 Hi, I'm Zor. Welcome to Unisor Education. I would like to talk today about function limit for compounded functions. Compound in terms of function from a function. Well, this lecture is part of the course, which is offered on Unisor.com. It's advanced mathematics for teenagers and high school students. I suggest you to watch this lecture from this website, because parallel to the video there is a very nice explanation of everything which I'm talking about. And in case of a problem, let's say, then some problems are actually solved in writing as well as during the lecture. And if we are talking about theorems, then most likely the proof of the theorem in writing is among the nodes. Alright, so we are talking about compounded functions. Alright, so let's assume that we have two functions. Function f which is defined for all real values x. So its domain is all real x, all real values x. And another function g which also is defined for all real argument x. Now, let's assume that both functions are continuous. It's important. And let me just remind you that function is called continuous if whenever it goes to certain limit if argument goes to some value, then continuous function has a value equals to l in this point. So whenever we are moving, okay, this is point r and this is l. So whenever we are moving towards point r from here or from here we have certain limit and limit should actually be equal to the value at that particular point. Then the function is continuous. We talked about this in one of the previous lectures. So both functions are continuous. Both functions are defined for all real arguments. Now let's assume that this function goes to value l as x goes to r. Now, we plan to do g of f of x. So whenever x goes to r, my f at x goes to l. So argument of g goes to l. So I would like to have that this is approaching the limit m is if x goes to l. So what does it mean? Now I'm taking some value x which I will use it to approach point r. Now my f at x will be going to l in this particular case because that's what I have suggested as a premise of this theorem. Now if f at x goes to l then under argument of function g I have something which is approaching l. And I know that if argument of g goes to l then the function goes to m. So that actually is supposed to approach m as x goes to r. So this is basically the theorem. So if this is given and the functions are continuous and defined for all real values of arguments then this must be the case. Well, how can I prove it? Well, let's just start from the definition of the limit. Now what does it mean that my function goes to limit m if its argument goes to r? Well, it means that for any positive epsilon however small which will measure our distance from this value to m I should find so there exists some kind of a delta such that if my argument is within delta neighborhood of r then my function would be within epsilon neighborhood of m. So that's what we have to do. So for any epsilon if exists such a delta for any epsilon exists delta such that from the closeness of the argument to the point of the limit we have the closeness of the function to its limit. The closeness is measured by epsilon. Then we have the function really has m as a limit. So our purpose is for any epsilon to find delta. Well, okay, how can we do it? Well, let's think about it this way. I know that g is a function which is continuous function and as x goes to l its argument as x goes to l then the function goes to m which means that for this particular epsilon so we just take any epsilon. I know that there is a delta such that if g of x minus sorry if x minus l is less than delta then g of x minus m would be less than or equal to epsilon. So I know that I can find this delta, right? Now I also know that this function is continuous and has a limit l if x goes to r. Now what does that mean? Well, it means that for any epsilon there is a delta such that if my argument is in the delta neighborhood of point r then function would be in the epsilon neighborhood of l. That's what I know, right? Now let me do it this way. Again, start from the beginning. For any epsilon greater than 0 I know how to find delta such that if my x is in the delta neighborhood of l then my g would be in epsilon neighborhood of m. Now I know that this delta I can use this delta instead of this and for this delta it doesn't matter which letter I use. So first I find this delta. Then for this delta I found something like gamma such that if x minus r less than gamma then f at x minus l would be less than delta, right? So basically I just change letters here. Now my delta is this one which I know how to find. Now I know how to find it because I know that g of x has a limit m as x goes to l. So for any epsilon, whatever epsilon I can choose I can find such a delta that from this follows this. Now how can I make sure that this is performed? This is true. Well, very easily. I take for this particular delta I can find gamma such that from this follows this. Now what happens? Now what happens with the function g of f or x of x? I know that for this epsilon we have found such a delta and for this delta we have found gamma. So if this is true then this is true. But if this is true then this actually follows from this that this minus this would be less than epsilon, which is exactly what we need. So again, let me just repeat again the steps. I choose any epsilon greater than zero. I know how to find delta for which this is a true statement. From this follows this. Then for this delta I find gamma based on the properties of the f function that from this follows this. And now I am using basically this gamma as the neighborhood of point r which results in the following. If my x minus r is less than gamma then what happens? Then first we have this and then from this this is the same, right? We have this instead of x, I just put f at x. And that's the end of the proof. So we found for any epsilon we found first delta and for delta we found gamma and that gamma is exactly the neighborhood of the point r. If we are within that gamma neighborhood of r then my g of f of x would be within the absolute neighborhood of m. Okay. Now intuitively it's kind of obvious but relatively rigorous proof is always the good thing to know. Now let me just make a couple of notes. First of all we started from the requirement that both functions are defined for all real values of argument. Basically it's not really necessary because we do need existence of this which means the main of function g should include co-domain of function f. Now when I say that all functions should actually be defined for all real values of argument that's kind of overshoot. I can actually concentrate on a little bit smaller kind of definition, smaller requirements rather and say that g should be defined wherever function f at x has values. But it's really such a minor point and it doesn't really matter quite frankly. In most cases which we will be dealing with we will definitely have situation when functions defined almost everywhere. That's number one. Another requirement which we were discussing here was requirement that the functions are continuous that is actually important. And here is why. Consider this case you have f of x is equal to 0 always for all x. Now g of x is equal to 1 if x is 0 and 0 if x is not 0. So the graph of the function g of x would be like this. It would be 0 everywhere here and 0 everywhere here but at point 0 it will be 1. So one point would be lifted from the line upwards. Now what happens if my g of x goes to 0 as x goes to 0. Because no matter from which side we approach 0 since the values are 0, 0, 0, 0, 0 whatever 0 limit would be 0. So this is my L. Now this is my M and this is my L. Now function f at x would go to 0 as x goes to 0. This is my L and this is my R in my previous. So what happens with the function g of f of x. What's interesting is that this function is always equals to 1. Why? Because f at x is always equals to 0 and function g is always equal to 1 if x is equal to 0. So I have this value always which means the limit of this will be also 1 as x goes to 0. But it's obviously not the same if you will use this one. You will have 0 here. So it looks like this continuity of the function g is very, very important. You will not have this nice things that the limit of function g of f at x would be equals to correspondingly you had f at x goes to L, you had g of x goes to M. And then I said that the g of f of x should go to M. Now this is not correct in this case because this goes to 1 and M is equal to 0. This is M is equal to 0. So this is example of where we really need the continuity. This kind of a one point lifted from the line, basically breaking continuity breaks the whole rule of relationship between the limits. So continuity is important. So all these nice properties of the compound function are good only if the functions are continuous. So that's very important. Okay, now how about a couple of limits for instance, where we can actually use this particular property. And why we started talking about this. So basically, I can just summarize it in a very kind of a short phrase that in some good cases limit of the function is equal to function of limits. Let me put it this way. Again, let me just say it again, if you have f at x goes to L as f goes to R, g of x goes to M as x goes to L. Now we are talking about g of f of x. So limit of function is equal to function of limit. Now function of limit is g of f of R. Now function f is continuous. So f at R is equal to the limit of f at x when x goes to R. So this is equal to L and this is g of L. And g is continuous function. So as x goes to L, function g goes to M and this is supposed to be M for continuous function. Again, we are using continuity. So as you see limit of this as x goes to R is equal to function. So limit of function is equal to function of limit. That's basically in two words what I was just trying to prove you. Now how can it be that's what I just wanted to exemplify. How can you use it as an example for instance. Well, here is one of the things. You remember among amazing limits I was talking about this limit as x goes to 0, right? You remember that. Okay, now let's do it slightly differently. Instead of some x goes to infinity, I can have some function which goes to 0. I can have some function which goes to 0. For instance 1 over x as this is my f as x goes to infinity. So my R would be infinity. My f at x would be 1 over x and as R goes to infinity it goes to 0. Now this is my g of x. So what is my g of f of x? Well, it's a sine of 1 over x divided by 1 over x, right? That's what it is. Which is equal to x sine 1 over x. And now x goes to infinity. So what can I say right now that this has the limit of 1 as x goes to infinity. So that's where I was using. I had a completely different function and completely different limit point. This limit point is 0. This limit point is infinity. But I have replaced argument to this g of x function with function f at x which is 1 over x. And now I can say that if R goes to infinity 1 over x goes to 0. Argument goes to 0. So the function goes to 1. So that's what basically I did. This is m in my terminology and this is l in my terminology. This is l. So argument to R, this function to l, which is 0 happened to be. And I know about what happens with this function if argument goes to 0. So I can say that the combined function has this same limit m. Now something like this can be used in many other cases. All you have to do, you have to recognize that if I'm for instance asking you what is the limit of this function when x goes to infinity, you should recognize that basically this is nothing but a composition of two functions. One is the familiar one from amazing limits. And another is 1 over x, which is continuous function. And you can use this compound function theorem about limits to basically reduce this limit to finding the limit of a compounding function which you know how to do based on whatever the theory might just prove. Okay. I do suggest you to read the notes because the notes might have some better examples or different examples than I was just saying. And other than that I think I will have certain exercises where I will try to put these examples of compounding as a good technique to find the limit whenever you really need to know what's the limit of the function. And you should be able to recognize how to make whatever the exercise actually says to have the limit of how to make it into a composition of functions which you already know how to take the limits off. Alright. That's it. Thank you very much and good luck.