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in the last video we took our first look at the epsilon-delta definition of limits which essentially says if you claim that the limit of f of X as X approaches C is equal to L then that must mean by the definition that if some if you were given any positive epsilon it essentially tells us how close we want f of X to be to L we can always find a delta that's greater than 0 it's essentially telling us our distance from C such that if X is within Delta of C then f of X is going to be within epsilon of L if we can if we can find a delta for any epsilon then we can say that this limit that this is indeed the limit of f of X as X approaches C now I know what you're thinking this seems all very abstract I want to somehow use this thing and what we will do in this video is use it and to rigorously prove that a limit actually exists so right over here I've defined a function f of X it's equal to 2x everywhere except for X equals 5 so it's 2x everywhere for all the other values of X but when X is equal to 5 it's just equal to X so I could have really just written 5 I could have just written 5 there it's equal to 5 when X is equal to 5 it's equal to itself and so we've got drawn the graph here everywhere else it looks just like 2 X at X is equal to 5 it's not along the line to X instead the function is defined to be that point right over there and if I were to tell ask you what is the limit of what is the limit of X f of X as X approaches 5 you might think of it pretty intuitively well let's see the closer I get to 5 the closer I get to 5 the closer f of X seems to be getting to 10 the closer I get to 5 the closer f of X seems to be getting to 10 and so you might fairly intuitively make the claim that the limit of f of X as X approaches 5 as X approaches 5 really is equal to 10 it looks that way but what we're going to do is use the epsilon definite the epsilon-delta definition to actually prove it and the way that most of these proofs typically go is we define Delta in the abstract and then essentially try to come up with a way that given any epsilon we can always come up with a Delta or another way is we're going to try to describe our Delta as a function of epsilon not to confuse you too much but maybe I shouldn't use F again but Delta equals function function of epsilon that is defined for any positive epsilon so you give me an epsilon I just kind of put it into our little formula or a little function box and I will always get you a delta and if I can do that for any epsilon that'll always give you a delta where this is true that if X is within that range of Delta is within Delta of C then the corresponding f of X is going to be within epsilon of L then the limit definitely exists so let's try to do that so let's let's let's think about let's think about being within Delta of our C so let's think about this right over here is 5 plus Delta this is 5 minus Delta so that's our range we're going to think about we're think about in the abstract at first and then we're going to try to come up with a kind of a formula for Delta in terms of Epsilon so how can we describe all of the X's that are in this range but not equal to 5 itself because we really care about the things that are within Delta a 5 but not necessarily equal to 5 this is just a strictly less than there within a range of C but not equal to C well that's going to be all of the X's that satisfy X minus 5 is less than is less than Delta that describes all of these X's right over here now what we're going to do and the way these proofs typically go is we're going to try to manipulate this let the left-hand side of the inequality so it starts to look something like this or it starts to look exactly like that and as we do that this the right-hand side of the inequality is going to be expressed in terms of Delta and then we can essentially say well look if the right-hand side looks like Delta the left-hand or is in terms of Delta and the left-hand side looks just like that that really defines how we can express Delta in terms of Epsilon if that doesn't make sense bear with me I'm about to do it so if we want if we want X minus 5 to look a lot more like this when X is is not equal to five and all of this this whole interval X is not equal to five f of X is equal to 2x our limit that our proposed limit is equal to 10 so if we could somehow get this to be 2x minus 10 then we're in good shape and the easiest way to do that is to multiply both sides of this inequality by 2 you multiply both sides of this inequality by 2 and 2 times the absolute value of something that's the same thing as the absolute value of 2 times that thing if I were to say 2 times the absolute value of a that's the same thing as the absolute value of 2a so on the left hand side right over here this is just going to be the absolute value of 2x minus 10 and it's going to be less than on the right hand side you just end up with a 2 Delta now what do we have here on the left hand side well this is f of X as long as X does not equal 5 and this is our limit so we can rewrite this as f of X minus L is less than 2 Delta and this is 4 for X does not equal 4 x does not equal 5 this is f of X this literally is our limit now this is interesting this statement right over here is almost exactly what we have want right over here except the right sides are just different this has in terms of Epsilon this has it in terms of Delta so how can we define Delta so that we can put so that 2 Delta is essentially going to be Epsilon well this is our chance we will just define we will we will make we will make and this is where we're we're we're defining Delta as a function of Epsilon we're going to make two Delta equal epsilon or if you divide both sides by 2 we're going to make Delta equal to epsilon over 2 so if you make Delta equal epsilon over 2 so if you make if you make let me switch colors just to ease the monotony if you make if you make Delta equal epsilon over 2 then this statement right over here becomes the absolute value of f of X minus L is less than instead of 2 Delta it'll be less than two times epsilon over two is just going to be is going to be less than Epsilon so this is the key if someone gives you any positive number epsilon for this function as long as you make Delta equal epsilon over two then any X within that range that corresponding f of X is going to be within epsilon of our limit so this tells you some would say say the epsilon and remember it has to be true for any positive epsilon but you could see how the game could go if someone gives you the epsilon Epsilon let's say they want to be within 0.5 of our limit so our limit is up here so our epsilon is 0.5 so it literally be the range I want to be between 10 plus epsilon would be 10.5 and then 10 minus epsilon would be 9.5 well we just came up with a formula we just have to make Delta equal to epsilon over 2 which is equal to 0.25 so that'll give us a range between four point seven five four point seven five and five point two to five and as long as we pick an x between four point seven five and five point two five then the corresponding f of X but not x equals five the corresponding f of X will be between nine point five and ten point five and so you give me any epsilon I can just apply apply this formula right over here to come up with the Delta this would apply for any real number I mean especially any positive number for any epsilon you give me I just get a delta and then a delta defined this way and then I can go through this arc if X minus five is less the absolute value of X minus five is less than Delta if Delta is defined in this way which I can define for any epsilon then it will be the case that f of X will be within epsilon of our limit

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