WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about the formal definition of a limit.
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In this lesson, we will explore the formal definition of a limit.
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Previously, we defined a limit with intuitive terms, like "approach" and "goes to."
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Now, we are getting serious, and it is time for some really technical stuff.
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However, I want to point out that very few students will have any use for what is in this lesson, let alone during a precalculus-level course.
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I want you to have the option to know the formal definition, if you are interested.
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But most people will not need it, even in math class.
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Problems based on the formal definition of a limit are extremely rare in calculus class--maybe as a bonus question, but probably no more.
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This stuff won't come up in science classes, and it will only be necessary for high-level college math.
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And when I say "high-level college math," I mean that this stuff is really going to only definitely start showing up
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by the second or third year of taking advanced college math courses.
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This stuff is not going to show up any time soon, so I really just want to be frank with you and honest:
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what I am going to teach here is really a totally optional lesson.
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It is not going to come up in the course that you are currently taking; I guarantee you that.
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And it may or may not come up in the next course you take, if you go on to take calculus.
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But it is really, really unlikely to be tested heavily.
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It will maybe be a bonus question, or maybe one real question; but for the most part, you are not going to see this formal limit definition.
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It just doesn't show up; it has very little direct applicability in the sciences; it is only used for some pretty advanced ideas in mathematics.
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It is really important for building advanced ideas in mathematics.
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But for day-to-day use of mathematics, we just never really need to see it.
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Unless you are particularly interested in mathematics, or you know that you really want to go into high-level physics,
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or you are interested in something like computer programming, where you will have to get enough mathematics learned,
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at some point, that you will end up needing to be able to understand those advanced math concepts,
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you are probably never going to end up encountering this formal definition, ever.
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And that is totally OK; wanting to learn this sort of thing--I know that it is not for everybody.
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But for me, I personally love this stuff; this is the reason I love math--because there are these really cool, complex,
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strange, puzzling logic ideas that make this really interesting structure that is just fascinating--to me, at least.
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So, I think this stuff is totally awesome.
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If you are not interested, don't worry about it; skip this lesson.
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There is going to be some really difficult stuff, and there is not really a whole lot of direct personal need to understand it.
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But at the same time, there is not really a whole lot of direct personal need to go to a museum; you do it because you are interested in the art.
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And that is the same reason that you might be interested in mathematics.
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It is the reason I am interested in mathematics--just because I find it personally satisfying and cool.
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If you are interested, let's go and check some stuff out.
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OK, and before we keep going, I just want to say, for those of you who decided to stay around, thanks.
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I think this stuff is really, really cool; I am glad to get the chance to share it with you.
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And I think it is really, really cool; I hope you end up liking it as much as I do.
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We few, we happy few, we mathematicians--we get to see this.
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All right, we are going to need some new Greek letters before we get started.
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That is right: this stuff is so cool that we need to bring out the new Greek letters.
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Our first one is Δ, and these are the lowercase versions; so Δ is lowercase Δ, and ε is lowercase ε.
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Here is Δ; and if you are going to draw it by hand, it is kind of hard to draw this thing exactly as it ends up getting typeset.
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So, I would recommend...you want something on the top, and then sort of a curved loop like that.
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You might end up seeing me write it like that, as I am writing it pretty quickly.
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But that is a pretty good idea of what you are trying for.
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This one is kind of not a very good one; but put some sort of curved loop on top, and then a sort of circular thing on the bottom.
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And that will be enough for most people who understand mathematics to realize that you are trying to write Δ.
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You don't have to ahve it absolutely perfect, because they will know what the symbol is, especially when they end up seeing it a bunch of times.
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ε is considerably easier to draw; it is just sort of a backwards 3; that is it.
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That is ε; that one is easier than Δ, I would have to say.
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All right, the formal definition that we are about to see involves these letters.
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So, it is sometimes called the (ε,Δ)-definition of a limit, or just the formal definition of a limit.
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We just talked about the intuitive idea of what a limit is; and that is a really great way to think about limits.
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But sometimes you really need to get to the basic facts and turn intuitive ideas into precise things, so that you can say "this is what it is."
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And that is what this definition is for.
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All right, let's get to it: the formal definition of a limit.
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Let f be a function defined on some interval a to b of the real numbers, where a is less than b.
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Let c be contained in a to b, and let l be a real number.
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Then, the limit, as x goes to c, of f(x) = l means that, for any real ε greater than 0,
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there exists some Δ greater than 0 such that, for all x where 0 is less than the absolute value of x - c,
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which is less than Δ, we have the absolute value of f(x) minus l is less than ε.
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What? Don't worry: if you found that definition confusing, that is exactly how I felt when I first heard it.
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The first time you see this, you think, "What is going on?"
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That is totally OK; it is something that you pick up and work through slowly, over time.
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If you opened a book for the very first time, you wouldn't expect to be able to understand every word on the page.
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It is something you have to work through; it is OK--you are building up these ideas of understanding it.
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It is not something that is going to make sense exactly the first time.
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It is something that makes sense over time; that is totally fine.
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So, we will start working through it, step-by-step, and we will see some ways to understand what is going on in this definition--
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to see how we can parse this thing and make sense of it.
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All right, first, let's get the basics out of the way, the ground work here.
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The first half of the definition is just setting up the limit; that is what we end up getting in this first half.
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Let f be a function defined on some interval a to b (they are real numbers) where a is less than b, and c is contained in a to b, and l is some real number.
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All right, let's break that down, piece by piece.
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First, let f be a function defined on some interval a to b.
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That just means that it is a function, and it is defined over some portion of the reals.
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There is some piece of it that it ends up working in.
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It is basically just saying that this function is defined somewhere; that is all that we are saying.
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There is some chunk of the reals where this function makes sense.
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This part where we say a < b is just a guarantee that we don't accidentally say a = b, and we end up having something that doesn't make sense.
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It is so that we know that we are going from some a, up until some b; that is our interval of where the function is defined.
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Next, the horizontal location c, contained inside of (a,b): remember, this symbol just means "inside of."
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So, (a,b) is a set of all possible values; it is the interval, and that symbol just means that c is inside of that set of possible values.
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So, the horizontal location c contained inside of (a,b) is the value that the limit will approach.
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That is the thing where we are getting closer and closer to that, and we are seeing where we end up going to.
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So, c is just a location that we are getting towards; that is the value that the limit will approach.
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And finally, l is some real number; l's real number is the vertical location that we end up going to.
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That is where the limit will go to; we are setting the l that we are saying our limit ends up going to as we approach this c.
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So, that is the basics that we are setting up: f is some function; it works on some chunk.
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c is where we are going to, and l is the value that we end up meeting up at, that we are headed towards.
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All right, the hard part is the second half of the definition.
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The limit as x goes to c of f(x) = l means that, for any real ε greater than 0, there exists some real Δ greater than 0,
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such that for all x where 0 < |x - c| < Δ, we have f(x) - l < ε; OK.
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To break this up, we want to be able to focus on this one piece at a time.
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So, what we are going to do is focus on the ε portion first.
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It will be easier to understand that; and then we will see the Δ part in that idea.
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So, for any real ε greater than 0: what that means is that, if we just grab some ε--
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we just say some number, like 1 or 1/2 or whatever number we feel like saying,
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we will end up having this be true later on: that the absolute value of f(x) - l is less than ε.
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Now, that might be kind of hard to understand: the absolute value of f(x) - l is less than ε...
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we are not used to working with absolute value that much; it is kind of confusing.
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But remember one of the things we talked about when we first talked about absolute values.
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Absolute value is a way of seeing the distance between two things.
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If we have the absolute value of m - n, if we have the absolute value of q - r, that is a way of saying the distance between those two points.
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The absolute value of some object minus some other object is just a way of saying how far those two objects are from each other.
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That is one of the ideas that we have from absolute value.
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When we say the absolute value of f(x) - l, what we are really saying is how far f(x) and l are from each other.
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Now, that means, with the combination of ε being greater than 0, and the distance between f(x) and l being less than ε...
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what we are doing is saying that there is this boundary around l.
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The maximum distance f(x) can be from l, the distance between f(x) and l, has to be less than ε, this thing right here.
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Since the distance between our f(x) and our l is less than ε, it means that our f(x) has to be within ε distance of l.
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We effectively create a boundary that we hold f(x) inside of.
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Whatever our f(x) ends up being, it has to be somewhere inside of this boundary.
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If it is outside of the boundary, like this, that is not allowed, because it would fail to be within ε distance of the l.
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We end up seeing...we have that our l is here, and then the ε's here are just the distance that we end up having for that boundary.
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We are setting a boundary that is ε distance out from the l.
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But notice: it has to be true for any ε that we set, so we will be able to create a boundary at whatever size we decide to make,
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whether it is a giant ε boundary or a really tiny ε boundary.
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It will always end up working for this next part that we are about to work towards.
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OK, the second half of the definition (continuing with that): previously, we saw ε as a boundary around the height l.
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The absolute value of f(x) - l is less than ε means that f(x) has to be, at most (has to be less than), the distance of ε from l.
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We know that we have to be somewhere contained inside of this boundary in here.
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OK, now let's move on to looking at Δ.
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What does Δ get us? What we did is went and set up for any real ε greater than 0; we had this boundary get set up.
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So, with that boundary in mind, we now have to go on to say that there exists some real Δ greater than 0,
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such that, for all x where 0 is less than x - c, which is less than Δ, we will end up having this |f(x) - l| < ε be true.
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So, what we are doing is putting a restriction on the allowed x-values.
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The Δ sets some restriction on it: you can't get any farther from c than Δ.
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So, what we are doing is saying, "Remember: the |x - c| is saying the distance between x and c."
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So, the distance between x and c has to be less than Δ.
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So, if the distance between x and c is less than Δ, that means that we have set up, once again, a boundary of how far our x is allowed to roam.
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And we know that what has to come out of this is: we have to have...
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if we set x's within that boundary, we will end up having the |f(x) - l| being less than ε.
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So, if you use x's from inside of this boundary here, we know that we end up having to be mapped into the f(x) that is within that vertical boundary.
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We are setting up some set of boundaries on what x is that we can have.
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And we know for sure that we have to have it so that they end up getting mapped inside of this ε.
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The other thing to notice is that it says that 0 is less than |x - c|, that 0 is strictly less than the absolute value of x - c.
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It means that x cannot be equal to c, because then we would have an absolute value of 0.
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We would have x = c; but since it is a limit, we are not concerned with x = c.
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Remember: we are not concerned with the destination; we are concerned with the journey.
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So, that means we care about the part with the boundary, but not the actual thing in the middle.
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That is what that 0 < |x - c| means: that we are guaranteed that we can't have x equal to c,
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and |x - c| < Δ means that we are bounded Δ distance, at maximum, from the c in that middle.
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And everything inside of it ends up getting mapped inside of this boundary here.
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It will be easier to see what is going on if we erase some of this.
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What we are doing here is: we can think of it as: some ε gets created around our l that sets up a boundary.
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We know that you have to be somewhere on this portion vertically.
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And then, we set some Δ radius around our c, so that it ends up mapping only to those vertical locations there.
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We set ε, and then we have to be able to always set some Δ from that ε
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that will end up mapping us inside of that vertical chunk--that we say that,
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if we are close enough to this value of c, we will always come out close enough to that value of l.
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OK, now the really important idea is that we can do this for any ε greater than 0 whatsoever.
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That is, for any ε greater than 0, there has to exist some Δ greater than 0,
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such that, for x within Δ of c, if we are close enough to c, f(x) will end up getting mapped within ε of l.
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If you are close enough to c, you will end up being close enough to l.
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We can imagine this process as a never-ending dialogue of ε challenges against Δ defenses.
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Someone says, "Is it possible to stay within this distance of our l?" and we say, "Yes, as long as we are within this distance of our c."
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And then, they say, "OK, is it possible to stay within this distance of our l, this ε of our l?"
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We say, "Yes, as long as we are within Δ distance of c."
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So, some ε gets named; for any ε named, there has to be a Δ that will cause us to stay within ε away from that l.
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Otherwise, if we can't do this, then the limit does not exist.
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But if we can do it, forever and ever and ever, then that means that the limit does exist.
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If for any ε that gets named, there is always a Δ, then that means that the limit does exist.
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Now, of course, we can't actually have this process of ε challenge and Δ defense; we can't do this forever.
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We can't do this for eternity, so we have to figure out another way to do this.
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If we wish to truly prove a limit, we have to prove that there will always be some Δ greater than 0,
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given some ε greater than 0--that if some ε gets named, there is always a method
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that will create the appropriate Δ, so that we can get within ε of that limit l.
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OK, now what we are going to go onto is: we are going to see both of these ideas.
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Both of these ideas will be explored in the next scene that will be played out by Sherlock Holmes and Dr. Watson.
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All right, the setting is late afternoon in Sherlock Holmes's apartment.
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He and Dr. Watson are lounging about.
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"Let us play a game, Watson!" "All right, Holmes, what type of game?"
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"A game of limits." "I do not believe I have ever..."
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"Do not worry: I shall teach it to you. We begin by setting up the game as follows."
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"One of us (I, in this case) claim the existence of a limit; I name some function f(x)
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and some value l that the function will approach as x approaches some number c."
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"Written out, we have the limit as x goes to c of f(x) is equal to l."
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"All right, that makes sense enough." "Good; after that, the game in earnest begins."
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"The opposing player (you, in this case) names an ε greater than 0."
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"It will then be the first player's job to name some Δ greater than 0, such that,
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for all values of x within Δ distance of c, except x = c (we don't care about x = c in this game),
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they will cause f(x) to be within ε of l."
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"So, if we use one of those x's that is within Δ of c, we know that the f(x) that will be produced by that x will have to be within ε of l."
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"I think I understand." "Let me write it out: I will name some ε greater than 0 that sets up some boundary around l."
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"So, we have some l - ε to l + ε boundary; we are going to be in that interval vertically."
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"Then, you name a Δ greater than 0 such that, for x contained within c - Δ to c + Δ,
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that horizontal bounding, we will end up getting f(x) contained in l - ε to l + ε;
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that is to say, the boundary that we set around l in the first place."
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"I will set some boundary vertically, and then you will respond with some horizontal boundary that will put our function inside of that vertical boundary."
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"Very good, except one important thing: we do not concern ourselves with x = c."
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"For this game, x is not equal to c; we never actually consider x equaling c, so x is not equal to c, generally."
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"All right, I understand: you set up a limit; I challenge it with some ε greater than 0, and then you defend with some Δ greater than 0."
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"One question, though: why can't ε equal 0?"
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"I understand that Δ must be greater than 0, because we must have x not equal to c;
00:17:39.200 --> 00:17:47.600
if Δ is greater than 0, then we know that we can't actually be on top of x = c, so x is not equal to c. But why not ε = 0?"
00:17:47.600 --> 00:17:54.200
"Because this is a game about limits! If you set ε = 0, we would be stuck on l."
00:17:54.200 --> 00:17:57.000
"We wouldn't have a boundary around it; we would be stuck there."
00:17:57.000 --> 00:18:01.700
"We need a boundary around l, so that we can talk about approaching it;
00:18:01.700 --> 00:18:06.300
the entire idea about a limit is to talk about going towards it, approaching it."
00:18:06.300 --> 00:18:11.900
"It is similar to why we must have x not equal to c; otherwise there is no approaching it, if we are stuck on top of it."
00:18:11.900 --> 00:18:18.500
"So, we can't be stuck on top of it, either vertically or horizontally; that is why we can't have ε or Δ equal to 0."
00:18:18.500 --> 00:18:23.600
"Makes sense; so what comes next?" "That is it; we stay with the same limit,
00:18:23.600 --> 00:18:29.400
but the process of ε challenges and Δ defenses repeats for as long as we care to play."
00:18:29.400 --> 00:18:34.900
"If you manage to give an ε that stumps me, you win! You will have shown the limit false."
00:18:34.900 --> 00:18:43.400
"Let us begin; what will your limit be?" "I choose the following limit to defend: limit as x goes to 3 of x² + 2 = 11."
00:18:43.400 --> 00:18:49.900
"All right, I begin by challenging with ε = 1." "I defend with Δ = 0.1."
00:18:49.900 --> 00:19:00.800
"Since x contained within the interval 2.9 to 3.1 (notice: that is a Δ of .1 away from our c of c = 3),
00:19:00.800 --> 00:19:08.200
if x is contained within 2.9 to 3.1, it will give f(x) contained within 10.41 to 11.61."
00:19:08.200 --> 00:19:13.500
"And that means I am easily within an ε value of 1 from our l of 11."
00:19:13.500 --> 00:19:19.600
"Thus, my defense stands: using Δ = .1, I stay within ε of 11."
00:19:19.600 --> 00:19:27.100
"OK, how about ε = 0.05." "I will go with Δ = 0.005."
00:19:27.100 --> 00:19:39.600
"Since x contained within 2.995 to 3.005 causes f(x) to be contained within 10.97 to 11.03, I am within ε of 11, so my defense stands again."
00:19:39.600 --> 00:19:50.200
"What if I tried larger...I set ε = 10!" "Nice try, old chap, but I already showed that Δ = 0.1 works for ε = 1."
00:19:50.200 --> 00:19:55.000
"Therefore, it must work for ε = 10 as well; if you started out small, and then you get bigger,
00:19:55.000 --> 00:20:00.000
well, I can just leave my Δ the same, and it will work just fine for the larger vertical boundary, as well."
00:20:00.000 --> 00:20:03.800
"Thus, there is no hope for going larger with your ε.
00:20:03.800 --> 00:20:09.600
Since I have already shown that it works for a small ε, it will automatically work for a larger ε, as well."
00:20:09.600 --> 00:20:20.400
"Hmm...fine, deal with this: ε = 0.000001." "Elementary, my dear Watson: Δ = 0.0000001."
00:20:20.400 --> 00:20:37.600
"Notice that, if x is contained within 2.9999999 to 3.0000001, we will have f(x) contained within 10.9999994 to 11.0000006."
00:20:37.600 --> 00:20:44.600
"That is to say, it will be contained within an ε of 11; thus, my defense is still standing."
00:20:44.600 --> 00:20:51.600
"All right, Holmes; you have me convinced--I believe that, no matter what ε I say, you will be able to defend with an appropriate Δ."
00:20:51.600 --> 00:20:57.000
"But technically, you haven't proven the limit; we can't play this game for eternity."
00:20:57.000 --> 00:21:00.700
"So, how can you prove that Δ will always exist?"
00:21:00.700 --> 00:21:06.400
"An astute question, Watson; the answer is by showing that there exists some method
00:21:06.400 --> 00:21:12.400
to create Δ from any given ε, and that the method never fails."
00:21:12.400 --> 00:21:15.700
"I will tell you the method I am using, and then show that it always works."
00:21:15.700 --> 00:21:21.600
"Of course, you already have a method." "Naturally; here it is: for any ε greater than or equal to 1,
00:21:21.600 --> 00:21:28.100
I will simply respond with Δ = 0.1; as we discussed earlier, Δ = 0.1 works for ε = 1,
00:21:28.100 --> 00:21:31.500
so it will work for any ε greater than or equal to 1."
00:21:31.500 --> 00:21:41.100
"On the other hand, for ε between 0 and 1 (that is, 0 < ε < 1), I will use Δ = ε/10."
00:21:41.100 --> 00:21:44.400
"Wait; how do we know that Δ is greater than 0?"
00:21:44.400 --> 00:21:49.700
"Well, we know that ε is greater than 0; and since we are creating Δ
00:21:49.700 --> 00:21:56.600
by dividing ε by 10, we know that Δ must be greater than 0, as well."
00:21:56.600 --> 00:22:02.100
"If ε is greater than 0, then we know that dividing something that is greater than 0 doesn't cause it to become 0,
00:22:02.100 --> 00:22:05.100
and doesn't cause it to become negative; it simply makes it smaller."
00:22:05.100 --> 00:22:08.700
"So we know that Δ must be greater than 0, as well, because we simply divided
00:22:08.700 --> 00:22:13.400
something that was already greater than 0 by some positive number--in this case, positive 10."
00:22:13.400 --> 00:22:19.600
"Now, let me prove to you that Δ = ε/10 always works."
00:22:19.600 --> 00:22:23.100
"Notice that, since we are working with the function x² + 2,
00:22:23.100 --> 00:22:32.000
the values farthest from c = 3 will produce the values farthest from l = 11, because 3² + 2 = 11."
00:22:32.000 --> 00:22:36.500
"Thus, we only have to concern ourselves with x = 3 - Δ and x = 3 + Δ.
00:22:36.500 --> 00:22:42.500
"If that didn't quite make sense, Watson, what we have here is that, at some horizontal location 3,
00:22:42.500 --> 00:22:45.400
we know that it puts out the vertical height of 11."
00:22:45.400 --> 00:22:51.100
"So, as we get farther and farther to either side, we will end up getting farther and farther to 11."
00:22:51.100 --> 00:22:57.800
"We know that this is a parabola; we are used to working with parabolas; so we see that, the farther we end up getting away from c = 3,
00:22:57.800 --> 00:23:04.400
the more extreme we are from c = 3, the more extreme we will be from our l = 11."
00:23:04.400 --> 00:23:08.200
"That is how 3² + 2 would work; as long as we don't go so incredibly far
00:23:08.200 --> 00:23:12.800
that we end up wrapping the parabola to the other side, we are safe from this; we don't have to worry about it."
00:23:12.800 --> 00:23:19.700
"So, as long as we can be absolutely sure of that fact, where we don't to -3 or go past 0, we will be fine."
00:23:19.700 --> 00:23:27.300
"And we know that Δ must be less than or equal to .1, because of our earlier restriction on it always being less than or equal to .1."
00:23:27.300 --> 00:23:34.200
"So, we know that we can't get too far; the most extreme values for our x will end up being using the whole of Δ--
00:23:34.200 --> 00:23:39.700
that is, we only have to concern ourselves with x = 3 - Δ and x = 3 + Δ."
00:23:39.700 --> 00:23:45.500
"Now, I want to point out, before we continue, that this logic is specific to the function we are working with, Watson."
00:23:45.500 --> 00:23:51.100
"It is specific to working with x² + 2, the fact that, as we get farther and farther away horizontally,
00:23:51.100 --> 00:23:58.000
we will get farther and farther away vertically, as well; so looking at maximum horizontal distance means maximum vertical distance."
00:23:58.000 --> 00:24:02.600
"A different function, though, might require different logic; so we have to think specifically
00:24:02.600 --> 00:24:05.300
about the function that we are working with, if we are going to prove something."
00:24:05.300 --> 00:24:08.500
"In this case, though, we see that this method will work."
00:24:08.500 --> 00:24:14.400
"Thus, we can show that our Δ = ε/10 works by showing that, if we use the above x
00:24:14.400 --> 00:24:20.000
(that is x = 3 - Δ and x = 3 + Δ--if they both work) in our function,
00:24:20.000 --> 00:24:27.100
for being within ε of l, if we have that they end up coming out like that,
00:24:27.100 --> 00:24:34.100
we know without a doubt that our Δ will always work, because it worked for the most extreme possible values."
00:24:34.100 --> 00:24:41.200
"We must show that the absolute value of 11 minus f(x)...in this case, if we are plugging in 3 - Δ,
00:24:41.200 --> 00:24:46.500
and our f(x) is x² + 2, we would have the absolute value of 11, our l, minus...
00:24:46.500 --> 00:24:55.400
plugging in our function's maximum, most extreme possible value for x, that is (3 - Δ)² + 2 must be less than ε."
00:24:55.400 --> 00:25:01.100
"Similarly, over here, for our other extreme value for x, x = 3 + Δ,
00:25:01.100 --> 00:25:11.300
if we plug in our f(x), we have the l that we are going to (11) minus the quantity (3 + Δ)² + 2, is less than ε."
00:25:11.300 --> 00:25:18.900
"Those are the most extreme values there; we are doing the absolute value of l minus f(x) must be less than ε."
00:25:18.900 --> 00:25:24.400
"That is what we want to show; so to show that this is the case, we will investigate the left sides of the expressions."
00:25:24.400 --> 00:25:30.900
"Let's work with the 3 - Δ part first; we have |11 - (3 - Δ)² + 2|."
00:25:30.900 --> 00:25:35.200
"Let's begin by expanding 3 minus Δ squared; we expand 3 minus Δ squared, and now,
00:25:35.200 --> 00:25:44.500
we have the absolute value of 11 minus (9 - 6Δ + Δ²); 3 - Δ, squared, gets us 9 - 6Δ + Δ²."
00:25:44.500 --> 00:25:51.500
"We have minus here, and we have 9 and 2 inside; so minus 9, minus 2, cancels out the 11."
00:25:51.500 --> 00:25:59.700
"We are left with -6Δ + Δ², but there is still this minus sign, so it distributes on,
00:25:59.700 --> 00:26:09.100
canceling that and making this negative; and we have the absolute value of 6Δ minus Δ², when we plug in x = 3 - Δ."
00:26:09.100 --> 00:26:16.200
"Over here, on the other one, we have |11 - (3 + Δ)² + 2|."
00:26:16.200 --> 00:26:21.000
"We expand 3 + Δ, squared; we get 9 + 6Δ + Δ²;
00:26:21.000 --> 00:26:27.800
so now we have |11 - (9 + 6Δ + Δ²) + 2|; once again, we are subtracting by what is inside of that."
00:26:27.800 --> 00:26:32.900
"So, the 9 and the 2 go together to combine and take out the 11, leaving us with 6Δ + Δ²."
00:26:32.900 --> 00:26:42.000
"The minus sign distributes, and we now have -6Δ - Δ² if we had plugged in x = 3 + Δ."
00:26:42.000 --> 00:26:50.800
"So, that is what the distance between l and our f(x) is when we plug in our most extreme possible x's, 3 - Δ and 3 + Δ."
00:26:50.800 --> 00:26:55.700
"Continuing from this, we can see that, since 0 is less than Δ, which is less than or equal to 0.1--
00:26:55.700 --> 00:26:59.400
remember: we set that Δ must be less than or equal to 0.1 at the beginning,
00:26:59.400 --> 00:27:03.700
because the largest ε that we were allowed to really care about was ε = 1,
00:27:03.700 --> 00:27:10.700
and we know that Δ = 0.1 works there, so we always kept our Δ's smaller than 0.1, and Δ must be greater than 0."
00:27:10.700 --> 00:27:15.800
"We can write the above as 6Δ - Δ² and 6Δ + Δ²."
00:27:15.800 --> 00:27:21.300
"If you are not quite sure about this, since 6Δ...well, we know that Δ must be greater than 0;
00:27:21.300 --> 00:27:28.500
thus, 6Δ must be positive, and minus Δ²...well, Δ² must be smaller than Δ,
00:27:28.500 --> 00:27:34.900
because Δ is less than or equal to .1; if we square something that is smaller than 1,
00:27:34.900 --> 00:27:39.000
it must end up being smaller than had it not been squared."
00:27:39.000 --> 00:27:45.800
"So, 6Δ - Δ²...well, 6Δ is positive; minus Δ² is negative, but smaller than 6Δ."
00:27:45.800 --> 00:27:52.800
"So, when we take the absolute value of this expression, we are left simply with 6Δ - Δ², even after the absolute value."
00:27:52.800 --> 00:27:59.000
"Over here, we have -6Δ - Δ²; well, that means the -6Δ and the -Δ² are both negative."
00:27:59.000 --> 00:28:06.800
"So, they combine forces and, after the absolute value is finished with them, they will end up coming out as positive, so we will have 6Δ + Δ²."
00:28:06.800 --> 00:28:10.600
"From this, we can create the following inequalities, the right side, in this case,
00:28:10.600 --> 00:28:15.800
being because Δ < 0.1 implies that Δ² < Δ, which we talked about very recently."
00:28:15.800 --> 00:28:26.500
"If we have 6Δ - Δ², well, if we had 7 - 5, that would clearly be less than simply 7, if we got rid of the thing subtracting from it."
00:28:26.500 --> 00:28:32.300
"So, we just get rid of the thing subtracting from it, and we have 6Δ - Δ² < 6Δ."
00:28:32.300 --> 00:28:36.600
"That is certainly true; furthermore, since Δ² is less than Δ,
00:28:36.600 --> 00:28:43.100
we have that 6Δ + Δ² < 6Δ + Δ, which is the same thing as 7Δ."
00:28:43.100 --> 00:28:47.500
"So, we now know that 6Δ + Δ² is less than 7Δ."
00:28:47.500 --> 00:28:55.700
"At this point, we can use our Δ = ε/10 that we originally set; and we have 6Δ = 6ε/10."
00:28:55.700 --> 00:29:00.900
"Remember: that is less than the most extreme value; and 6ε/10 is clearly less than ε."
00:29:00.900 --> 00:29:06.200
"Furthermore, the other most extreme value was going to be less than 7Δ, and we have 7ε/10,
00:29:06.200 --> 00:29:11.300
which is indeed less than ε; thus, we have shown that plugging in our most extreme values
00:29:11.300 --> 00:29:15.900
will show that the distance between f(x) and l is always less than ε."
00:29:15.900 --> 00:29:20.900
"We plug in our most extreme possible value; that gives us our most extreme possible vertical distance,
00:29:20.900 --> 00:29:25.100
and it still ends up being less than ε when we work the whole thing through."
00:29:25.100 --> 00:29:34.000
"So, the method of choosing Δ always works; we see that this Δ = ε/10 will always work; it will never fail."
00:29:34.000 --> 00:29:42.500
"Very good, Holmes; my only question is how you decided on setting Δ equal to ε/10 in the first place."
00:29:42.500 --> 00:29:45.600
"How did you think, 'ε/10; that is the thing for me'?"
00:29:45.600 --> 00:29:49.300
"Indeed, that is the most difficult part of proving a limit."
00:29:49.300 --> 00:29:57.100
"In general, once you have a sense of how the function works, it helps to set it up as if you knew what Δ was."
00:29:57.100 --> 00:30:00.400
"As you work forward, you will find its requirements."
00:30:00.400 --> 00:30:05.200
"For example, with this limit, I realized that whatever Δ was, whatever I ended up choosing for Δ,
00:30:05.200 --> 00:30:12.100
it would end up having to get to the point where we would see that 6Δ + Δ² had to be less than ε."
00:30:12.100 --> 00:30:17.400
"So, if 6Δ + Δ² had to be less than ε, I could restrict Δ < 0.1,
00:30:17.400 --> 00:30:22.800
because a smaller Δ will never cause any harm, and see that 6Δ + Δ²
00:30:22.800 --> 00:30:30.200
would be the same thing as saying that it is less than 7Δ; and 7Δ would certainly work if Δ was ε/10."
00:30:30.200 --> 00:30:33.700
"I could have gone with ε/7, but I figured, 'Why not have a little bit of extra room?'"
00:30:33.700 --> 00:30:36.500
"And ε divided by 10 just seems so nice and round;
00:30:36.500 --> 00:30:42.200
so I saw that Δ = ε/10 would do fine, at which point I was prepared to work through the proof."
00:30:42.200 --> 00:30:48.100
"Ah, I am feeling a mite peckish; nothing like a good proof to develop the appetite--what are you in the mood for, Watson?"
00:30:48.100 --> 00:30:53.800
"Somehow, I find myself craving Greek." "Excellent--good! A brace of gyros it will be!"
00:30:53.800 --> 00:30:56.700
And they make their exit for the neighborhood gyro shop.
00:30:56.700 --> 00:31:00.500
All right, with Sherlock Holmes and Watson having helped us see what is going on here,
00:31:00.500 --> 00:31:04.400
the idea of an ε challenge and a Δ defense, and how to prove this stuff in general,
00:31:04.400 --> 00:31:10.500
by showing that this Δ method will always give us a Δ that works, we are ready to start working through some examples.
00:31:10.500 --> 00:31:16.600
Our first example: Below is the graph of f(x) = xsin(1/x).
00:31:16.600 --> 00:31:22.700
First, we need to determine if the function has a limit, as x goes to 0, and if so, find it and prove it.
00:31:22.700 --> 00:31:27.000
We can see, right from the beginning, that it looks like it is going to 0.
00:31:27.000 --> 00:31:32.900
We can see from this nice, convenient graph that it looks like it is going to end up having a limit.
00:31:32.900 --> 00:31:40.400
It would appear to be the case; however, we notice that we can't simply plug in x = 0.
00:31:40.400 --> 00:31:50.200
If we did that, we would have 0sin(1/0); we are dividing by 0, so f(0) does not exist.
00:31:50.200 --> 00:31:54.800
We can't simply go down that method; we can't simply plug in a number and figure out what is going to come out.
00:31:54.800 --> 00:32:00.800
So, we have to show this by proving it directly; we have to go and see that this is going to always work.
00:32:00.800 --> 00:32:06.200
We have to figure out a way; so we can see from the graph that, yes, it makes sense that it is going to end up coming out to be 0.
00:32:06.200 --> 00:32:11.200
But it will be our job to now prove it through a good proof.
00:32:11.200 --> 00:32:13.900
Let's try to get an understanding of why this thing looks like this.
00:32:13.900 --> 00:32:17.300
We can see, from the graph, that it looks like this; but let's see what is going on.
00:32:17.300 --> 00:32:20.600
What we can do is break this down into multiple pieces.
00:32:20.600 --> 00:32:31.600
We can first see this as sine of...well, let's look, actually, first, at 1/x; what does 1/x graph as?
00:32:31.600 --> 00:32:40.100
Well, 1/x is going to end up graphing something like this; we are going to go asymptotic as we approach that vertical axis.
00:32:40.100 --> 00:32:45.200
Then, how does sine of some value graph?
00:32:45.200 --> 00:32:55.300
Well, sine of some value ends up graphing like this; it has that nice repeating nature.
00:32:55.300 --> 00:32:59.600
If you watched the last lesson, you probably ended up seeing this in the previous lesson.
00:32:59.600 --> 00:33:03.200
We talked about just sin(1/x), and how it went crazy and came really, really close.
00:33:03.200 --> 00:33:06.800
That is the same idea so far; we have that part going on so far.
00:33:06.800 --> 00:33:12.000
If we plug these two things in--if we plug 1/x in for sine of t, then what we are concerned with,
00:33:12.000 --> 00:33:17.800
as we get close to 0, as x goes to 0, is that part that is getting close to that.
00:33:17.800 --> 00:33:27.900
If we end up looking at that, if we look at sin(1/x), we are going to end up getting this part where,
00:33:27.900 --> 00:33:31.400
as it is far away from 0, it is going to end up going sort of slowly.
00:33:31.400 --> 00:33:35.100
But as it gets closer, it is going to speed up, and it is going to start to go crazy.
00:33:35.100 --> 00:33:41.300
And it is going to bounce back and forth between 1 and -1 really, really fast, because it has to go through all of infinity.
00:33:41.300 --> 00:33:49.600
We shoot off to infinity as we get towards 0; so since it is shooting off towards infinity, 1/x is shooting off towards infinity inside of the sine.
00:33:49.600 --> 00:33:55.500
So, the sine is now going to have to speed up faster and faster and faster, because it has to go through all of infinity by the time it makes it to 0.
00:33:55.500 --> 00:34:00.300
We see the same thing from both sides: it shoots up and gets really, really, really, really fast, bouncing up and down.
00:34:00.300 --> 00:34:01.600
And so, that is the behavior we see.
00:34:01.600 --> 00:34:06.200
If it was simply sin(1/x), it wouldn't have a limit, as we talked about in the previous lesson,
00:34:06.200 --> 00:34:10.800
because it is never going to agree on a single location that it is working towards.
00:34:10.800 --> 00:34:19.300
However, in this case, we also have x showing up; well, if we graph just x, what does just x look like?
00:34:19.300 --> 00:34:30.600
Well, that looks just like this; so x times something is going to end up expanding it or shrinking it by whatever x's current value is.
00:34:30.600 --> 00:34:36.400
As x gets closer and closer to 0, it is going to end up squishing it and squishing it and squishing it, closer and closer to that x-axis.
00:34:36.400 --> 00:34:49.500
Now, notice: sine of anything always produces values that are going to be between -1 and +1,
00:34:49.500 --> 00:34:55.200
based on how the unit circle works and what we can see from the graph of just sine.
00:34:55.200 --> 00:34:58.700
So, what we are going to have is either a positive x, effectively, or a negative x.
00:34:58.700 --> 00:35:13.900
So, if we graph in that -x...here is -x in green...then as we plug in x times sin(1/x), it is going to get...
00:35:13.900 --> 00:35:17.300
the maximum value that they can have is whatever the x is that it is next to.
00:35:17.300 --> 00:35:20.500
So, as it gets far away, it ends up bouncing between these things.
00:35:20.500 --> 00:35:24.700
But as it gets closer and closer and closer, it ends up bouncing faster and faster and faster and faster.
00:35:24.700 --> 00:35:30.000
But because there is that x, it shrinks it down and crushes it to go down to 0.
00:35:30.000 --> 00:35:37.400
So, it is not actually going to be defined at x = 0; but it will end up being crushed into this point right here.
00:35:37.400 --> 00:35:49.800
So, with all of these ideas in mind, we can now say that the limit as x goes to 0 of x times sin(1/x) equals 0.
00:35:49.800 --> 00:35:55.000
We figured out that, yes, it does have a limit; it makes sense; we understand what is going on
00:35:55.000 --> 00:35:57.400
in a thing that now makes sense with this graph.
00:35:57.400 --> 00:36:04.900
And notice: you can also see that the x is going out like this in the graph.
00:36:04.900 --> 00:36:12.400
And then, -x goes out here; so this nice, computer-generated graph--we end up seeing the same behavior
00:36:12.400 --> 00:36:20.700
of the sine bouncing infinitely quickly...it gets trapped between +x and -x, because it has to multiply x times sin(1/x).
00:36:20.700 --> 00:36:23.900
So now, we see that this is what our limit ends up being.
00:36:23.900 --> 00:36:27.300
And at this point, we can try to work out a proof here.
00:36:27.300 --> 00:36:34.800
So, if we are going to prove that this is the case, then it must be that, for any ε greater than 0,
00:36:34.800 --> 00:36:44.200
there exists some Δ that will end up making us within that appropriate distance of our vertical ε.
00:36:44.200 --> 00:36:48.000
For any ε greater than 0, what Δ do we choose--what is the appropriate method?
00:36:48.000 --> 00:37:00.800
Well, notice: if we go back to our x (here is a graph of x, once again; and I will also map the -x here, as well...
00:37:00.800 --> 00:37:09.000
sorry; that should be dots all the way through); if we have -x and x mapped here, well, what is the biggest thing that x allows?
00:37:09.000 --> 00:37:14.900
Well, the biggest thing that x is going to allow that sine to get expanded out to is whatever x currently is.
00:37:14.900 --> 00:37:18.600
For any ε greater than 0, if we want to talk about the maximum vertical size we can allow,
00:37:18.600 --> 00:37:24.200
then we can allow x to go out as far as that ε, and that will be the maximum allowed vertical size.
00:37:24.200 --> 00:37:31.500
So, we can simply say that our Δ (since that is how far out our x can go from 0) is simply equal to ε.
00:37:31.500 --> 00:37:40.400
We set Δ = ε now, it is going to be our job...what we want to show now is that,
00:37:40.400 --> 00:37:56.100
for any f(x) - l, it is going to be less than ε if we have our x within Δ distance of our c.
00:37:56.100 --> 00:37:57.600
All right, so let's start working through this.
00:37:57.600 --> 00:38:06.000
We can swap out what f is and what l is, and we can swap things out, and we can start working and showing that this is true.
00:38:06.000 --> 00:38:17.400
f(x) - l...what is the left side there? f(x) is this thing right here, x times sin(1/x), minus l...
00:38:17.400 --> 00:38:24.000
well, l is just 0 in this case, so minus 0...it just closes up, and we have that it is less than ε.
00:38:24.000 --> 00:38:28.600
So, at this point, we can look at this and say, "Well, x times sin(1/x)...the absolute value of all of that
00:38:28.600 --> 00:38:33.000
is just going to be the same thing as the absolute value of x, times the absolute value of sin(1/x)."
00:38:33.000 --> 00:38:36.100
If we strip any possible negative signs after they have multiplied,
00:38:36.100 --> 00:38:39.900
or strip any possible negative signs before they multiply, it is not going to have an effect on the product.
00:38:39.900 --> 00:38:41.500
So, we have that this thing right here...
00:38:41.500 --> 00:38:46.100
Oh, we don't know that it is less than ε; what we want to show is that it comes out to be less than ε.
00:38:46.100 --> 00:38:48.400
We want to work towards showing that; I'm sorry about that.
00:38:48.400 --> 00:38:58.000
The absolute value of xsin(1/x) is just the same thing as |x| times |sin(1/x)|.
00:38:58.000 --> 00:39:01.000
Well, what is the absolute value of sin(1/x)?
00:39:01.000 --> 00:39:15.100
Well, the absolute value of sin(1/x)...the absolute value of sine of anything at all is going to be less than or equal to 1.
00:39:15.100 --> 00:39:20.500
You can't get larger than 1 or larger than -1 when you plug something into sine.
00:39:20.500 --> 00:39:25.100
The absolute value of sin(1/x) is going to still be less than or equal to 1,
00:39:25.100 --> 00:39:29.200
except for that specific value of x = 0, which would cause the whole thing to break.
00:39:29.200 --> 00:39:34.800
But we don't have to worry about that, because it is x going to 0; so we have 0 < x - c.
00:39:34.800 --> 00:39:40.600
So, we don't have to worry about x = 0; so we know that sin(1/x) ≤ 1 will always be true,
00:39:40.600 --> 00:39:43.300
because we don't have to worry about actually plugging in x = 0.
00:39:43.300 --> 00:39:51.300
So, we can now use this information up here; and we have that |x| times |sin(1/x)| is less than or equal to...
00:39:51.300 --> 00:39:57.600
we swap out that fact that sin(1/x) is less than or equal to 1, and we have |x| times 1.
00:39:57.600 --> 00:40:03.600
So, |x| times |sin(1/x)| is less than or equal to |x| times simply 1.
00:40:03.600 --> 00:40:07.300
We have that that is just the same thing as saying x; great.
00:40:07.300 --> 00:40:12.000
Now, at this point, we also know that x - c is less than Δ.
00:40:12.000 --> 00:40:16.100
Let's use yet another color here; so...well, what is our c?
00:40:16.100 --> 00:40:19.900
Our c is equal to 0 in this case, because that is what we are going towards.
00:40:19.900 --> 00:40:26.700
So, we have that the absolute value of x - 0...well, we will just leave it as the absolute value of x...has to be less than Δ.
00:40:26.700 --> 00:40:30.900
The absolute value of x is less than Δ; we can now swap that in here.
00:40:30.900 --> 00:40:36.800
And we have that the absolute value of x is less than Δ.
00:40:36.800 --> 00:40:42.600
What did we set our Δ to? We know that Δ is equal to ε, because that is what we decided to make our method.
00:40:42.600 --> 00:40:49.800
We have that that is equal to ε; so at this point, we have now shown that the absolute value of x,
00:40:49.800 --> 00:40:59.200
times the sine of 1/x, gets to being less than or equal to x, which we know is less than Δ, which...
00:40:59.200 --> 00:41:02.200
Δ is equal to ε, but at this point we can stack together our signs.
00:41:02.200 --> 00:41:06.500
The most extreme sign that we ended up having was that direct strictly less than;
00:41:06.500 --> 00:41:17.700
so at this point, we now see that the absolute value of x times sin(1/x) is, indeed (close that absolute value), less than ε.
00:41:17.700 --> 00:41:21.200
So, that always ends up being true, if we use this method.
00:41:21.200 --> 00:41:25.300
We have now completed our proof; this limit does, indeed, work out.
00:41:25.300 --> 00:41:30.500
We have shown that whatever ε we end up choosing, there will always be an appropriate Δ,
00:41:30.500 --> 00:41:35.800
if we simply set Δ = ε it will end up satisfying any and every ε.
00:41:35.800 --> 00:41:39.700
Thus, we have completed the proof; the limit does exist; the limit is, indeed, 0.
00:41:39.700 --> 00:41:43.500
Awesome; all right, the second example (and there are only going to be two of these examples,
00:41:43.500 --> 00:41:46.200
total, because they clearly take a little while to get through):
00:41:46.200 --> 00:41:52.300
consider the piecewise function f(x) = 1 when x is rational, -1 when x is irrational.
00:41:52.300 --> 00:41:54.900
Prove that f(x) has no limit anywhere.
00:41:54.900 --> 00:42:04.400
The very first thing we want to do here, before we try to prove anything, is understand just what f(x) = 1 when x is rational, -1 when x is irrational, means.
00:42:04.400 --> 00:42:07.800
First, we need to remember what it means for a number to be rational.
00:42:07.800 --> 00:42:19.800
A number is rational when it can be put as some a/b, where a and b are integers; that is what it means for a number to be rational.
00:42:19.800 --> 00:42:24.100
A number is irrational when it can't be expressed as a rational.
00:42:24.100 --> 00:42:31.100
A rational means any decimal number that has a fixed length to it, and also decimals that end up having repeating patterns.
00:42:31.100 --> 00:42:35.000
But it will actually be enough, with just a fixed length, to understand what is going on here.
00:42:35.000 --> 00:42:41.900
2.000000001, stop, is a rational number.
00:42:41.900 --> 00:42:45.900
An irrational number is a decimal number, or something that can be expressed as a decimal number,
00:42:45.900 --> 00:42:49.900
where the decimals continue going on forever; they never establish a single pattern,
00:42:49.900 --> 00:42:52.400
and they just constantly are changing forever and ever and ever.
00:42:52.400 --> 00:43:04.500
For example, π is 3.1415...stuff going on forever; √2, e...we have seen some irrational numbers that are pretty important.
00:43:04.500 --> 00:43:10.200
But they are also everywhere total; imagine...we had that rational at 2.000000001,
00:43:10.200 --> 00:43:15.400
but we could also have an irrational that is right next to it, practically, at 2.000000001...
00:43:15.400 --> 00:43:28.300
and then a random string forever and ever, like 158234671111583...it just keeps changing and changing and changing.
00:43:28.300 --> 00:43:32.500
So, we can have an irrational number right next to any rational number.
00:43:32.500 --> 00:43:35.000
We can get as close as we want to any rational number.
00:43:35.000 --> 00:43:41.500
And we can get as close as we want to any irrational with a rational, by just putting as many decimals as we want, and then stopping at a certain point.
00:43:41.500 --> 00:43:46.700
So, the rationals and the irrationals are right next to each other, everywhere total.
00:43:46.700 --> 00:43:52.500
Every single place on the number line has a rational and then an irrational, effectively, infinitely close to it;
00:43:52.500 --> 00:43:58.000
and then, a rational right next to that and an irrational right next to that...everything is just infinitely mashed together.
00:43:58.000 --> 00:44:02.400
2 isn't both a rational and an irrational; 2 is simply rational.
00:44:02.400 --> 00:44:08.600
But right next to it is an irrational; you can get as close to it as you want, with 2.0000000...random pattern.
00:44:08.600 --> 00:44:17.800
And then, you can get as close as you want to 2.000000...random pattern with a rational, like 2.0000000000000, stop.
00:44:17.800 --> 00:44:23.700
In matching the random pattern 2.00000000....match the first 50 thousand digits of the random pattern,
00:44:23.700 --> 00:44:28.200
and then stop, and that is still going to be a rational number, because it is technically an integer divided by an integer.
00:44:28.200 --> 00:44:36.600
It will be a large integer divided by a large power of 10, but it is still technically an integer over an integer, so it is going to be a rational number.
00:44:36.600 --> 00:44:41.700
What that means is that rationals and irrationals are right next to each other everywhere.
00:44:41.700 --> 00:44:45.200
If we tried to graph this, what would f(x) end up looking like?
00:44:45.200 --> 00:44:52.700
Well, we would end up having some that looked like this, where, let's say, we use blue when one x is rational;
00:44:52.700 --> 00:44:59.700
so we are going to end up having...everything that ends up being a rational will come out at 1.
00:44:59.700 --> 00:45:03.300
There are always going to be points, but between any point and the next point over,
00:45:03.300 --> 00:45:06.700
it is going to end up having a little, tiny hole for the irrationals.
00:45:06.700 --> 00:45:09.300
And where do the irrationals show up? Well, they show up at -1.
00:45:09.300 --> 00:45:14.500
So, we end up having the same thing here; down below them are our irrationals.
00:45:14.500 --> 00:45:20.200
And so, there are a bunch of irrationals; there are infinitely many irrationals down there; there are infinitely many rationals up here.
00:45:20.200 --> 00:45:22.300
And they both are just right next to each other.
00:45:22.300 --> 00:45:28.000
You are constantly jumping back and forth and back and forth and back and forth and back and forth, if we were to try to plot this thing out.
00:45:28.000 --> 00:45:31.400
It is certainly not continuous, because it is two things simultaneously.
00:45:31.400 --> 00:45:37.400
So, why can't we have a limit? Now let's try to think about why there will be no limit anywhere.
00:45:37.400 --> 00:45:46.300
Well, consider: if we tried to talk about some location c, then that means that that location c would have to be in either the top or the bottom.
00:45:46.300 --> 00:45:50.000
And if it ended up being in either the top or the bottom--let's just say the top for argument's sake--
00:45:50.000 --> 00:45:56.700
then that means that we can put some ε boundary that is small enough to contain just one of these.
00:45:56.700 --> 00:46:05.000
We make ε equal to a half or something, so that it can just contain one of these height strings, one of these lines.
00:46:05.000 --> 00:46:10.600
So, since it can only contain one of these lines, no matter what Δ we end up choosing,
00:46:10.600 --> 00:46:19.100
if we choose any Δ, then we are going to end up having points that are both on the bottom and on the top.
00:46:19.100 --> 00:46:23.400
So, if we were in the top, we can make ε only contain the top;
00:46:23.400 --> 00:46:27.200
but whatever Δ size we choose, since the rationals and the irrationals are right next to each other,
00:46:27.200 --> 00:46:30.700
if we have some rationals, we have to have some irrationals with it, as well.
00:46:30.700 --> 00:46:34.600
If we have some irrationals, we have to have some rationals with it, as well.
00:46:34.600 --> 00:46:40.500
If you catch this stuff, it has both types, no matter what, if you take any interval chunk of it.
00:46:40.500 --> 00:46:46.400
So, since Δ will always produce some interval chunk, it has to have both rationals and irrationals in that interval chunk.
00:46:46.400 --> 00:46:50.400
We grab a chunk, and it is going to have both the top and the bottom,
00:46:50.400 --> 00:46:54.600
even though we have restricted our ε to only contain either the top or the bottom.
00:46:54.600 --> 00:47:00.200
And this is the idea of why there can be no limit anywhere: because we can always choose an ε
00:47:00.200 --> 00:47:04.900
that is going to only take either the top or the bottom line, but no matter what Δ we end up choosing,
00:47:04.900 --> 00:47:07.400
we are going to have to be in both the top and the bottom lines.
00:47:07.400 --> 00:47:15.500
So, we won't be stuck just inside of that ε it will fail to end up having a limit.
00:47:15.500 --> 00:47:18.900
If we are going to actually prove this formally, that will be a little bit difficult.
00:47:18.900 --> 00:47:21.200
But understanding the idea is 90% of the battle.
00:47:21.200 --> 00:47:25.900
I would much prefer that you understand this idea, and the proof doesn't make sense,
00:47:25.900 --> 00:47:29.500
than that you really, really work hard through the proof and have a vague understanding of doing the proof,
00:47:29.500 --> 00:47:31.600
but you don't understand the idea of the limit.
00:47:31.600 --> 00:47:38.200
At this point (and probably ever), the most important thing, by far, is understanding the general idea,
00:47:38.200 --> 00:47:41.700
and having a sense of what is going on, rather than being able to do the specific mechanics.
00:47:41.700 --> 00:47:45.600
The specific mechanics can always come later; understanding the idea is the goal.
00:47:45.600 --> 00:47:48.600
All right, but let's work through those specific mechanics.
00:47:48.600 --> 00:47:51.400
OK, proof by contradiction; how do we show this?
00:47:51.400 --> 00:47:59.500
We will show by assuming it is possible; and then we will see that, if we were to assume that it is possible, crazily impossible things will occur--
00:47:59.500 --> 00:48:04.700
that if this were true, that we did have a limit, then it won't work--it just doesn't make any sense.
00:48:04.700 --> 00:48:09.800
And so, we will have contradicted it, and we will know that it must be the case that the limit cannot exist.
00:48:09.800 --> 00:48:20.000
So, what we do is begin by assuming (there will be a lot of writing for this; I'm sorry)
00:48:20.000 --> 00:48:29.100
that there exists some limit that as x goes to c; our f(x) is equal to some value of l.
00:48:29.100 --> 00:48:37.300
Then, for any ε greater than 0, by the definition of how a limit works,
00:48:37.300 --> 00:48:45.600
we know that there exists some Δ greater than 0 such that...
00:48:45.600 --> 00:48:50.600
and from here on, when I write "such that," I will probably just use s.t. if I have to write that again...
00:48:50.600 --> 00:49:03.500
such that, for any 0 < |x - c| < Δ--that is to say, an x that is less than Δ distance
00:49:03.500 --> 00:49:11.600
away from whatever our c happens to be--we know that it must be the case that the absolute value of f(x) - l,
00:49:11.600 --> 00:49:19.500
that is, the distance between the f(x) that gets mapped from that x and l, what our limit goes to--has to be less than ε.
00:49:19.500 --> 00:49:23.700
So, that has to be true; that is what it means for it to have a limit.
00:49:23.700 --> 00:49:29.200
If there exists a limit, then it must be that for any ε greater than 0, there will always exist a Δ greater than 0,
00:49:29.200 --> 00:49:38.000
such that if x is within Δ of c, f(x) must be within ε of l; that is what the formal definition is.
00:49:38.000 --> 00:49:43.900
Now, let us consider (let me actually change colors, because now we are changing ideas)...
00:49:43.900 --> 00:49:55.300
that was the setup to this thing: the next thing is: Consider ε = 1/2.
00:49:55.300 --> 00:50:03.500
Therefore, there must exist, by this thing up here (for any ε greater than 0, there exists a Δ greater than 0)...
00:50:03.500 --> 00:50:10.300
there exists a Δ greater than 0 such that it does this thing here.
00:50:10.300 --> 00:50:16.300
There exists Δ greater than 0, such that it follows this idea right here.
00:50:16.300 --> 00:50:22.200
OK, now, let's continue to talk about this ε equal to 1/2.
00:50:22.200 --> 00:50:41.900
Notice: by the absolute value of f(x) - l being less than ε, being less than 1/2, it must be either...
00:50:41.900 --> 00:50:46.700
well, what are the possible values that can come out of f(x)? 1 and -1.
00:50:46.700 --> 00:50:58.300
It must be either |1 - l| < 1/2, or |-1 - l| < 1/2.
00:50:58.300 --> 00:51:06.500
The distance that our f(x) is from l must be less than 1/2; so either 1 - l is less than 1/2, or -1 - l is less than 1/2.
00:51:06.500 --> 00:51:11.400
That is what |f(x) - l| < 1/2 has to come out meaning.
00:51:11.400 --> 00:51:25.500
But also notice: only one of these can be true.
00:51:25.500 --> 00:51:34.100
These two ideas right here, |1 - l| < 1/2 and |-1 - l| < 1/2, can't both be true,
00:51:34.100 --> 00:51:41.600
because if our l is within 1/2 of -1, it can't possibly also reach to being 1/2 of 1.
00:51:41.600 --> 00:51:47.500
And if our l is within 1/2 of 1, it can't possibly also be within 1/2 of -1.
00:51:47.500 --> 00:51:55.100
It has to pick one of the two bands; our l can't be 1/2 away from both bands, because 1 and -1 are two apart.
00:51:55.100 --> 00:52:03.000
If you are within one of them, you can't be near both of them; so it can't be near both of them.
00:52:03.000 --> 00:52:09.200
Only one of these can be true; we are going to use that fact very soon.
00:52:09.200 --> 00:52:21.200
Now, we go on to say, "Well, there existed some Δ greater than 0 such that it made 0 < |x - c| < Δ, which means that |f(x) - l| < ε."
00:52:21.200 --> 00:52:35.300
So, by our Δ, we know that 0 is less than x - c, which is less than Δ,
00:52:35.300 --> 00:52:46.000
and that that ends up having to mean that the absolute value of f(x) - l has to be less than our 1/2, this thing right here.
00:52:46.000 --> 00:52:53.200
We know that our Δ has to end up making our f(x) within 1/2 of l, since we set our ε as 1/2.
00:52:53.200 --> 00:53:20.400
But 0 < |x - c| < Δ--that is to say, x is within Δ of c--must contain rational and irrational x.
00:53:20.400 --> 00:53:25.600
Since we know that Δ has to be greater than 0, and our x is within Δ of c--
00:53:25.600 --> 00:53:44.600
it is within some interval of c - Δ to c + Δ--well, any interval...remember: rationals and irrationals are right on top of each other;
00:53:44.600 --> 00:53:55.300
then that means our interval, this interval here, the one set up by 0 < |x - c| < Δ, must contain both rational and irrational x-values.
00:53:55.300 --> 00:54:06.100
If it contains both rational and irrational x-values, we know, because it has to be true for any x in this interval,
00:54:06.100 --> 00:54:12.600
and since we contain both rational and irrational, we now have 1 and -1 popping out of our f(x).
00:54:12.600 --> 00:54:29.500
Thus, the absolute value of 1 - l is less than 1/2, and the absolute value of -1 - l is less than 1/2 must be true simultaneously,
00:54:29.500 --> 00:54:35.300
because we know that, within our Δ bound, we are going to have both rationals and irrationals.
00:54:35.300 --> 00:54:39.400
And anything within that Δ bound, by the definition (for any ε greater than 0,
00:54:39.400 --> 00:54:44.100
there exists Δ greater than 0, such that, if you are contained within that boundary of Δ,
00:54:44.100 --> 00:54:48.700
it must map within that boundary of ε) that we are contained within that boundary of Δ;
00:54:48.700 --> 00:54:53.500
we are contained within c - Δ to c + Δ; but there are going to be rationals and irrationals in there.
00:54:53.500 --> 00:54:58.700
So, that means that our ε boundary must contain both 1 and -1.
00:54:58.700 --> 00:55:03.900
So, the absolute value of 1 - Δ must be less than 1/2, and the absolute of -1 - Δ must be less than 1/2.
00:55:03.900 --> 00:55:11.800
But only one can be true; this and this are mutually exclusive, because you can't be in both bands at once,
00:55:11.800 --> 00:55:17.200
if the band can only be a total distance of 1 wide around positive 1 or around -1.
00:55:17.200 --> 00:55:25.900
1/2 down and 1/2 up don't manage to reach each other; they can't reach across, so you can't end up being in both bands at once with that ε of 1/2.
00:55:25.900 --> 00:55:30.400
So, it says that they must be true simultaneously; but we know that only one can be true.
00:55:30.400 --> 00:55:36.100
Since only one can be true, and we know that they have to be true simultaneously, those are impossible statements to be true at the same time.
00:55:36.100 --> 00:55:40.500
We have a contradiction; there is a contradiction in our logic.
00:55:40.500 --> 00:55:47.500
And because we know everything else in our logic was flawless, except for our assumption that there existed some limit,
00:55:47.500 --> 00:56:05.300
we know, therefore, that no Δ can exist that will satisfy ε equal to 1/2; so no limit can exist.
00:56:05.300 --> 00:56:07.300
And we have completed our proof.
00:56:07.300 --> 00:56:12.000
All right, I think that is really cool; it is not for everyone, but I think it is so beautiful.
00:56:12.000 --> 00:56:14.500
So, I hope you enjoyed it; I hope you get a cool sense of this.
00:56:14.500 --> 00:56:20.800
And remember: what we talked about here is not something that is necessary for virtually any students--
00:56:20.800 --> 00:56:23.000
certainly no student currently at a precalculus level.
00:56:23.000 --> 00:56:26.100
If you are just in the level of this course, you don't need to know this stuff.
00:56:26.100 --> 00:56:28.800
I am honest; I won't have to use this stuff at all.
00:56:28.800 --> 00:56:35.600
It is going to maybe, maybe, maybe end up showing up when you take calculus, but probably not even then.
00:56:35.600 --> 00:56:38.000
It is really something that you only need if you are getting into advanced math.
00:56:38.000 --> 00:56:42.400
However, I think that this stuff is so cool; and there is no reason that, if you end up liking this stuff...
00:56:42.400 --> 00:56:49.100
it is like going and checking out art for me; if you end up liking this stuff, you might as well check it out any time that you are able to enjoy it.
00:56:49.100 --> 00:56:52.800
It is really cool stuff; it gets your brain thinking in all of these totally new ways.
00:56:52.800 --> 00:56:55.700
And you can end up exploring this stuff for years and years and years.
00:56:55.700 --> 00:56:58.000
You can make an entire career out of exploring this stuff.
00:56:58.000 --> 00:57:05.000
If you think this is kind of cool, you can study mathematics in college, go on to become a mathematician, and just study math for the rest of your life.
00:57:05.000 --> 00:57:08.400
There is a whole bunch of other stuff to explore; math is really, really cool.
00:57:08.400 --> 00:57:11.000
All right, see you at Educator.com later--goodbye!