 Hi, I'm Zor. Welcome to Unizor Education. Today we will talk about continuity of the function. First of all, we are talking about real functions, which means their argument is real and their value are real. And this is basically within the framework of function limits, which is in turn part of the course of advanced mathematics for teenagers and high school students presented on Unizor.com. I suggest you to watch this lecture from this particular website because it has explanation and you can read it in the textbook. So, talking about continuity. Well, first of all, let's think about what is your intuitive understanding of what a continuous function is supposed to be. Well, it looks like the graph of such a function should be smooth, right? Which means you just continually drawing the graph of this function without lifting your pen. Well, it can be something like this. This is also continuous, but this is not. Because you have to lift the pen, put it in another point and then continue graphing the function. So, our intuitive understanding of continuity is basically drawing the graph of the function without lifting your pen from the paper or marker from the whiteboard. Alright, so let's try to explain this in more mathematical terms. First of all, let's talk about the main of these functions. Well, if I want to draw the graph without lifting my pen or marker, whatever, it means that it's supposed to be defined on a contiguous area, contiguous interval on the x-axis, right? Otherwise, I will have to lift my, if it's defined here and then here, then I have to draw the graph here, then lift my pen and start drawing there. So, if I don't lift my pen, it means my area where the function is defined, the domain of the function is a contiguous area. It can be finite, which means like from here to here, or it can be infinite from here to plus infinity or from here to minus infinity or from minus infinity to plus infinity. So, finite or infinite domain is the first requirement of this function. Alright, now, we finally have agreed that the domain is supposed to be a contiguous area on the x-axis. Now, how can I express mathematically, more rigorously, that the graph of the function is supposed to be drawn without lifting my pen? Well, as I approach any point on my graph, the next point is supposed to be exactly where my old points start, right? Which means basically that if I take two points, x1 and x2, and these are correspondingly f at x1 and f at x2. If my x1 and x2 are coming closer and closer, then my functions, my these points also supposed to come closer and closer. So, as I'm decreasing the distance between these, the distance between these two also is supposed to be getting closer and closer. So, if my x1 minus x2 goes to zero, which is an infinitesimal value, from this should follow that my difference between these two also is supposed to be infinitesimal. And what's also important, it should be actually for any position of my two points, x1 and x2, which I'm going to express slightly differently, but which means exactly the same thing. I will do the following. I'm saying that for any point r, which is a real number, which belongs to domain, which means that the function is defined, I can take x which tends to this r in any way, and it might be expressed as this. So, that's what I'm going to do. So, I'm fixing one point, but basically it can be any point. So, as soon as I fix this point r, then I can say that no matter how my x within the domain is approaching r, my function of x should approach function of r. So, first of all, it's defined at r because r belongs to domain, and the value is exactly the limit of the value of f at x. I will give you a couple of examples when it's not true, but my continuous function should actually have this particular property. And finally, let me express it in epsilon delta language, which is kind of customary for all these function limit kind of things. What does it mean? Well, it basically means that for any r, which belongs to domain, and for any epsilon, which is a positive number, however small, which basically dictates how close to each other will be the value of the function of point x and the value at point r. There exists such delta so that if my x minus r less than or equal to delta, then my f at x minus f at r by absolute value would be less than or equal to epsilon. So, that's what basically this mean on epsilon delta language. So, for any fixed r from domain and for any epsilon neighborhood around this, for any fixed r and for any value of the function, my epsilon neighborhood would be here from this function f at r up and down. There is always a delta neighborhood of r, so whenever my x within this delta neighborhood, my function values would be within epsilon. So, that's what actually mean in mathematical terms that you can draw the graph without lifting your hand. Quite frankly, this is nice and looking very impressively mathematically. I kind of prefer this type of notation. From this follows this, where r is any. And this is just basically an explanation what this arrow means. Epsilon delta, that's what it is. So, this is basically the definition of continuous function. Now, the function which usually students deal in school and universities maybe, they are mostly continuous. There are some examples. I will give you a couple of examples which is not the case. But generally speaking, what people are dealing with in mathematics and in mathematical applications, usually the functions are continuous. But anyway, obviously there are some exceptions. And what I'm going to do right now is just I'll give you a couple of examples of continuous and non-continuous functions. So, you basically have a feeling of how to prove the continuity of the function. My first example is function f at x equals to x cube. Well, the graph of this function, as you know, is something like this. Obviously continuous, right? So, how can I prove that this is a continuous function? Well, let's just use my epsilon delta definition. Choose any point r and any degree of closeness epsilon positive, which I would like to have between the value of the function at point r and the value of the function at point x. So, that's supposed to be less than or equal to epsilon. I have to find delta such that if my x minus r less than delta, then this would follow. That's what I have to do. I have to find delta. So, based on r and epsilon, I have to find delta. So, that this is from this follows this. Okay, let's just find it. I mean, how can I prove that this is continuous function? Well, I choose any r and any epsilon and they'll just come up with exact value of delta, which basically cause from this inequality to be this one to be true. Alright, so, x minus f of r is what? It's x cubed minus r cubed, right? By absolute value. Now, x cubed minus r cubed is basically, we all know this. Again, if you don't believe, just multiply it. But we did it many times. Now, and that means absolute value of the product is the product of absolute value less than or equal to. Now, this is sum of certain numbers. Now, some of them may be positive. Now, this is positive. This is positive. But if I will replace it with individual absolute value, I will only increase, right? The sum. This is greater or equal to this. If r and x are positive, then there is an equality. If r and x are negative, it's also equality because this is positive and this is the same as this one. But if r and x are of different signs, then this is definitely greater than this. So, that's true. Now, here is what I'm suggesting to do. Let's choose delta 1, something close to r, let's say equals to r. Why do they choose it? Because I would like this thing to be bounded from above, not exceeding certain value. Because this is obviously going to zero, right? So, I would like to exploit this, but I would like this to be relatively bounded. Because if this is an infinitesimal and this is bounded value, then their multiplication will be infinitesimal, right? So, if I choose, let's say, delta 1 equals to r, what does it mean? From this, now, this is what does it mean? It means that x is in delta neighborhood of r. So, if this is r and this is r minus delta and r plus delta. Now, if my delta 1 is equal to r, so it's from r minus r to r plus r. So, I definitely know that I am less than 2r by absolute value, at least, right? So, my x by absolute value would be less than or equal to 2r from this, right? If my x minus r less than delta 1, then I know this. So, let's choose it this way. Then I know that this is x minus delta times, now, x square would be 4r square plus 2r square plus 1r square. So, it's 7r square. I don't really need absolute value because this is square. Now, what I will do is I will choose delta 2, which is equal to epsilon divided by 7r square. Now, then there are multiplication if my x minus r is less than delta 2, which is equal to this. Then I definitely have this would be less than epsilon, right? That's what I need. Now, what I am going to do, I will choose the minimum between delta 1 and delta 2. Then I know that both actually will be true. This will be true and this will be true. And that's why it's less than epsilon. So, I explicitly found it's a minimum between r and epsilon divided by 7 divided by r square. This is exactly the delta which I am looking for. So, for any r and epsilon I found exactly the delta which is needed. Delta is minimum of r comma epsilon divided by 7r square. That's it. So, I found and from this follows this, which is this. Okay. So, I have proven that x cube is a continuous function. Let's talk about something else. Let's have another function, another example. Sin of x. Okay. Again, we know that the graph looks like this, right? Which is a continuous function. Now, we would like to prove it. Okay. So, let's first analyze what do we have to prove? Well, again, we choose r and choose epsilon. Now, what we have to do? We have to find such delta that this would be less than or equal to epsilon if x minus r delta. Okay. So, what is my delta? Let's just think about it. Well, my analysis actually is to transform this into something more palatable. Now, the difference between two signs. Well, if you remember trigonometry what we can do in this particular case is the following. We can have x is equal to x plus r divided by 2 and x minus r divided by 2. r is equal to x plus r divided by 2 minus r divided by 2, right? x would cancel and r would be r, r2 and r2. This is x2 and x, yes. And now, I will transform sin of x as sin of sum of these and sin of r as sin of difference between two these. And you will see that the final formula will be, well, let me just, sin of sum is sin cosine, cosine sin. So, sin x plus r cosine x minus r plus cosine sin. Cosine x plus r2 sin x minus r2. Now, that's my sin of x. Now, minus sin of r, sin of this one, which is sin x plus r cosine x minus r minus sin x minus. Sorry, sin cosine cosine x plus r sin x minus r. Now, why did they do all this? Because of this. And this is plus, by the way, because minus and minus. We have to subtract this minus and this minus would be plus. So, the answer is 2 cosine x plus r2 and sin x minus r2. Okay. Now, why is it better? For a very simple reason. Because now, I can say that this thing is less than or equal to absolute value of this is product of absolute value. Cosine is always from minus one to one. So, the cosine has absolute value of maximum absolute value of one. So, I can put two and then sin x minus r2. Right? So, instead of cosine, I took it maximum of its absolute value, which is one. Now, let's talk about this one. Sin phi less than phi. How about this for inequality? And I think you should remember it from definition of the sin. Remember, with unit circle, if this is my angle phi in regions, then the length of this is also phi. Because the circle has reduced r. Right? 2 pi r is the length of an entire circle. r is equal to 1. So, it's the length is 2 pi. And the angle phi obviously is and the total angle is 2 pi. Right? So, it's the length of 2 pi and the total angle is 2 pi. So, if the angle is phi, then the length is phi in regions. And what is sin? Sin is this one. This is sin. Now, the length of the arc is greater than the perpendicular. Because the perpendicular is the shortest distance from the point to the line. And this is not the perpendicular, so it's always greater. than sin phi. So, I can say in this particular case that this is less than 2. And instead of angle, I will put sin of angle, put angle itself. x minus r divided by 2, which is x minus r. So, as a result, what I can say is the following. I will just choose delta equals to epsilon. And since I have this inequality, this is less than delta. Therefore, this would be less than epsilon. That's how we have proven this. And final example which I would like to present is a very simple example of a function which does not have this property of continuity. So, I have to express graphically, I have to lift my pen. So, let me just show you how I can define this function. So, let's have a function which is always equal to 0, except at one point. Here, it has the value of 1. So, f at x is equal to 0 if x is not equal to 0. And f at x is equal to 1 if x is equal to 0. So, my function is as follows. I have a graph here, but I do not go into this point 0. I have to jump, put a point here. And then I start again and put it. So, I put little errors here. And that's how usually we know that there is no point in between and the point is somewhere else. So, this is a typical example of the non-continuous function. How can I prove it? Well, very simply. Remember, we have two major requirements. Number one, domain should be contiguous. Now, in this case domain is all the real values from minus infinity to plus infinity. So, it is contiguous. The first requirement is satisfied. The second requirement is if x goes to r, therefore f at x goes to f at r. Now, let's take r is equal to 0. So, x goes to 0 as x goes to 0. My function at x should go to f of 0 which is equal to 1, right? But f at x as x getting closer and closer to point 0 is always equal to 0, you see? So, it is 0, 0, 0, 0 and the limit of 0 is 0. So, this is actually it goes to 0, not to 1. So, that's basically the proof. The second requirement of continuity is not satisfied. We have a sequence which goes to a point 0 where the limit is not equal to the value of the function at this point. By the way, if you will take, for instance, any other point, that would be a point of continuity. So, I have only one point where the function is not continuous. Just as a, well, mathematical joke, if you wish. I would like to find the function which is not continuous at any point. Can that be done? Well, apparently, yes. Consider function which is equal to 0 if x is rational and 1 if x is irrational. So, at any point, wherever you are, if you go by rational points, if you approach to this point, then you will have 0s and the limit will be 0. If you approach by irrational points to the same point, you will have 1, 1, 1, 1, 1 and the limit will be 1. So, basically, there is no limit. Whenever you approach this point and any point, there is actually no limit. So, we are not talking about function tending to any value at all. Which means, at every point, wherever you point, with every point, it is a point of non-continuous, not continuity. Non-continuous, whatever. It is a non-continuous point. All right, so, I suggested to read the notes for this lecture. They are basically expressing the same ideas and the same approach I did in the lecture. They are unizord.com. And, well, that's it. Thank you very much and good luck.