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It is not unusual in calculus textbooks to see something-- to

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see problems along the lines of, prove that the limit as x

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approaches-- I don't know, say, as x approaches 1.

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Let me think of an interesting function when x approaches 1.

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Let's see.

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3x times x minus 1 over x minus 1.

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And you might say, hey Sal, why did you do this x minus 1 in

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the numerator and denominator?

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Well, I did that because obviously you can't

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have a 0 down here.

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So this function is actually not defined as x is

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equal to 1, right?

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If I had just put a 3x, this cancels out for

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all the other values.

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So this is essentially the line 3x, except it has a

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hole at x is equal to 1.

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And that's why I did it.

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Because I want to take the limit, and it's interesting to

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take a limit a to a point where the function doesn't exist.

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But the books will say, hey, prove that this limit is equal

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to-- well, it would actually be equal to 3, right?

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Prove that.

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And you essentially have to use the epsilon delta

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definition of a limit.

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So to prove this, you're essentially proving

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what this means.

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And this means-- proving this is the same thing as

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proving-- so you could just prove that given--.

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So if you give me some epsilon greater than 0-- and remember,

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that's how close you want f of x to get to the limit.

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Given some epsilon greater than 0, in order for this to be

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true, I need to prove that I can give you-- that there

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exists-- that, you know, I can give you a delta

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greater than 0.

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Where as long as the distance between x and our limit point,

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x and 1, is less than delta-- remember, we don't want x to

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ever get right on top of the limit point.

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Because a function might not be defined there.

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So the distance between x and 1 has to be

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greater than 0, right?

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That's what this tells us.

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So I have to prove, you give me an epsilon, that

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I can give you a delta.

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And as long as x is within delta of 1-- and this is all

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this is saying-- as long as the distance between

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x and 1, right?

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That's when you subtract 2 things and take

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the absolute value.

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That's just distance.

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So as long as the distance between x and 1 is less than

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delta then the distance between 3x-- let me write that up

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there-- the distance between 3x times x minus 1 over x minus 1,

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that's the function, right?

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The distance between that and the limit point-- and 3--

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is going to be less than the epsilon.

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A lot of calculus text books, you know, they'll

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write it generally.

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They're like, the limit of f of x as x approaches a of equals

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l, and so this would be a.

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This would be your f of x, and this would be

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your l, right there.

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But anyway, when someone says they want you to prove this,

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they essentially want you to prove this.

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I'll give you an epsilon.

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You have to prove that any epsilon that you're given, or

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that you give me, that I-- let's say I'm the one who has

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to prove it-- any epsilon you give me, I have to be able to

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give you a delta where, as long as x is within delta of 1,

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then the function is within epsilon of the limit point.

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And how do we do that?

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And I'll just say, you know, a lot of proofs, it's just kind

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of-- there's not a-- well, there's not a systematic

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way to do proofs.

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And that's kind of what makes them so interesting

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on a lot of levels.

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But these, there kind of tends to be a systematic way.

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And this problem I've given you, to some degree, is kind of

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the easy one that they'll give when they ask you

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to prove these.

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But they usually involve starting from this point right

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here and algebraically manipulating it until you get--

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until this expression right here looks something

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like this expression.

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And then you'll have this expression is less

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than, you know, epsilon divided by something.

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Or, you know, some function of epsilon.

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And it's like, oh, well as long as delta is that, you give me

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an epsilon, I can just apply that, you know, take epsilon

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and divide it by 4, or whatever.

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And use that as my delta.

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Then I've proved the theorem.

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And so we'll do that here.

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So let's start with this.

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Let's start with kind of where we want to get to.

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And then figure out a delta that's essentially a

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function of epsilon.

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So that whenever I'm given an epsilon, I can say, hey,

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whatever number you give me, I'm going to give you back

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a number that's, you know, that divided by whatever.

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And then this will all work.

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So let's do that.

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So immediately we can simplify this.

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We can cross these out.

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And when you cross that out, you still have to say, well, we

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can't do that as long as x-- you can only do this as long as

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you assume that x does not equal 1.

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Because if x equals 1, this function is undefined, right?

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And that's OK because we only care about the intervals

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as x approaches 1.

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We only care about the intervals as x approaches 1.

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We don't care exactly when x is equal to 1.

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So that's fine.

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So then this simplifies to-- this right there simplifies

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to-- the absolute value of 3x minus 3 is less than epsilon.

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And then we could factor out a 3, so that's, you know, we

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could say the absolute value of 3 times x minus 1 is

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less than epsilon.

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And you might already see, we already have an x minus 1

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there and an x minus 1 there.

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If we can get this 3 over on this side, then we'll

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be done, essentially.

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So the absolute value of 3 times some other number, it's

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going to be the same thing as the absolute value of 3 times

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the absolute value of x minus 1.

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And if you don't believe me, I mean, try it out with

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a bunch of numbers.

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This is going to be positive no matter what.

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This is going to be positive no matter what.

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It's going to have the same magnitude, right?

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And of course, that is less than epsilon.

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And this is what we say-- you know, a lot of times a calculus

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teacher will say, if this is true, this is true, if and

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only if this is true.

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And that's just a fancy way of saying.

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So sometimes they'll do a two way arrow.

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Sometimes they'll write IF and only IF, kind of an IFF.

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An IF with 2 f's.

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But that just means if that's true, that's true.

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And if this is true, then that's true.

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And that's generally the case, whenever you just do some

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algebraic manipulation.

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You can manipulate this back to get that, or that to get that.

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And then they'll say, well, if this is true, this is true

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if and only if this is true.

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But this is just algebraic manipulation.

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Right?

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Because you can go back and forth between these steps.

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And that's true, if and only if that is true.

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And then the absolute value of 3 is just 3, right?

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We know that.

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Let's divide both sides by 3 and you get-- well, I can

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just get rid of the absolute value signs right there.

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Because we know that's 3.

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If you divide both sides of the equation by 3, you get the

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absolute value of x minus 1 is less than epsilon over 3.

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So what have we just done so far?

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We've just proved that the absolute value-- I need to

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write this in another color.

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We've just proved that the absolute value of essentially

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3x times x minus 1 over x minus 1 minus 3 is less than epsilon,

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if and only if x minus 1 is less than epsilon over 3.

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Right?

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So it's like, this is true, as long as the distance between x

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and 1 is less than epsilon over 3.

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So we can just use epsilon over 3 as our delta.

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Remember the whole point of this was like, you give me

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an epsilon, you give me a distance, or you say, I want

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to be within this distance of my limit point.

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I want the function to be that close to the limit point.

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And this is just the distance between the function and

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the limit point, right?

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This is f of x.

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And this is the limit that it approaches.

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So you want to be that close.

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And I say, well, you're going to be that close if and only if

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x-- the distance between x and 1, or the distance between x

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and the value it's approaching-- is

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less than this.

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So we just prove this algebraically.

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So whatever number you give me-- if you give me a-- if

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you say, Sal, I want to be within 1 of the limit point.

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I want f of x to be no further than 1 from the limit point,

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then I'll say, OK, and I'll give you 1/3, right?

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Because this is true.

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Let me write that out.

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So let's say you pick 1 as an epsilon.

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So you say, hey, you need to prove to me that there's some x

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value where, as long as the x value is no more-- there's

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some value of delta.

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So as long as we're no more than delta away from 1, then

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the function itself will be no more than 1 away

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from the limit value.

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Right?

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And then I say, well, you, gave me an epsilon

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that's equal to 1.

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So I could just make-- I can just say this is, you know, we

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just proved, if and only if absolute value of x minus 1 is

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less than whatever number you gave me divided by 3.

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And that actually makes a lot of sense if you

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graph this equation.

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Let me see.

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Let me draw the x-axis, y-axis.

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We already established.

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This thing looks just like the equation of 3x except it has

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a hole at x is equal to 1.

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Let me draw the graph.

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It'll have slope.

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It'll have the slope of 3.

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So it'll be pretty steep.

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It'll look something like that.

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And when we're at 1-- this point right

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here-- there's a hole.

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Right?

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This is 3.

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So all we're saying, we just proved that-- so if you said

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that you wanted to be within 1 of our limit point, that's like

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you giving me an epsilon of 1.

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So this distance right here is 1.

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Which would be a pretty big number.

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So you want the function to be within 1 of our

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limit point, right?

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Because this is the limit point.

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Then all we're saying is, you just have to take 1/3 of that.

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So we just have to be within 1/3-- so this distance right

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here is 1/3 away from 1.

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Or we could have said it more generally, this is epsilon.

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This is epsilon.

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This would be epsilon over 3.

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This is epsilon over 3.

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And the reason why this is a proof, is because it shows no

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matter what epsilon you give me, I can always find a delta.

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Because I can just take whatever number you gave

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me, and divide it by 3.

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And I can divide any real number greater than 0.

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I can divide any real number by 3.

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So I can always-- no matter what you give me-- I can always

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find you something else.

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And therefore, I have proven by the epsilon delta definition of

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limits that the limit as x approaches 1, of

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this, is equal to 3.

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