episode 8 the precise definition of the limit so we've talked about the limit using just the intuitive concept that if f of X is defined on some open interval containing a except possibly at a if f of X is arbitrarily close to some number L is it can be made as close to L as we want for all X sufficiently close to but not equal to a then we say that the limit of f of X as X approaches a is L and then we have the notation that you're familiar with now so we haven't really defined what this thing is we've talked a little bit about the definition of an infinite limit or a limit at infinity but we didn't really nail that down precisely we'll borrow some concepts from that today and then find the actual definition of a limit using more precise mathematical language let F be a function defined on some open interval that contains the number a except possibly at a itself and we say the limit of f of X as X approaches a is L and we write this notation here this is the thing that we're defining if and only if for every number epsilon greater than 0 there exists a corresponding number Delta greater than 0 such that if 0 is less than the absolute value of X minus a is less than Delta then the absolute value of f of X minus L is less than Epsilon this is the definition of the limit it uses the Greek letters epsilon and Delta so this is sometimes known as the epsilon Delta definition of the limit but it really is the only definition the first question on the first test and the first question on the final exam will be to state the definition of the limit of f of X as X approaches a and this is what I'm talking about because this is the definition of a limit so let's step through this again let's try to figure out exactly what it means to say that the limit as X approaches a of f of X equals L if and only if for every epsilon greater than zero there is a corresponding Delta greater zero such that if zero is less than the absolute value of X minus a less than Delta then absolute value of f of X minus L is less than Epsilon let's start with that last piece that last inequality the f of X minus L in absolute value is less than Epsilon the absolute value of f of X minus L less than Epsilon represents the open interval L minus epsilon less than f of X less than L plus Epsilon we have a neighborhood centered at L with the radius of Epsilon what this inequality is saying is that we need to make sure that f of X is in that interval for every epsilon we're doing this for every positive epsilon so we're getting as close to L this is what the arbitrarily close to L part means I can make f of X as close to L at 1 because I've made f of X closer to L than every positive number now how do I do that the control that I have is over X so that's what the other inequality here the 0 less than the absolute value of X minus a less than Delta so the absolute value of X minus a is definitely non-negative because the absolute value is never negative but the only way it could be 0 is if X was equal to a and we don't care what happens when X is equal to a we have this open interval centered at a with radius Delta except that we've removed the middle a and this is what we need to find we need to find a Delta so that if we construct this interval here for that Delta and then it will make f of X for every one of those values of x be less than epsilon away from L I think this is easier to see in a graph right so we have y equals f of X and then we have a particular value of a and as I've drawn it here this function is not defined to a so we have to first identify a candidate for the limit right so this is what L would be and so we think L is gonna be the limit we're gonna prove it with the diff so L if it actually is the limit than we know that it would be the Y value of that little hole there where the x value is a so we have to show that for every epsilon greater than 0 there is a delta greater than 0 such that if we make X closer than Delta to a f of X will be closer than epsilon to L so we identify epsilon that gives us a range of values of our function L minus epsilon to L plus Epsilon so then we pick our Delta so we make an interval on the x-axis where everything in between a minus Delta and a plus Delta except for a we'll put our function values into that interval and as I've drawn it here we've done that everything between a minus Delta and a plus Delta will map to an F value that's in between L minus epsilon and L plus Epsilon and since we've done this for every epsilon we can shrink that interval of the y-axis down as small as we want now it would force Delta to get smaller and it would make the interval and the x-axis get smaller that's the arbitrarily close to L part and the sufficiently close to a part so understanding this definition is possibly the hardest thing I'm going to ask you to do this semester it's very much unlike everything we'll do in the rest of the semester it's getting at the higher language of math the more formal math that you would see in a class that you would take as a junior senior math major you know so in that respect I don't necessarily need you to understand it fully now or things like it but it's really something that you don't understand the first time you see it and so I want you to see it now where it's not so crucial that you don't understand it just try to get a little bit of the understanding of it and it'll help you understand the more formal language of mathematics we'll come back to a couple of things you're kind of like this semester but most of the things we'll do are very different than this we'll do an example I want to use this definition to prove that the limit as X approaches 2 of 3x plus 1 is equal to 7 we've looked at this women before we know that the limit is 7 because we just plug it x equals 2 in and we get 7 all right let's use that knowledge of how this function works to get a greater understanding of what the definition means so here's the structure of the proof I let epsilon be greater than 0 that means it's any number greater than 0 because I have to do it for all of them I don't specify a particular value of Epsilon I have to do it for every value of Epsilon what I need to do now is find a value of Delta that's positive such that if we impose the absolute value of X minus 2 to be less than Delta then it will follow that the absolute value of our f of X 3x plus 1 minus 7 or L is less than Epsilon so it turns out that Delta should be epsilon over 3 we'll talk about how I knew that in just a second listen but for now let's just see how it works in the proof so I need to prove an if-then statement so I need to start with the if and then show that the thing that follows the then must be true right so I start with supposing that the absolute value of x minus 2 is less than Delta but Delta I chose to be epsilon over 3 I need to find the absolute value of f of X minus L is less than Epsilon well it's not too hard to do right now we'll just multiply both sides by 3 see that will give me something less than Epsilon that's something that I need so I have 3 times the absolute value of X minus 2 I can bring that 3 inside the absolute value because 3 is positive so that's the absolute value of 3 minus 6 less than Epsilon what I need is the absolute value of 3x plus 1 minus 7 well that's too hard to get it right minus 6 is equal to 1 minus 7 and so there's the thing that I have my absolute value of f of X minus L is less than Epsilon I've shown that if X minus 2 is less than epsilon over 3 then and the absolute value of 3x plus 1 minus 7 is less than Epsilon and that's the definition of the limit as X approaches 2 of 3 X plus 1 equals 7 now how did I know to choose Delta equals epsilon over 3 well basically I just worked backwards from what I wanted to know I started with the thing that he needed to show and worked backwards to find the absolute value of X minus 2 to be less than something and then I'll call that something Delta right so the thing that I need to show is that the absolute value of 3x plus 1 minus 7 is less than Epsilon I need to manipulate this to get the absolute value of X minus 2 1 minus 7 is minus 6 and then I can factor a 3 out of this factor that 3 out inside the absolute value and then realize that the absolute value of 3 is 3 hey there's my absolute value of X minus 2 so I just divide by 3 so I have the absolute value of X minus 2 the thing that I've wanted less than epsilon over 3 so that means Delta should be epsilon over 3 let's try this again so use the definition of the limit to prove that the limit as X approaches negative 1 of 1/3 X plus 4 equals 11 thirds again I plug in negative 1 I get minus 1/3 plus 4 not equals 11 thirds so that is my limit now how do I prove this using the definition before we do the proof we have to pick Delta that's where we start we can't start the proof until we know what Delta it is so we need to start with the thing that we need to prove we need to prove that the absolute value of one-third X plus 4 minus 11 thirds is less than Epsilon this is the absolute value of f of X minus L is less than Epsilon okay well 4-11 thirds is one third so I have the absolute value of one third X plus one third and that's less than Epsilon what I need is the absolute value of X minus -1 so hey if i factor out the 1/3 I get 1/3 the absolute value of x plus 1 less than Epsilon so I'm almost there right X plus 1 is X minus -1 and then if I just divide both sides by 1/3 I get that the absolute value of X minus minus 1 is less than 3 Epsilon that's the thing that I need Hey so I should choose Delta to be 3 Epsilon now I'm ready to begin the proof and so the proof is just this work that we did backwards right so this is really scratch work but it gives us the ability to write the proof so here's the proof let epsilon be greater than 0 and choose Delta to be 3 epsilon then we need to suppose that the absolute value of x minus minus 1 is less than 3 Epsilon well if that's true we can divide both sides by 3 and realized that X minus -1 is X plus 1 and then I can bring that 1/3 inside the absolute value so I have 1/3 X plus 1/3 and absolute value is less than epsilon that can write 1/3 as 4 minus 11 thirds so I have the absolute value of 1/3 X plus 4 minus 11 thirds is less than Epsilon hey that's the definition of the limit right so I've I have satisfied the limit I've supposed that if the absolute value of x my name is minus 1 is less than 3 epsilon then I must have that the absolute value of 1/3 X plus 4 minus 11 thirds is less than Epsilon and that's the sufficiently close to negative 1 to get arbitrarily close to 11 thirds and then let's do one works is possibly the hardest example I'll ask you to do this semester let's do this for a quadratic function so let's use the definition of the limit to prove that the limit as X approaches 2 of 2x squared minus 3x minus 5 is equal to negative 3 so the first thing I have to do is choose my Delta so the first thing I'm going to do is let epsilon be greater than 0 and then I need to find a delta such that if the absolute value of x minus 2 is less than Delta then that forces the absolute value of 2x squared minus 3x minus 5 minus minus 3 to be less than Epsilon so I start with the thing that I need to show absolute value of f of X minus L is less than Epsilon and I simplify it a little bit and I get that the absolute value of 2x squared minus 3x minus 2 is less than Epsilon now what I did before was I needed to find the absolute value of x minus 2 less than Delta and I needed to do the exact same thing here now that quadratic I can factor I can factor it into 2x plus 1 times X minus 2 hey that X minus 2 is something that I need and then I have the absolute value of this product so it's equal to the product of their absolute values so I have something very similar to the ones that I did in the previous cases with one very big difference before I had a constant times the absolute value of x minus my a less than Epsilon and here I have something different I have a function of x times the thing that I want the absolute value of X minus 2 so I can't proceed exactly like I had done I can't divide by the absolute value of 2x plus 1 because it be dividing by something it has an in it and I can't do that Delta can depend on epsilon but it cannot depend on X so we need to figure out some way of bounding the absolute value of 2x plus 1 near x equals 2 so let's just pick an interval that contains 2 so we have the interval 1 less than X less than 3 right that's an open interval that contains 2 and I want to see what this bound here does to the absolute value of 2x plus 1 so I just need to construct 2x plus 1 right so I multiply by 2 and then I add 1 so I get the 2x plus 1 on the interval 1 less than X less than 3 2x plus 1 is greater than 3 and less than 7 so an absolute value it would be greater than 3 or less than 7 but what I really want is 1 over that quantity right because I want to divide by that and so this means that 1/7 is less than 1 over the absolute value of 2x plus 1 which is less than 1/3 so therefore on the interval 1 to 3 the inequality that I was concerned with the absolute value of 2x plus 1 times the absolute value of X minus 2 less than Epsilon so I'll do like I did before I'll divide by the absolute value of 2x plus 1 but that's got an X in it so this is something I can't use but I just need it to be an inequality right so I'll choose absolute value of X minus 2 to be less than epsilon over 7 all right epsilon over 7 will be less than epsilon over absolute value of 2x plus 1 on this interval and I could always make this closer to 2 and still have it work right so this is my Delta except that I could be outside that interval so I need to think about Delta being the smaller of 1 or epsilon over 7 if epsilon is bigger than 7 then X minus the absolute value of x minus 2 less than epsilon over 7 is outside of the interval 1 2 3 so that's why I picked the the minimum all right so I'm gonna be either in the interval 1 2 3 or I'm gonna be in an interval smaller than that of with epsilon ok so now we're ready to begin the proof so we let epsilon be greater than 0 and then Delta is the minimum of 1 or epsilon over 7 so we'll suppose that absolute value of X minus 2 is less than Delta so we have to think about this in two cases either epsilon is less than 7 or it's not so if it is less than 7 then the absolute value of x minus 2 is less than epsilon over 7 I can multiply on the right by 7 and the left by absolute value of 2x plus 1 because the absolute value of 2x plus 1 is less than 7 on the interval 1 to 3 and I'm inside that interval and now I just repeat what I did before right I'm bringing the product inside the absolute value I file it out and I get x squared minus 3x minus 2 and then I realized that I could rewrite that as 2x squared minus 3x minus 5 minus minus 3 hey that's my excellent value of f of X minus L less than excellent on the other hand if X alone is greater than 7 then my Delta is 1 so I have the same basic idea I've restricted now X to being inside the interval one two three and so I know that the absolute value of 2x plus 1 is less than 7 on that interval so if I multiply on the left by the absolute value of 2x plus 1 and on the right by seven I still have my inequality and so if epsilon is greater than 7 I've increased the right-hand side so I still have my inequality and then they do the same thing so this is not the easiest thing to show but it's something you need to see you know this is not something I'm going to ask you to do a lot but it's something I want you to follow me through you understand the formal language of mathematics by C people use it first and then using it on your own so I want to make sure that you've seen someone use it again please do not be afraid of this the rest of the semester is not like this this is a very different topic than the rest of the semester most of the semester the answer is is going to be a function with a box around it or something like that right but here the answer is a paragraph and you need to be able to string together multiple steps at some point and so here's an example of that