 Hi, I'm Zor. Welcome to Unizor Education. Well, it's time to talk about properties of the function limits. Well, these properties are very much analogous to properties of limits of the sequences, and I will definitely rely on that material. But at the same time, we have two different definitions of the function limits, one of them through sequences and another through epsilon-delta language. So I will try to prove whatever the properties we will discuss using both ways. Okay, so limits of the functions have certain properties. This is a lecture which is part of the advanced course for high school students and teenagers of mathematics presented on Unizor.com. I suggest you to watch this lecture from the website because the website has very detailed notes for each lecture. Plus, you have the ability to take exams, which is like a very good self-testing kind of thing. The site is free, so properties of the limits. As I said, properties of the function limits are very much analogous to properties of the sequence limits. And the first one is if my function converges to some number a as x converges to some number r, then function which is a times f of x converges to a times a as x converges to r. Now, let's consider that a is positive. If a is negative, it's exactly the same, so it's just easier to consider a as a positive thing. Now, I will explain it the first time in a little bit more details about how to prove this using the sequences. But for all other properties, it's exactly the same, so I might actually be much more brief in this case. So, the first definition of the function limit deals with any sequence of arguments which is converging to r should result in sequence of the function of these arguments converging to a. That's what it means from the definition number one of the function limit. This is supposed to be for any sequence of arguments which is converging to r. Okay, fine. So, what do we have to prove using this definition if we want to prove this? Well, we want to prove that if you take any sequence of arguments which is converging to r, then sequence of a times f of x n converges to a times a. Now, since we know that for any sequence of arguments converging to r, if you have a sequence converging to a from the properties which have been proven for the sequence limits, if you have a sequence which is converging to some number, a times this sequence converging to a times that number. So, that's basically, this theorem is basically direct consequence from the corresponding theorem for limit of the sequences. And the property is correspondingly related to the properties of the sequence limits. Okay, so that's according to my definition of the limit of the function through sequences. I spend a little bit more time maybe this first time because all other will be exactly the same for all other properties. I'll just refer to the corresponding property of the sequences. But how about the second definition? That's more important and I think it's just more interesting as a nice logical exercise. So, what if I would like to prove this using epsilon delta definition of the limit? So, let's remind that this thing that the function has limit a as x converges to r means that for any epsilon greater than 0 exists such delta that as long as my x is within delta neighborhood of its limit point from this immediately follows that f at x would be within epsilon neighborhood of limit a. That's what's given. Now, what do I have to prove in this case? I have to prove that for any epsilon greater than 0 exists delta such that as long as my x is within neighborhood within delta neighborhood of its limit point r my function would be within epsilon neighborhood from a times a. This is needed to be proven. So, this is given and this is to be proven. Okay, so for any epsilon. Alright, so let's just take any epsilon first. Take any epsilon. Now, how can I find this delta? Well, in this case it's really very very simple because if I will choose delta equals to epsilon divided by a we have agreed that a is positive, right? So, let's take this. Take any epsilon and then, or I'm sorry, it's not delta, it's just epsilon prime. Now, for this epsilon prime number two, we find delta such that using this one. So, instead of epsilon, we will use epsilon prime. So, for this epsilon prime there is a delta such that this implies this epsilon prime. Now, what is epsilon prime? Again, this is epsilon divided by a, right? So, it's f of x minus a less than epsilon divided by a. So, now we have proven this. Well, multiplied by a, a is positive, so it will be here, a, a and not this one. And we have proven this particular inequality which we need, right? So, again, how did we prove this thing? Well, we actually physically found that delta using what? First, from our epsilon, which is completely anything whatever we want, we have to prove it for any epsilon. We have derived this new epsilon prime and then using this instead of epsilon, epsilon prime. We know basically that this is given. And that's how we got this and from this we got this. That's it, very simple. Next. Next property is, so the first one is function times constant converges to its limit times this constant. Now, the second property is limit of sum of two functions, each of them converging to something as x converging to r. Then I am stating that sum is converging to sum of the limits. And I will prove exactly the same way I actually built for any epsilon I will build delta. All right, by the way, as far as definition using the sequences, again, it's obvious because this means if I would choose any sequence of x, x1, x2, xn, etc., which is converging to r, then this would be right. So for any sequence of x, xn, which is converging to r, this would be this, which is a sum of two converging sequences. And we have already proven the theorem for sequences. So the sequence of the sum converges to the limit, which is sum of limits of each one of them. So with sequences, it's very simple. Now with epsilon delta, that's a little bit different, but as simple actually. Let's just think about it. Let's just fix any delta, any epsilon first. What we have to prove is, well, first what's given. What's given is that exists such delta that as long as x is within delta neighborhood of r, my f at x would be within delta neighborhood of a, epsilon neighborhood of a. Let's put it delta 1. Why? Because for the same epsilon, we have delta 2 for g. So as long as we know some kind of an epsilon, using the convergence of x, of f at x to a, we have found delta 1. As long as x within delta 1 neighborhood of r, this is true. And then we found, again, using the convergence of the g, we found delta 2. As long as x within delta 2 neighborhood of r, this is correct. Well, how about if I will take delta equal to minimum of delta 1 and delta 2. So if this means that there is such a delta 1 neighborhood, this is delta 2 neighborhood. And what is delta neighborhood? Well, if this is r, now this might be my delta 1 neighborhood, r plus delta 1. And this is r minus delta 1. Now, this might be r minus delta 2 and r plus delta 2. But if delta is minimum of them, then it's this interval, which is smaller than any of these. And since it's smaller, then both conditions are true. So basically, we have found such delta for any epsilon when both conditions are true. Well, if both conditions are true, then obviously we can write the following. f at x plus g at x minus a plus b, right? We have to prove that for any epsilon, for any epsilon, there is such a delta that this is less than of epsilon if my x minus r within delta neighborhood. How can I find this delta? Okay, here is a very simple solution. I will use an axiom, which everybody knows, right? So absolute value of sound is less than equal than absolute value of... Basically, if a and b both are positive, then this is equality. If a and b are negative, this is also equality. But if they are a different sign, like minus 2 and plus 5, now this is 3, but this is 2 and this is 5, it would be greater. So in this case, I will do the following. I will do f at x minus a plus g of x minus b, right? Now, I can make for any delta, for any epsilon, I can find such delta that both of them are true, right? So let me just choose, for this epsilon, I will choose epsilon divided by 2 and find such a delta that will be delta 3 already, right? That if x minus r is less than delta 3 by absolute value, then this is true. And for the same epsilon 2, I will find delta 4 such that if x is within delta 4 neighborhood, this is true, that this is less than epsilon divided by 2. And then I will take, again, the minimum of delta 3 and delta 4 and that would be my final delta when both of them will be true. So this is less than epsilon over 2 and this is less than epsilon over 2, which is epsilon. So for any epsilon, I will find such a delta when this is true. So first, from epsilon, I go to epsilon divided by 2. From epsilon divided by 2, I find this one, this delta 3, so that this is less than epsilon divided by 2 if x is in delta 3 neighborhood. Then for the same epsilon divided by 2, I find delta 4 neighborhood of r as long as x is in there, this is true. And that's where the whole thing is true. So I can always use all these manipulations to find for any epsilon, I can find delta. The third property is product, as you probably have guessed already. Product is just a little bit trickier. So this is given as x converges to r. What I have to prove is this. Again, speaking about definition of the product in terms of sequences, it has already been proven for sequences. So if I take any sequence of arguments which converges to r, any, then the corresponding sequence of function f is converging to a, corresponding sequence of g of xn converges to b, and the product of two converging sequences has a limit which is a product of two limits. So with sequences, it's done. Now, epsilon delta is more interesting, I would say. Well, let's do some analysis. f of xg of x minus ab. That's what we have to somehow prove that this is a very, very small, any kind of degree of smallness, infinitesimal actually, if x is converging to r. So we have to prove that this is infinitesimal as x converging to r. How can I prove it? Here's a little trick. I subtract it and add it a times gx. Well, how did I come up with this? Difficult to say because probably some very, very long time ago somebody showed it to me and since I remember this trick. But anyway, it is a little trick and why did I do it? Here is why. Because now I can separate f and g. Here is how. Remember this? So this is my a and this is my b. So it's less than or equal to f of x times g of x minus a times g of x, which is this. Right? I took g of x factor out. Plus, and here I will factor out absolute value of a. All right? Now, why did I do it? Because here this is really infinitesimal and this is infinitesimal as x goes to r. This is given. Now, this is the constant. So infinitesimal times constant is infinitesimal, right? And the only thing which is kind of strange thing is I don't know what to do with this one. Right? Well, here is what I can suggest. G goes to b. Right? So let's just assume that b is not equal to zero. Because if b is equal to zero, it's slightly different but very simple as well. Now, if b is not equal to zero, then g, since g is converging to b, eventually it will be, let's say, in this interval. Right? Well, not exactly. It would be b plus b over two and b two, which is b minus b over two. Right? That's what I can definitely say. So if g is converging to b, then it will be greater than b minus b two and less than b plus b two. Eventually. So it will be in the b two neighborhood. Why? Well, that's quite obvious since I know that for any epsilon greater than zero exists such delta that as long as x minus r is less than or equal to delta, it immediately follows that g of x minus b would be less than epsilon. Right? So choose epsilon equal to b over two. It's for any epsilon. Right? So I choose this one. And now I have this. This is equivalent of this. Now, why did I do it? Well, primarily for this particular reason. Since I know that g of x right now is some kind of a value in between these two as long as my delta is sufficiently small. I can actually say that with sufficiently small, with x sufficiently close to r, with x sufficiently close to r, this is less than or equal to, let's say, f of x minus a times b plus b two. I just replace, which is 3b over two. Now, instead of b over two, I can choose anything actually here. I can choose one, for instance. For epsilon equal to one, I can always find sufficiently close x to r that it would be between b minus one and b plus one. Now, if I choose, instead of this b one, b plus one, it will be b plus one. So any kind of a b plus something would work here. Right? And the sign is less than or equal plus a g of x minus b. Right? Now, I have constant here and constant here. And knowing this, now this is analysis. Now, how can I actually find out exactly whatever I need, which is for any epsilon, find delta such and such that this is less than delta? Well, very simple. If I choose any epsilon, then we will find a different epsilon one, let's say equals to epsilon divided by two b plus one. Something like this. That would be even better. Right? Now, whenever my x is sufficiently close to r, I can find such delta one that f of x minus a would be less than this. But since we are multiplying by b plus one, as a result, I will have less than epsilon over two. Right? So there is such a delta one when this is less than this. On the other hand, there is for the same, let's say let's take epsilon two equals to epsilon divided by absolute value of a. I will find delta two such that if x is sufficiently close, closer than delta two to b to r, then this would be smaller than this. And multiply by a. Yeah, I should put two here as well. And also epsilon over two. So this will be epsilon over two less than or equal. And this will be less than epsilon over two. So to make both of them satisfied, they just have to take minimum between delta one and delta two. When both of them are chosen. Again, for any epsilon, I choose epsilon one and find from convergence of f of x to a, find delta one when this is smaller than this. Then I choose using convergence of g of x to b, I choose for epsilon two equals to this one delta two such that this is true. And then minimum between delta one and delta two is the delta which I need because then this is less than epsilon over two. And this is epsilon over two. So the sum would be equal to epsilon whatever I need. So what's difficult about this? Just this trick. Everything else is just trivial inequalities. And the last property. Now the last property is, now we have multiplied, now we have to divide, right? Now I'm not going to divide two functions. I will just have function this. So if f of x goes to a, I need one over f of x goes to one over a. Obviously a not equal to zero as x converges to r. Now, again, speaking about sequences, it's obvious it just follows from the property for the sequence limit. Now talking about epsilon delta language, so what we have to find out is this thing an infinitesimal if x converges to r. But let's think about this is equal to, right? Well a minus f of x or f of x minus a, the same thing because it's an absolute value. If I will prove that this is just some kind of a limited value bounded on both sides, then I'm fine. As long as this is not, because this is infinitesimal, right? As x goes to r. So all we need is to prove that this is not infinitesimal as well. Because if it's bounded, if it's far from zero, then infinitesimal divided by this would be infinitesimal. How can I prove it? Well, very simply, again, let's have epsilon equals to a over two. Now, I find, obviously I can find delta one such that if x is closer than delta one, f at x minus a is closer than n a over two. Now, what is this? It means that a minus a over two, f at x, a plus a over two, which is a over two, and this is three a over two, right? So that's how we bound it. As long as f at x converges to something which is not equal to zero, eventually it will be close enough to separate it from zero, right? So if this is zero and this is a, obviously if it converges to a, after a certain point when argument is significantly close to r, I will be separated from zero, a over two and three a over two. Or I can choose a over four, it doesn't really matter. As long as this is not equal to zero, I'm far from zero. Now, what does it mean? It means that for this particular delta one, as long as my x is within delta one neighborhood, this thing, how can I increase the fraction? I will decrease the denominator. So I will take the smaller one, right? I will take the left, left border, the smaller one. A over two, within delta one neighborhood. A over two is smaller than epsilon square over f at x. Now, now I have to prove that this is infinitesimal, but this is the constant, right? So it's quite simply to prove that this is infinitesimal, because for any epsilon I can find delta such that this is, I will choose epsilon, any epsilon, right? Then I will choose epsilon prime equals to epsilon divided by a square over two, right? No, the other way around. Multiply it. Now, for this epsilon, epsilon prime, I can find such delta that this thing is less than this. And then if I divide it by this, I will have, so if this is less than epsilon, epsilon, then f at x minus a divided by a is less than epsilon. So I need this delta. Now, combined with the previous, now this is not just the first delta which we were using. Before that, we have found delta such that f at x minus a was basically, that f at x was close enough to a, so it's separated from zero, right? So this is separated from zero. So now I have one delta and that delta, so I'll probably take the smallest, the minimum of these two deltas, and then I will get the minimum delta is the one where both. This is true and this is true. And that's why the whole thing would be less than epsilon, whatever I'm supposed to prove. So in this and all other cases, it's a little bit maybe tricky to find the proper delta. You have to really separate, separately satisfy a few different conditions. But if you want to satisfy different conditions and different conditions require different deltas, you just take the minimum of these deltas. So whenever x is closer and closer and closer than any other requirement, then all the requirements will be satisfied. My f at x would be greater than a over two. And in the previous cases, we have similar conditions, etc. Now, a little bit more detailed notes are on unizord.com for this particular lecture. And it's a little bit more like a textbook. I'm not taking any shortcuts. I'm really proving something. Okay, this is delta one. This is delta two. This is delta three. And we're taking minimum of all of them, etc. I don't want to go to all these scruples details during the lecture because my purpose was for you to understand really how this thing is supposed to be proven. So it's divided into smaller pieces. Each small has its own requirement for the delta. And then we take the minimum of these delta to satisfy all the requirements. Okay, so basically that's it. I strongly suggest you to read the notes for this lecture. Go to unizord.com to derivative and function limits. This is lecture called properties of the limits. And read it very thoroughly. It's relatively easy reading, but a little bit more, I would say, mathematically rigorous than whatever I was just trying to explain. All right. Thank you very much and good luck.