43.780 x is skeptic picks, the prover picked a delta such that the function

0:22:43.780,0:22:48.100 is trapped. It's completely trapped, okay? It's not an issue

0:22:48.100,0:22:51.130 of whether the skeptic picks the stupid x. Do you think that this

0:22:51.130,0:22:52.130 proves the statement?

0:22:52.130,0:22:59.130 KM: No, I mean in this case it still depended on the epsilon that the

0:23:01.030,0:23:01.820 skeptic chose.

0:23:01.820,0:23:04.980 Vipul: It's still dependent on the epsilon that the skeptic chose? So,

0:23:04.980,0:23:05.679 yes, that's exactly the problem.

0:23:05.679,0:23:09.370 So, we proved that the statement -- we prove that from this part onward

0:23:09.370,0:23:12.500 but it still, we didn't prove it for all epsilon, we only prove for

0:23:12.500,0:23:16.309 epsilon is 2, and 2 is a very big number, right? Because the

0:23:16.309,0:23:19.970 oscillation is all happening between minus 1 and 1, and if in fact the

0:23:19.970,0:23:26.970 skeptic had pick epsilon as 1 or something smaller than 1 then the two

0:23:27.030,0:23:32.169 epsilon strip width would not cover the entire -1, +1

0:23:32.169,0:23:35.490 interval, and then whatever the prover did the skeptic could actually

0:23:35.490,0:23:39.530 pick an x and show that it's not trapped. So, in fact the reason why

0:23:39.530,0:23:43.110 the prover could win the game from this point onward is that the

0:23:43.110,0:23:45.900 skeptic made of stupid choice of epsilon. Okay?

0:23:45.900,0:23:52.289 In all these situation, all these misconceptions, the main problem is,

0:23:52.289,0:23:58.919 that we're not ... keeping in mind the order which the moves I made

0:23:58.919,0:24:04.179 and how much information each claim has at the stage where that move

0:24:04.179,0:24:04.789

is being made.

Conceptual definition and various cases

Formulation of conceptual definition

Below is the conceptual definition of limit. Suppose $f$ is a function defined in a neighborhood of the point $c$ , except possibly at the point $c$ itself. We say that:

$lim_{x \to c} f (x) = L$

if:

For every choice of neighborhood of $L$ (where the term neighborhood is suitably defined)
there exists a choice of neighborhood of $c$ (where the term neighborhood is suitably defined) such that
for all $x \neq c$ that are in the chosen neighborhood of $c$
$f (x)$ is in the chosen neighborhood of $L$ .

Functions of one variable case

The following definitions of neighborhood are good enough to define limits.

For points in the interior of the domain, for functions of one variable: We can take an open interval centered at the point. For a point $c$ , such an open interval is of the form $(c - t, c + t), t > 0$ . Note that if we exclude the point $c$ itself, we get $(c - t, c) \cup (c, c + t)$ .
For the point $+ \infty$ , for functions of one variable: We take intervals of the form $(a, \infty)$ , where $a \in R$ .
For the point $- \infty$ , for functions of one variable: We can take interval of the form $(- \infty, a)$ , where $a \in R$ .

We can now list the nine cases of limits, combining finite and infinite possibilities:

Case	Definition
$lim_{x \to c} f (x) = L$	For every $ε > 0$ , there exists $δ > 0$ such that for all $x$ satisfying $0 < \| x - c \| < δ$ (i.e., $x \in (c - δ, c) \cup (c, c + δ)$ ), we have $\| f (x) - L \| < ε$ (i.e., $f (x) \in (L - ε, L + ε)$ ).
$lim_{x \to c} f (x) = - \infty$	For every $a \in R$ , there exists $δ > 0$ such that for all $x$ satisfying $0 < \| x - c \| < δ$ (i.e., $x \in (c - δ, c) \cup (c, c + δ)$ ), we have $f (x) < a$ (i.e., $f (x) \in (- \infty, a)$ ).
$lim_{x \to c} f (x) = \infty$	For every $a \in R$ , there exists $δ > 0$ such that for all $x$ satisfying $0 < \| x - c \| < δ$ (i.e., $x \in (c - δ, c) \cup (c, c + δ)$ ), we have $f (x) > a$ (i.e., $f (x) \in (a, \infty)$ ).
$lim_{x \to - \infty} f (x) = L$	For every $ε > 0$ , there exists $a \in R$ such that for all $x$ satisfying $x < a$ (i.e., $x \in (- \infty, a)$ ), we have $\| f (x) - L \| < ε$ (i.e., $f (x) \in (L - ε, L + ε)$ ).
$lim_{x \to - \infty} f (x) = - \infty$	For every $b \in R$ , there exists $a \in R$ such that for all $x$ satisfying $x < a$ (i.e., $x \in (- \infty, a)$ ), we have $f (x) < b$ (i.e., $f (x) \in (- \infty, b)$ ).
$lim_{x \to - \infty} f (x) = \infty$	For every $b \in R$ , there exists $a \in R$ such that for all $x$ satisfying $x < a$ (i.e., $x \in (- \infty, a)$ ), we have $f (x) > b$ (i.e., $f (x) \in (b, \infty)$ ).
$lim_{x \to \infty} f (x) = L$	For every $ε > 0$ , there exists $a \in R$ such that for all $x$ satisfying $x > a$ (i.e., $x \in (a, \infty)$ ), we have $\| f (x) - L \| < ε$ (i.e., $f (x) \in (L - ε, L + ε)$ ).
$lim_{x \to \infty} f (x) = - \infty$	For every $b \in R$ , there exists $a \in R$ such that for all $x$ satisfying $x > a$ (i.e., $x \in (a, \infty)$ ), we have $f (x) < b$ (i.e., $f (x) \in (- \infty, b)$ ).
$lim_{x \to \infty} f (x) = \infty$	For every $b \in R$ , there exists $a \in R$ such that for all $x$ satisfying $x > a$ (i.e., $x \in (a, \infty)$ ), we have $f (x) > b$ (i.e., $f (x) \in (b, \infty)$ ).

Limit of sequence versus real-sense limit

Fill this in later

Real-valued functions of multiple variables case

We consider the multiple input variables as a vector input variable, as the definition is easier to frame from this perspective.

The correct notion of neighborhood is as follows: for a point $\bar{c}$ , we define the neighborhood parametrized by a positive real number $r$ as the open ball of radius $r$ centered at $\bar{c}$ , i.e., the set of all points $\bar{x}$ such that the distance from $\bar{x}$ to $\bar{c}$ is less than $r$ . This distance is the same as the norm of the difference vector $\bar{x} - \bar{c}$ . The norm is sometimes denoted $| \bar{x} - \bar{c} |$ . This open ball is sometimes denoted $B_{r} (\bar{c})$ .

Suppose $f$ is a real-valued (i.e., scalar) function of a vector variable $\bar{x}$ . Suppose $\bar{c}$ is a point such that $f$ is defined "around" $\bar{c}$ , except possibly at $\bar{c}$ . In other words, there is an open ball centered at $\bar{c}$ such that $f$ is defined everywhere on that open ball, except possibly at $\bar{c}$ .

With these preliminaries out of the way, we can define the notion of limit. We say that:

$lim_{\bar{x} \to \bar{c}} f (\bar{x}) = L$

if the following holds:

For every $ε > 0$
there exists $δ > 0$ such that
for all $\bar{x}$ satisfying $0 < | \bar{x} - \bar{c} | < δ$ (i.e., $\bar{x}$ is in a ball of radius $δ$ centered at $\bar{c}$ but not the point $\bar{c}$ itself -- note that the $| \cdot |$ notation is for the norm, or length, of a vector)
we have $| f (\bar{x}) - L | < ε$ . Note that $f (\bar{x})$ and $L$ are both scalars, so the $| \cdot |$ here is the usual absolute value function.