AP®︎/College Calculus AB
Explore the epsilon-delta definition of limits in calculus, as we rigorously prove a limit exists for a piecewise function. Dive into the process of defining delta as a function of epsilon, and learn how to apply this concept to validate limits with precision. Created by Sal Khan.
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- the video was really awesome, thank you so much
just a little question:
is this "ε = 2∂" true for any kind of function or just this example??(81 votes)
- Just this function! Because he defined f(x) to be 2x when x is not 5. Another function f(x) will yield a different function for ε.(160 votes)
- At3:20-3:35, Sal says that x is not equal to 5 as if the expression |x-c| < ∂ exactly implies that x is within range of c but not equal to c. For the latter proposition, shouldn't there also be a 0 < at left-hand side of the inequality ? ( 0 < |x-c| < ∂ ). Because we are trying to approach c, not to get there, given the definition of limit.(31 votes)
- the inequality shown @3:25will include c. The inequality should read:
0<|x-c|<delta. When you remove the absolute value braces to evaluate
this expression, you will create 2 inequality expressions: one describing
the range x<c, the other the range x<c. The value x=c is excluded.(3 votes)
- where did the 5 come from in |x-5| ?(10 votes)
- In the last video, Sal wrote down the equation |x - c| < ∂, which means x is within ∂ of c. |x - 5| < ∂ is the same thing, but we are defining c to be 5.(24 votes)
- What would it look like if the limit you were trying to prove was actually wrong? For example, if you tried to prove that for f(x) = x^2, that as x => 3 that L = 10.(11 votes)
- To disprove a limit, we can show that there is some ∈>0 such that there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.
Let's say ∈<1 (because 3^2=9 and |9-10|=1).
We can always pick c=3 so that |x-c|<δ (because |x-c|=0 and 0<δ), but |f(c)-L|>∈ (because |f(c)-L|=|9-10|=1 and 1>∈).
This means that if ∈<1, there is no δ>0 such that for all c such that |x-c|<δ, |f(x)-L|<∈.
Thus, we have disproved that the limit of x^2 as x approaches 3 is 10.
I hope this helps you disprove limits with the epsilon-delta definition!(17 votes)
- What if it's a function that grows in a non linear way, like in an exponential function?
For exemple, in the function f(x) = 2^x. Given E > 0, would the delta in the left side of a value c be different from the delta in the right side of c?(16 votes)
- What does he mean when he writes, '|x - c| < delta'?(3 votes)
- This is a formal way of writing that the difference between x and c is less than some extremely small number, delta.(15 votes)
- Why should we prove that for all epsilon if we have a delta then the limit at that point (at which we have to prove the limit) is going to be equal to L(Here L =limf(x) x->a). We can just take the casewhen delta->0 and see whether the epsilon->0.
If epsilon->0 then we can say that within the range of a+delta and a-delta for every x(such that x not equal to a) the value of f(x) approaches L or is approximately equal to L. Isn't that what a limit is?(2 votes)
- Well, if x<y and x<z how y becomes equal to z. Such as if 2<3 and 2<4 than 3<4. Shouldn't the equation be y=kz.(4 votes)
- I'll try to explain it another way.
We want to choose such an y so that x < y. We know that there is a z for which x < z. Therefore we can choose y = z, because z fits the description.(8 votes)
- I understand the relationship/back and forth between epsilon ranges and delta ranges, but I'm a little bit confused about how this definitively proves the limit of a function? Maybe I'm missing something very obvious and this question may be vague. Thanks.(3 votes)
- It proves it in a clever way. Here's how I think about it.
When you were first introduced to limits, you were taught that they were numbers a function approached when the input came very close to another number. For example, if f(x) = x+3, as x gets closer to 3, f(x) gets closer to 6. Now, the problem is that "getting very close" isn't very rigorous. Like, how close? 1 unit? 1/10th of a unit? 1/10000th of a unit? That's where epsilon-delta comes in.
Epsilon-delta essentially says that for any range of epsilon values you choose, I can give you a range of delta values such that any x within the "delta range" will give an f(x) within the "epsilon range". See that this completely solves our problem! Instead of having to prove the limit as we get closer to the input, we've proven it for every number in the range. You want to switch the epsilon range? No problem, because I can give a corresponding delta range.
See if this made sense.(8 votes)
In the last video, we took our first look at the epsilon-delta definition of limits, which essentially says if you claim that the limit of f of x as x approaches C is equal to L, then that must mean by the definition that if you were given any positive epsilon that it essentially tells us how close we want f of x to be to L. We can always find a delta that's greater than 0, which is essentially telling us our distance from C such that if x is within delta of C, then f of x is going to be within epsilon of L. If we can find a delta for any epsilon, then we can say that this is indeed the limit of f of x as x approaches C. Now, I know what you're thinking. This seems all very abstract. I want to somehow use this thing. And what we will do in this video is use it and to rigorously prove that a limit actually exists. So, right over here, I've defined a function, f of x. It's equal to 2x everywhere except for x equals 5. So it's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. So I could have really just written 5. It's equal to 5 when x is equal to 5. It's equal to itself. And so we've drawn the graph here. Everywhere else, it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. And if I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. Well, let's see. The closer I get to 5, the closer f of x seems to be getting to 10. And so, you might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. But what we're going to do is use the epsilon-delta definition to actually prove it. And the way that most of these proofs typically go is we define delta in the abstract. And then we essentially try to come up with a way that given any epsilon, we can always come up with a delta. Or another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. But maybe I shouldn't use f again. But delta equals function of epsilon that is defined for any positive epsilon. So you give me an epsilon, I just put into our little formula or little function box. And I will always get you a delta. And if I can do that for any epsilon that'll always give you a delta, where this is true, that if x is within delta of C, then the corresponding f of x is going to be within epsilon of L. Then the limit definitely exists. So let's try to do that. So let's think about being within delta of our C. So, let's think about this right over here is 5 plus delta, this is 5 minus delta. So that's our range we're going to think about. We're going to think about it in the abstract at first. And then we're going to try to come up with a formula for delta in terms of epsilon. So how could we describe all of the x's that are in this range but not equal to 5 itself? Because we really care about the things that are within delta of 5, but not necessarily equal to 5. This is just a strictly less than. They're within a range of C, but not equal to C. Well, that's going to be all of the x's that satisfy x minus 5 is less than delta. That describes all of these x's right over here. And now, what we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality, so it starts to look something like this, or it starts to look exactly like that. And as we do that, the right-hand side of the inequality is going to be expressed in terms of delta. And then we can essentially say well, look. If the right-hand side is in terms of delta and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it. So, if we want x minus 5 to look a lot more like this, when x is not equal to 5-- in all of this, this whole interval, x is not equal to 5-- f of x is equal to 2x, our proposed limit is equal to 10. So if we could somehow get this to be 2x minus 10, then we're in good shape. And the easiest way to do that is to multiply both sides of this inequality by 2. And 2 times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. So on the left-hand side right over here, this is just going to be the absolute value of 2x minus 10. And it's going to be less than on the right-hand side, you just end up with a 2 delta. Now, what do we have here on the left-hand side? Well, this is f of x as long as x does not equal 5. And this is our limit. So we can rewrite this as f of x minus L is less than 2 delta. And this is for x does not equal 5. This is f of x, this literally is our limit. Now this is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. In terms of epsilon, this has it in terms of delta. So, how can we define delta so that 2 delta is essentially going to be epsilon? Well, this is our chance. And this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon. Or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. Let me switch colors just to ease the monotony. If you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2. It's just going to be less than epsilon. So this is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going to be within epsilon of our limit. And remember, it has to be true for any positive epsilon. But you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. So our limit is up here, so our epsilon is 0.5. So it would literally be the range I want to be between 10 plus epsilon would be 10.5. And then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25. So that'll give us a range between 4.75 and 5.25. And as long as we pick an x between 4.75 and 5.25, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. And so, you give me any epsilon, I can just apply this formula right over here to come up with the delta. This would apply for any real number. I mean, especially any positive number. For any epsilon you give me, I just get a delta defined this way, and then I can go through this arc. If the absolute value of x minus 5 is less than delta, if delta is defined in this way, which I could define for any epsilon, then it will be the case that f of x will be within epsilon of our limit.