I discovered your channel when I was in Sixth Form (16-18 year olds in the UK) about 5 years ago. At the time, I had only just started getting into maths and I thought the crazy integrals you did were really funny and interesting (I remember the cube root of tan(x) in particular). I showed some of the teachers the DI-table method you use for integration by parts, and one was so impressed she said she would use it too. I went on to do maths at uni and I've just found out I will graduate with a First. Even though I ended up not doing much like what you do on your channel (I preferred algebra, in particular ring theory, Lie algebras and algebraic geometry), you undoubtedly played a big role in my early interest in maths. Thank you for inspiring me.
@Ehud Kotegaro I assume you meant to make this a separate comment lol. But yes, I was also thinking that the result immediately follows from the theorem that a function is continuous at a point if and only if its limit at the point is just its value at the point. On the other hand, you could think of this as proving 1/x is continuous at 2.
@@finlay5789 I assume you're doing maths at uni? If so, good luck, but be aware that maths at uni isn't about calculating limits and integrals like most of this channel (this is geared towards Americans), you're learning theorems and proofs. This particular video however is a good example of what you might do in Analysis.
Lol i remember my teacher giving me LESS points on an integral i solved with the DI method on a test because he'd never heard of that method. The function was something like x^3*sinx and i just couldnt be bothered solving that with standard IBP
If you wanna know how crazy calculus gets, goes even further. I know we only did limits here, but if you do differentiation and integration, all of it fits under the topic of differential forms. Basically, if you want to differentiate and integrate over arbitrary smooth manifolds (that are not R^n), then you just need to introduce the concept of a tangent space, the ideas of a smooth manifold (charts, atlases, partitions of unity), and some concepts from abstract algebra like exterior derivatives, wedge products. This defines differentiation and integration both intrinsically and extrinsically. And now you can now obtain the fundamental theorem of calculus in any dimension using something called Stokes’ Theorem. Then beyond this, you can study the algebra behind the exterior derivative, wedge product, and the types of algebraic structures that admit these (exterior algebras). Then you can go even further from that and learn about how category theory unifies a lot of these types of algebras that admit similar types of structure. Mathematics is one deep rabbit hole (:
Then you can exclude the law of excluded middles and get sets of numbers in [0, 0] that are not {0} and also have that every function over domain R be continues and infinitely differentiable. Seriously.
@@xinpingdonohoe3978 It's CollegeBoard's grading system for AP exams, which are standardized exams for high schoolers to prepare for college. I got a 5 on AP Calc BC as well
I did my calculus classes long time ago. I wasn’t a bad student nor a genius, but I also liked math in general. I’m over 30 now and just realized how calculus works de facto. Man I just wish I could redo my classes with the insights you give us. 14 years ago I would be totally addicted to your channel hahahaha.
What is the difference between Newton and Leibniz calculus? Newton's calculus is about functions. Leibniz's calculus is about relations defined by constraints. In Newton's calculus, there is (what would now be called) a limit built into every operation. In Leibniz's calculus, the limit is a separate operation.
Good thing I will be a legal adult by the time I take my AP Calc BC exam. I can use the epsilon-delta theorem for limits without having to worry about committing fraudulent behavior!
Here is why we need it. You and I have seen the graph of y = 1/x, so we know what it looks like. But how do we really know it's continuous? How do we know that it doesn't have a discontinuity at x=3.2493 for example? Epsilon-delta is a way to prove that a function really is continuous. Calculus requires that we work with continuous functions, so it is necessary to prove that functions are continuous. (Fortunately, most functions are.) The goal of epsilon-delta is to show that, as f(x) gets closer and closer to "1/a", x must also get closer and closer to "a". Think of the graph of y = 1/x, and imagine a rectangle whose center is at (a, 1/a). Make this rectangle tall enough that the function never touches the rectangle's top or bottom edges. This rectangle has a height of 2*epsilon, and a width of 2*delta. Now, can you shrink this rectangle smaller and smaller (so it becomes less tall and also less wide), all the way to zero, and the function still never touches the top or bottom edges? If you can construct rectangles that do that, then it means that the limit really does approach 1/a. So, the epsilon-delta method is really about defining the proportions of the rectangles: if you can mathematically prove that you can create rectangles that operate as described, then the function must be continuous. That means establishing a relationship between epsilon and delta. It may be that the relationship works only when x is near a; that is fine, since limits are about a function's behavior near a specific point.
In our university they are currently teaching real analysis and I have had problems with Limits and continuity. Really great to find this video, it helped me a lot! Thanks BPRP
Ive had to do similar proofs like this in my proofs lesson, but the problems were much simpler than this. Still, I find real analysis terrifying. (The problem was proving that for all epsilon > 0, if 0
Everyone has Real Analysis PTSD. Although I would say back when I took Complex analysis, that course truly killed me, but strangely, abstract algebra, number theory, and linear analysis was an easier pill to swallow xD.
@@xghoulxx I can only speak for me and my classmates, but at least in Germany, real analysis 1 and 2 are probably the easiest classes in the early years at university. Abstract algebra is a little harder, others even worse..
I remember using that result for some questions without ever proving it while wondering if I was allowed to or not. Though the proof is so trivial that it was indeed allowed to be used.
0:39, no es necesario ya que la función es continúa en un intervalo abierto que contiene a x=2. ¿porqué es continúa?. Toda función polinomica es continúa, además el cociente de dos funciones continuas es continuo siempre y cuando el denominador no sea nulo. Así que en este caso la función f(x)=x es continúa para todo x, por otro lado g(x) = 1/f(x), es continúa para todo x que diferente de cero.
Te estás liando un poco. El ejercicio pide demostrar que el límite de 1/x cuando x tiende a 2 es ½. Lo que hemos probado, (dado que 1/2 = ½) es que la función 1/x es continua en x=2. Sí, los cocientes de funciones continuas son continuos cuando el denominador no es nulo, pero eso también hay que probarlo, y se probaría de este mismo modo
It’s hard to see why it might seem necessary for this example because it’s continuous at that particular point. But how do you show lim x->2 (x²-2x)/(x-2)=2. You probably recognize the common and will cancel it out. Doing a proof like this is the reason WHY we’re allowed to certain things like this.
I think the point is, how do you really know that a function is continuous? I mean, we can all envision what y = 1/x looks like on a graph, but we know that from graphing a whole bunch of points and connecting the dots. How do you really know that there aren't any discontinuities, other than the one at x=0 of course? Can you prove it? Sure you can say "I don't see why it WOULDN'T be continuous", but that's not really a proof. I get the feeling that, 99.9% of the time, these proofs are an abstract consideration; usually you and I can tell whether a function is going to be continuous just by looking at it. Even so, mathematical rigor is its own reward.
@@kingbeauregard thank you Joe and kingbeauregard. I now understand why we need to go through all these difficult proofs but it is necessary if we want to be rigid. And now I can see the beauty of this proof now. Thank you.
It’s less a ‘proof’ and more the actual definition of a limit. Now, we have further proofs which say if a function is of a certain form (e.g. for this one, rational functions where the denominator is not zero at this location) then the function is continuous. And we know that continuous functions have limits that are relatively easy to evaluate.
Thanks so much for posting today. Down with the flu or maybe even COVID, and feel horrible right now. But your video made me forget my sickness for a little bit🖤♥️🖤♥️
@@blackpenredpen Using calculus, I predict that the virus is dying off at a rate of 0.67% every hour. Using this, I predict that I will be fully recovered by [100/0.67 hours, too tired to do calculations] from now! (Totally a joke, just trying to hang in there)
I am working on a loose set of mental rules for Always Getting Epsilon-Delta To Work. Here's what I've got so far, assuming we're trying to prove the limit of f(x) = L at x = a: 1) Your first job is to figure out how to peel off a multiplier of |x - a|. You'll probably get there by applying some identity or technique that puts the f(x) and L together; for example, in this video, the secret was cross-multiplying. If we were doing sin(x), it would involve combining sin(x) and sin(a), and the trick there would be an identity that takes us into sin(x - a), and then doing an arcsin. 2) Your next job is to get all the "x" terms out of the expression that's left after you have peeled off the |x - a|. You have two main tools for this and they are usually used in conjunction with one another: you can replace any terms in that expression with terms guaranteed to make the whole expression bigger (or at least no smaller), and you can also set an arbitrary range of x-values over which that replacement is valid. That will allow you to replace terms involving "x" with a constant. 2a) You might need to use the two tools from #2 to peel off the |x - a| in #1. But it's more likely you can get there with straight algebra, so try to algebrize it into submission. 3) About that arbitrary range: I've noticed that people get confused with the step where we say "delta = min { 1, ..." because they don't get why we're picking 1. Would it help un-confuse people to start with "delta = min { TBD, ..." and then later change the "To Be Decided" to "1" at the point we have to employ it? I don't know if it would be clearer that way. It might be. I think part of the problem is, it's not clear why "1" is a reasonable pick until you need to use it.
... about "your first job" and "your next job", I guess you can approach them in whatever order you need to. Sometimes, you might need to do #2 before you can do #1.
The idea with using 1 in the delta is just that you need to pick some random number, and for that case it's usually slightly nicer to use 1 than it is to use 0.459683 or whatever other number you'd pick. As long as that's emphasized, it's fine. The times I did an epsilon-delta as a class presentation, I didn't include the min at all. I just did "let delta = [awkward space] sqrt x" or whatever the function is, and then after I got to the point I needed to use a minimum, I scrolled back up to add the min{1, to it after explaining specifically why what I wrote wouldn't work if delta was more than 1.
Well, we just need to say that 1/x is continuous in the interval [1;3], which means by definition that for a in [1;3]: lim x->a of f(x) = f(a), from there, lim x->2 of 1/x = 1/2 !
We should talk about the arbitrary values for a second. When you do epsilon-delta proofs, you're going to need to rework | f(x) - f(a) | into a form that is something like |x - a|*(some expression without any x's in it). You're usually going to reach a point where you can't get rid of any lingering x's through sheer algebra, and | f(x) - f(a) | has turned into |x-a|*g(x). That's when you cheat. Since we're really only concerned with values in the vicinity of x=a, we can restrict ourselves to a region that is as arbitrarily small as we choose. Then we can do some math and determine that, in that region, g(x) has a maximum value "M", so we can swap out g(x) and instead work with |x-a|*M. Why are we allowed to do this cheat? Well, it's a squeeze proof. If we are saying that |x - a|*M is bigger than |x - a|*g(x) over the entire region in question, and |x - a|*M has a limit at (a, L) (and of course it does, it's a straight line), then it follows that |x - a|*g(x) must likewise have a limit at (a, L). So then, why a minimum of "1" specifically? Primarily for mathematical ease, but also, it doesn't make us trip over that discontinuity at x=0. A minimum of "1" is a fine choice if we're concerned with x=2. It's a terrible choice if we're concerned with x=0.5.
It doesn't have to be a nightmare. IMHO there are two parts to think about: the concept, and the technique. The concept: Suppose you are trying to prove the limit of f(x) = L at x=a. So, imagine a rectangle centered at (a, L) that has proportions such that f(x) never touches the top or bottom edges of the rectangle. Now, can you shrink that rectangle down to nothing, such that the function never touches the top or bottom edges? If you can mathematically prove that such a rectangle exists, then the limit must exist too. "delta" is all about the width of the rectangle, and "epsilon" is all about the height of the rectangle. Sooooo, all of this math is about figuring out whether such a rectangle exists, and if you pick a given epsilon, what size does delta have to be? The technique: You start with | f(x) - f(a) | < epsilon, and you want to wrestle with it until you get to the form |x-a| < (some function of epsilon). That |x-a| will become our delta. So you have to do a lot of algebra, and you can use one special trick: you can say that, if we limit our x values to a small distance from a, then within that range, the function will never cross the line |x-a|*(some constant that you determine with some side math). At that point, you've switched over to determining your epsilon against that line rather than the original function, but that's fine: since that line has the limit you want, so will the original function.
I recall my first week or so of high school calculus being taught this "adult" nonsense before we really got to the good stuff like derivatives. I say "nonsense" because basically it was teaching how to prove the obvious, which could have been done a different way, especially with the invention of the calculator. (Also, even test makers didn't waste time by having it on the AP exam, which by the way I scored a 5 on which allowed me to place out of Calc I and II in college.) In this example, you would have one student input 1.99 into the equation 1/x where he gets .5025 on his calculator, and see if another student can come up with a closer number to 2 (such as 1.999) on his calculator and see if his answer is closer to .5 than .5025 (which of course it is, .50025), and then ask if anyone can come up with an x value, but not 2, which results in an answer even closer to .5. The point of the exercise is that when you are dealing with limits, someone can always come up with a number closer to the limit than someone else. So, for example, one student posits a 1. with a billion 9's behind the point, another comes along and counters with a 1. with a trillion 9s behind the point. All that could be done in less than one day in class. Maybe for those students interested in pursuing a degree in math, which thrives on proofs, the teacher could provide a few homework exercises for them to prove the obvious just for their enjoyment.
@@someon3 that wasnt my point. Even then, the only way to show that kx^a is continuous is through epsilon-delta proofs. You get a better understanding of continuity, and calculus overall if you take your time to figure out why epsilon-delta proofs make sense and why they're important
@@nestorv7627 of course, I'm just saying that following already proven theorems you can easily solve limits knowing it will he 100% true. For a better understanding of the theory u're right, for this reason in calculus you study both theorems/demonstrations and exercises assuming true things like continuous functions and so on
I am confused, what about the f(x)-L bit why wasn't that explicitly mentioned in the proof? Why does simply substituting epsilon for delta work so neatly when they are addressing two related butt different inequalities?
I am not sure at all that this was what you asked for - it’s probably not. “For any ε there exists a δ such that…” means that δ is a function of ε, at least in this context, not sure if this is always the case. |f(x)-L| < ε and 0 < |x-a| < δ(ε) the proof is complete once δ is found. At least in this case, probably in most, it is easiest to find some other function h(•), such that |f(x)-L| < h(δ(ε)) in this case he found h(x) = x/2 |1/x-1/2| < δ(ε)/2 and 0 < |x-2| < δ(ε) Now you can just notice ( maybe ) that if δ(ε) = ε, nothing breaks, and the proof would thus be complete.
This is a hard question to answer because there are many real analysis books to be made and they are a bit subjective in terms of how you learn the subject. Some books offer no pictures and are all text, and some can have various illustrations but gloss over the fine details through writing, It really depends on what type of learning works best for you. When I took analysis back in 2011, my instructor was a co-author for the textbook Elementary Classical Analysis Jerrold E. Marsden, Michael J. Hoffman. My instructor was close to my top 5 most humble mathematicians I ever met and this book was very approachable to the subject. By the time I was graduating, he was giving us drafts of the 3rd edition for free for us to find any errors or input, so my natural bias is to recommend this book, lol.
@@xghoulxx i am at first year in university for electrical engineer and I want to work on this subject that I have . But our book here is , lets say "weak " ., thats the reason why I asked this question
@@dlrmfemilianolako8 Hopefully my recommendation helps, real analysis is a tough beast! I'm glad I'm not in school anymore, haha. I should add, that my book recommendation can be found online most likely and a solutions manual if you're internet savvy enough. Maybe an older edition, but it's good to have plenty of references.
can anyone do this? They give three prime numbers, p, q, r. Solve that ³√p, ³√q, ³√r are not three characters (not necessarily consecutive characters) of any arithmetic suite.
There is a theorem that is quite easy to prove which states that power functions are continuous in every point of their domain. If a function is continuous in a certain point by the definition of continuity of a function the limit can be evaluated by evaluating the function in that point. Who knows the proof of this theorem and uses simple substitution should than be considered as part of the “adult calculus” set.
Oooh, Ima take a stab at it. If we're trying to prove the limit of x^n at x=a, then: | x^n - a^n | < epsilon | (x - a) * (x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1)) | < epsilon |x - a| * | x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1) | < epsilon So from there, set an arbitrary limit on delta, figure out what the maximum value of the second absolute value is over that set of x-values (let's call it "M"), and we're left with |x - a| * M < epsilon.
@@kingbeauregard Well done! But instead to take the maximum it is better to take a majorant of that set, because to apply Weierstrass theorem f needs to be continuous in the closed interval [a-h;a+h], where h is a real number. The proof would than be circular. You can also prove it, considering Lim |x^(n+h)-x^n| for h->0. It becomes than Lim|x^n||x^h-1|=0 because x^h-1∼(x-1)h for h->0.
BPRP’s proof is like the Calc 1, week 1 homework assignment. Your version is like what the tutor tells the student to turn in because it’s more efficient and fits the topic though the student may not directly understand it, haha.
@@stephenbeck7222 After doing a LOT of thinking about epsilon-delta (and I'm not saying it's high-quality thought, just the best I'm capable of), I feel like the practical approach to arriving at a delta is, start with | f(x) - f(a) |, find some way to peel off an |x - a| term, and remove any x's from whatever's left. It just so happens it's real easy to pull an |x - a| out of | x^n - a^n |. I'll take it!
This seems unnescessrily complicated both in definition and in proof. The way I was taught the definition of limit didnt have the delta part it just had that there exist a such that |f(a)-L|
no we do not have to prove it using the epsilon delta definition as the definition already proves the existence of limits. this is a finite limit, so by the proof of that we already have the proof in place and we can simply turn the mathematical handle.
Youre using a circular argument. What prevents anyone from saying that the left limit is 1/3 and not 1/2? It may seem trivial to you, but he showed the more rigorous way of doing it because later on if you continue down the math route, you will need to do epsilon-delta proofs a lot
@@nestorv7627 computer programmer so I am only interested in turning the mathematical handle. I am not saying that the epsilon delta proof is invalid or unnecessary. I am saying that the limit of finite series is already been proven. It is a valid argument to use already existing proofs without having to keep proving the existing proofs. Prove that 1+1=2. You see we do not always prove everything in a proof.
I don’t know what a 1 point indicator function is, but he wasn’t defining continuity at any point. He was only defining the limit. For continuity you would remove the x=/= a condition.
You are right that in the usual definition of continuity, there is no requirement that x != a. That doesn't make his definition wrong! He's defining a different, but related concept. You can actually give an equivalent definition of continuity in terms of limits as "f is continuous at a if lim_{x -> a} f(x) = f(a)", which you may check does indeed exclude indicators of singletons. The reason you can't have x = a in the definition of a limit is that you need to be able to evaluate limits like (sin x)/x at 0, where the function you are taking a limit of might not even be defined. The most famous example is the difference quotient in the definition of the derivative.
Slightly disappointing for me since I would never discover the trick of choosing a delta that is the min of a two element set that includes 1(or some other convenient constant) Delta is a function of Epsilon as your limit proof showed. Thanks for the mild criticism of the term arbitrary in conjunction with phrase “arbitrary epsilon” Epsilon > 0 is all that is necessary 😀
So all you have to do is prove that x-a is smaller than something and f(x)-L is also smaller than something?? How does that prove anything. Where does the epsilon delta definition come from
The epsilon delta definition can seem obtuse since he didn't really explain what epsilon and delta actually were in this video. I'm pretty sure he has other videos on it, but let me try to explain: The reason we use absolute values is because both delta and epsilon are _distances_ from some point. Delta is a distance in the input space (usually the x-axis), and epsilon is a distance in the output space (usually the y-axis). So when we say that |x - a| < d, what that means is we can pick a point x in the input space which is within a distance d of the point a. Then, putting this point into the function we want that f(x) will be within the distance e of the point L in the output space. And make sure this holds for any epsilon! So, what this means is if we are given some small area around a point in the output space, we should be able to pick a small area in the input space which gets mapped entirely within the area in the output space if the function is continuous. If we can't do that, the function is discontinuous and our limit won't necessarily converge.
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Okay
Time traveling mathematician?!
Do you sell those t shirts?
There is Taylor series for the Lambert W function (we^w)?... HOW DID U COMMENTED 3 DAYS AGO????
*When you look at the board:* 0:31
*When you look outside of the window for lim x->2 (1/x) milliseconds:* 11:24
😂
"this is a horrible mistake, but it's not a big problem"
😆
I discovered your channel when I was in Sixth Form (16-18 year olds in the UK) about 5 years ago. At the time, I had only just started getting into maths and I thought the crazy integrals you did were really funny and interesting (I remember the cube root of tan(x) in particular). I showed some of the teachers the DI-table method you use for integration by parts, and one was so impressed she said she would use it too. I went on to do maths at uni and I've just found out I will graduate with a First. Even though I ended up not doing much like what you do on your channel (I preferred algebra, in particular ring theory, Lie algebras and algebraic geometry), you undoubtedly played a big role in my early interest in maths. Thank you for inspiring me.
This is awesome. I just finished my a-levels, and can only hope I can experience a similar situation :D
@Ehud Kotegaro I assume you meant to make this a separate comment lol. But yes, I was also thinking that the result immediately follows from the theorem that a function is continuous at a point if and only if its limit at the point is just its value at the point. On the other hand, you could think of this as proving 1/x is continuous at 2.
@@finlay5789 I assume you're doing maths at uni? If so, good luck, but be aware that maths at uni isn't about calculating limits and integrals like most of this channel (this is geared towards Americans), you're learning theorems and proofs. This particular video however is a good example of what you might do in Analysis.
Lol i remember my teacher giving me LESS points on an integral i solved with the DI method on a test because he'd never heard of that method. The function was something like x^3*sinx and i just couldnt be bothered solving that with standard IBP
Did you do further maths a level?
Baby Calculus : limit as x --> 2 of 1/x = *1/2*
Adult Calculus : limit as x --> 2 of 1/x = *0.5*
"50% of unit natural number"
“25% of a sec^2 x - tan^2 x multiplied by 2”
0:10
If you wanna know how crazy calculus gets, goes even further. I know we only did limits here, but if you do differentiation and integration, all of it fits under the topic of differential forms. Basically, if you want to differentiate and integrate over arbitrary smooth manifolds (that are not R^n), then you just need to introduce the concept of a tangent space, the ideas of a smooth manifold (charts, atlases, partitions of unity), and some concepts from abstract algebra like exterior derivatives, wedge products. This defines differentiation and integration both intrinsically and extrinsically. And now you can now obtain the fundamental theorem of calculus in any dimension using something called Stokes’ Theorem.
Then beyond this, you can study the algebra behind the exterior derivative, wedge product, and the types of algebraic structures that admit these (exterior algebras). Then you can go even further from that and learn about how category theory unifies a lot of these types of algebras that admit similar types of structure. Mathematics is one deep rabbit hole (:
("one deep rabbit hole") ... whose denumerable branches aren't singularities, mostly ; ]
Then you can exclude the law of excluded middles and get sets of numbers in [0, 0] that are not {0} and also have that every function over domain R be continues and infinitely differentiable.
Seriously.
What
does this fall under Differential Geometry?
I got a 5 on AP calc BC, all thanks to you sir! 😊
I made a 4 on ap physics
Is that good or bad?
@@xinpingdonohoe3978 very good, half of people fail which would be a 2 or lower, I made a 4 and the max is 5
@@nuts5388 okay. Which country's/place's grade system is this?
@@xinpingdonohoe3978 It's CollegeBoard's grading system for AP exams, which are standardized exams for high schoolers to prepare for college. I got a 5 on AP Calc BC as well
Finally some adult content i dont have to close in panic when someone enters my room randomly.
You could pick delta=epsilon instead of 2epsilon, since epsilon/2
Yup 😃
I did my calculus classes long time ago. I wasn’t a bad student nor a genius, but I also liked math in general. I’m over 30 now and just realized how calculus works de facto. Man I just wish I could redo my classes with the insights you give us. 14 years ago I would be totally addicted to your channel hahahaha.
hahahaha 😂😂😏😒🙄🙁☹😤😠😡🤬
Baby calculus was so short, I totally forgot we were comparing them by the end of the video.
11:30 "Our sponsor today, BLYAT"
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cyka bLYAT
feed mid gg ez
man im dead
What is the difference between Newton and Leibniz calculus?
Newton's calculus is about functions.
Leibniz's calculus is about relations defined by constraints.
In Newton's calculus, there is (what would now be called) a limit built into every operation.
In Leibniz's calculus, the limit is a separate operation.
Good thing I will be a legal adult by the time I take my AP Calc BC exam. I can use the epsilon-delta theorem for limits without having to worry about committing fraudulent behavior!
Math fraud is SERIOUS!
how would that b commuting fraud
11:08 why is it that you wrote "less than or equal to epsilon" but also wrote "equals epsilon"?
in one line its like writing |f(x)-1/2| < delta*(1/2)
The best explanation for epsilon delta proof I have seen so far, and I am in my last year of math.
Thanks
BPRP has several helpful videos going through this. I wonder if there is a playlist?
0:11, great answer
Please can you make a video and explain the concept behind Epsilon Delta and why we need it when carrying out proofs involving calculus
Here is why we need it. You and I have seen the graph of y = 1/x, so we know what it looks like. But how do we really know it's continuous? How do we know that it doesn't have a discontinuity at x=3.2493 for example? Epsilon-delta is a way to prove that a function really is continuous. Calculus requires that we work with continuous functions, so it is necessary to prove that functions are continuous. (Fortunately, most functions are.)
The goal of epsilon-delta is to show that, as f(x) gets closer and closer to "1/a", x must also get closer and closer to "a". Think of the graph of y = 1/x, and imagine a rectangle whose center is at (a, 1/a). Make this rectangle tall enough that the function never touches the rectangle's top or bottom edges. This rectangle has a height of 2*epsilon, and a width of 2*delta. Now, can you shrink this rectangle smaller and smaller (so it becomes less tall and also less wide), all the way to zero, and the function still never touches the top or bottom edges? If you can construct rectangles that do that, then it means that the limit really does approach 1/a.
So, the epsilon-delta method is really about defining the proportions of the rectangles: if you can mathematically prove that you can create rectangles that operate as described, then the function must be continuous. That means establishing a relationship between epsilon and delta. It may be that the relationship works only when x is near a; that is fine, since limits are about a function's behavior near a specific point.
@@kingbeauregard nicely explained via plain text.Now imagine how smoothly you can explain concepts using video. Thank you so much.
In our university they are currently teaching real analysis and I have had problems with Limits and continuity. Really great to find this video, it helped me a lot! Thanks BPRP
i solved this problem like 5 hours ago, and now you uploaded the same problem !! 🤣🤣
Ive had to do similar proofs like this in my proofs lesson, but the problems were much simpler than this. Still, I find real analysis terrifying.
(The problem was proving that for all epsilon > 0, if 0
Everyone has Real Analysis PTSD.
Although I would say back when I took Complex analysis, that course truly killed me, but strangely, abstract algebra, number theory, and linear analysis was an easier pill to swallow xD.
@@xghoulxx I can only speak for me and my classmates, but at least in Germany, real analysis 1 and 2 are probably the easiest classes in the early years at university. Abstract algebra is a little harder, others even worse..
I remember using that result for some questions without ever proving it while wondering if I was allowed to or not. Though the proof is so trivial that it was indeed allowed to be used.
This is a great introduction to epsilon delta proofs... I will have to watch again with some pen and paper - thank you!
0:39, no es necesario ya que la función es continúa en un intervalo abierto que contiene a x=2. ¿porqué es continúa?. Toda función polinomica es continúa, además el cociente de dos funciones continuas es continuo siempre y cuando el denominador no sea nulo. Así que en este caso la función f(x)=x es continúa para todo x, por otro lado g(x) = 1/f(x), es continúa para todo x que diferente de cero.
Te estás liando un poco. El ejercicio pide demostrar que el límite de 1/x cuando x tiende a 2 es ½. Lo que hemos probado, (dado que 1/2 = ½) es que la función 1/x es continua en x=2. Sí, los cocientes de funciones continuas son continuos cuando el denominador no es nulo, pero eso también hay que probarlo, y se probaría de este mismo modo
And here I was, thinking I understood limits. Now I feel like a baby! Thank you, now I’ve nightmares, I understand nothing 😱😵💫😂
More short: |x-2|/|2x| |x-2|/4 |x-2|2 => |2x|->4). 😀😉
Can you do the indefinite integral of x^x dx please? Thank you!
That intrgral cannot be expressed with elementary functions
Hi.... You watch this for indefinite integral of x^x dx by BPRP... ^_^
ua-cam.com/video/tIGnbH4qIjY/v-deo.html
4:19 x分之1 vs. 1 over x 😆
😆
i love your shirts!
My background is physics. Can anyone explain to me why a long proof is necessary for proofing this limit?
It somehow gives deep internal satisfaction to the mathematicians.
It’s hard to see why it might seem necessary for this example because it’s continuous at that particular point. But how do you show lim x->2 (x²-2x)/(x-2)=2. You probably recognize the common and will cancel it out. Doing a proof like this is the reason WHY we’re allowed to certain things like this.
I think the point is, how do you really know that a function is continuous? I mean, we can all envision what y = 1/x looks like on a graph, but we know that from graphing a whole bunch of points and connecting the dots. How do you really know that there aren't any discontinuities, other than the one at x=0 of course? Can you prove it? Sure you can say "I don't see why it WOULDN'T be continuous", but that's not really a proof.
I get the feeling that, 99.9% of the time, these proofs are an abstract consideration; usually you and I can tell whether a function is going to be continuous just by looking at it. Even so, mathematical rigor is its own reward.
@@kingbeauregard thank you Joe and kingbeauregard. I now understand why we need to go through all these difficult proofs but it is necessary if we want to be rigid. And now I can see the beauty of this proof now. Thank you.
It’s less a ‘proof’ and more the actual definition of a limit. Now, we have further proofs which say if a function is of a certain form (e.g. for this one, rational functions where the denominator is not zero at this location) then the function is continuous. And we know that continuous functions have limits that are relatively easy to evaluate.
Thanks so much for posting today. Down with the flu or maybe even COVID, and feel horrible right now. But your video made me forget my sickness for a little bit🖤♥️🖤♥️
Oh no. Sorry to hear that. I wish you feel better soon!
@@blackpenredpen Using calculus, I predict that the virus is dying off at a rate of 0.67% every hour. Using this, I predict that I will be fully recovered by [100/0.67 hours, too tired to do calculations] from now! (Totally a joke, just trying to hang in there)
Same, got the COVID and it sucks real bad. Hang in there, take the meds and you're going to be fine.
I think it's absolutely dope how you have a pair of last shot jordan retro 14's just sitting nice in the background!!
I usually just say that the function looks nice enough to me around 2 and put a little box
I am working on a loose set of mental rules for Always Getting Epsilon-Delta To Work. Here's what I've got so far, assuming we're trying to prove the limit of f(x) = L at x = a:
1) Your first job is to figure out how to peel off a multiplier of |x - a|. You'll probably get there by applying some identity or technique that puts the f(x) and L together; for example, in this video, the secret was cross-multiplying. If we were doing sin(x), it would involve combining sin(x) and sin(a), and the trick there would be an identity that takes us into sin(x - a), and then doing an arcsin.
2) Your next job is to get all the "x" terms out of the expression that's left after you have peeled off the |x - a|. You have two main tools for this and they are usually used in conjunction with one another: you can replace any terms in that expression with terms guaranteed to make the whole expression bigger (or at least no smaller), and you can also set an arbitrary range of x-values over which that replacement is valid. That will allow you to replace terms involving "x" with a constant.
2a) You might need to use the two tools from #2 to peel off the |x - a| in #1. But it's more likely you can get there with straight algebra, so try to algebrize it into submission.
3) About that arbitrary range: I've noticed that people get confused with the step where we say "delta = min { 1, ..." because they don't get why we're picking 1. Would it help un-confuse people to start with "delta = min { TBD, ..." and then later change the "To Be Decided" to "1" at the point we have to employ it? I don't know if it would be clearer that way. It might be. I think part of the problem is, it's not clear why "1" is a reasonable pick until you need to use it.
... about "your first job" and "your next job", I guess you can approach them in whatever order you need to. Sometimes, you might need to do #2 before you can do #1.
The idea with using 1 in the delta is just that you need to pick some random number, and for that case it's usually slightly nicer to use 1 than it is to use 0.459683 or whatever other number you'd pick. As long as that's emphasized, it's fine.
The times I did an epsilon-delta as a class presentation, I didn't include the min at all. I just did "let delta = [awkward space] sqrt x" or whatever the function is, and then after I got to the point I needed to use a minimum, I scrolled back up to add the min{1, to it after explaining specifically why what I wrote wouldn't work if delta was more than 1.
@@calvindang7291 I like that!
I'm gonna have to watch this one a few times. I'm not even sure what we did at the end
Second day of calculus class, we did epsilon-delta proof already
well explained the demonstration of that limit
When you learn what s.t. stands for, you have passed baby calculus.
I'm 72 and I have passed b.c.
Well, we just need to say that 1/x is continuous in the interval [1;3], which means by definition that for a in [1;3]: lim x->a of f(x) = f(a), from there, lim x->2 of 1/x = 1/2 !
Need a limit proof for that too
You only know 1/x is continuous in that interval via epsilon/delta proof of its continuity
Proofs relying on other proofs, perfect!
start is gold
wow, you are getting young day by day.
Hi would you mind this integral 1/((x-1)(3-x))
no u were correct with add2 add2 add2
I saw this in your 101 limit video 7 hour 28 minute)
Why did we choose delta to be 1 for the proof? Can I just pick out any number and check whether it works?
yeah he said choose any number we want as long as it's > 0. He chose 1 to make it simple. Right everyone?
@@Purplesjh Yes
We should talk about the arbitrary values for a second. When you do epsilon-delta proofs, you're going to need to rework | f(x) - f(a) | into a form that is something like |x - a|*(some expression without any x's in it). You're usually going to reach a point where you can't get rid of any lingering x's through sheer algebra, and | f(x) - f(a) | has turned into |x-a|*g(x). That's when you cheat. Since we're really only concerned with values in the vicinity of x=a, we can restrict ourselves to a region that is as arbitrarily small as we choose. Then we can do some math and determine that, in that region, g(x) has a maximum value "M", so we can swap out g(x) and instead work with |x-a|*M.
Why are we allowed to do this cheat? Well, it's a squeeze proof. If we are saying that |x - a|*M is bigger than |x - a|*g(x) over the entire region in question, and |x - a|*M has a limit at (a, L) (and of course it does, it's a straight line), then it follows that |x - a|*g(x) must likewise have a limit at (a, L).
So then, why a minimum of "1" specifically? Primarily for mathematical ease, but also, it doesn't make us trip over that discontinuity at x=0. A minimum of "1" is a fine choice if we're concerned with x=2. It's a terrible choice if we're concerned with x=0.5.
@@kingbeauregard wow thanks a lot 😊
Ha ha awesome thanks a lot Tomorrow is my birthday so can you please do a surprise video?
if epsilon < 1/2 then delta = 2/(1+2*epsilon) it more precise))
How can we solve when x tends to -3
شكرا باجر امتحاني
Me first few seconds within the video: Yeah I know that...
Me 2 minutes in: What.
Thanks for teaching us sir!, Can you make a content about how you master MATH? Hope you notice this sir! Thank you in advance 💙💙💙
Oooooh no, epsilon-delta definition is a pure NIGHTMARE.
It doesn't have to be a nightmare. IMHO there are two parts to think about: the concept, and the technique.
The concept: Suppose you are trying to prove the limit of f(x) = L at x=a. So, imagine a rectangle centered at (a, L) that has proportions such that f(x) never touches the top or bottom edges of the rectangle. Now, can you shrink that rectangle down to nothing, such that the function never touches the top or bottom edges? If you can mathematically prove that such a rectangle exists, then the limit must exist too. "delta" is all about the width of the rectangle, and "epsilon" is all about the height of the rectangle. Sooooo, all of this math is about figuring out whether such a rectangle exists, and if you pick a given epsilon, what size does delta have to be?
The technique: You start with | f(x) - f(a) | < epsilon, and you want to wrestle with it until you get to the form |x-a| < (some function of epsilon). That |x-a| will become our delta. So you have to do a lot of algebra, and you can use one special trick: you can say that, if we limit our x values to a small distance from a, then within that range, the function will never cross the line |x-a|*(some constant that you determine with some side math). At that point, you've switched over to determining your epsilon against that line rather than the original function, but that's fine: since that line has the limit you want, so will the original function.
I recall my first week or so of high school calculus being taught this "adult" nonsense before we really got to the good stuff like derivatives.
I say "nonsense" because basically it was teaching how to prove the obvious, which could have been done a different way, especially with the invention of the calculator. (Also, even test makers didn't waste time by having it on the AP exam, which by the way I scored a 5 on which allowed me to place out of Calc I and II in college.) In this example, you would have one student input 1.99 into the equation 1/x where he gets .5025 on his calculator, and see if another student can come up with a closer number to 2 (such as 1.999) on his calculator and see if his answer is closer to .5 than .5025 (which of course it is, .50025), and then ask if anyone can come up with an x value, but not 2, which results in an answer even closer to .5. The point of the exercise is that when you are dealing with limits, someone can always come up with a number closer to the limit than someone else. So, for example, one student posits a 1. with a billion 9's behind the point, another comes along and counters with a 1. with a trillion 9s behind the point. All that could be done in less than one day in class.
Maybe for those students interested in pursuing a degree in math, which thrives on proofs, the teacher could provide a few homework exercises for them to prove the obvious just for their enjoyment.
Next he’s going to tell me I can’t chew on my field extensions any more 😭
Extreme explanation. U don't need it to prove the limit is 1/2. 1/x is continuous, then you can just evaluate lim f(x) for x->c = f(c) for the theorem
How do you know that 1/x is continuous?
@@nestorv7627 cause it's x^-1, any kx^a is continuous
@@someon3 that wasnt my point. Even then, the only way to show that kx^a is continuous is through epsilon-delta proofs. You get a better understanding of continuity, and calculus overall if you take your time to figure out why epsilon-delta proofs make sense and why they're important
@@nestorv7627 of course, I'm just saying that following already proven theorems you can easily solve limits knowing it will he 100% true. For a better understanding of the theory u're right, for this reason in calculus you study both theorems/demonstrations and exercises assuming true things like continuous functions and so on
Is there any example of some limit of f(x) that cant be proven because of this equation?
Can you put infinity on minimum 🤔
I am confused, what about the f(x)-L bit why wasn't that explicitly mentioned in the proof? Why does simply substituting epsilon for delta work so neatly when they are addressing two related butt different inequalities?
I am not sure at all that this was what you asked for - it’s probably not.
“For any ε there exists a δ such that…” means that δ is a function of ε, at least in this context, not sure if this is always the case.
|f(x)-L| < ε and 0 < |x-a| < δ(ε)
the proof is complete once δ is found.
At least in this case, probably in most, it is easiest to find some other function h(•), such that
|f(x)-L| < h(δ(ε))
in this case he found h(x) = x/2
|1/x-1/2| < δ(ε)/2 and 0 < |x-2| < δ(ε)
Now you can just notice ( maybe ) that if δ(ε) = ε, nothing breaks, and the proof would thus be complete.
your professor your teachers will be really happy when they see pf for proof
this got me laughing lmao
4:17
老師錄中文英文影片已經錄到快錯亂了
對 😂
besides maths you have an excellent skill of interchanging markers 😂
Very very interesting !
Please can you help me . I want to know that which is the best mathematical analysis 1 and 2 book ?
There is not. The best best book of mathematical analysis is the one you write.
@@paolo_benda thank you for your reply
This is a hard question to answer because there are many real analysis books to be made and they are a bit subjective in terms of how you learn the subject. Some books offer no pictures and are all text, and some can have various illustrations but gloss over the fine details through writing, It really depends on what type of learning works best for you.
When I took analysis back in 2011, my instructor was a co-author for the textbook Elementary Classical Analysis
Jerrold E. Marsden, Michael J. Hoffman. My instructor was close to my top 5 most humble mathematicians I ever met and this book was very approachable to the subject. By the time I was graduating, he was giving us drafts of the 3rd edition for free for us to find any errors or input, so my natural bias is to recommend this book, lol.
@@xghoulxx i am at first year in university for electrical engineer and I want to work on this subject that I have . But our book here is , lets say "weak " ., thats the reason why I asked this question
@@dlrmfemilianolako8 Hopefully my recommendation helps, real analysis is a tough beast! I'm glad I'm not in school anymore, haha.
I should add, that my book recommendation can be found online most likely and a solutions manual if you're internet savvy enough. Maybe an older edition, but it's good to have plenty of references.
I'm good at baby calculus
can anyone do this?
They give three prime numbers, p, q, r. Solve that ³√p, ³√q, ³√r are not three characters (not necessarily consecutive characters) of any arithmetic suite.
you should have done examples with epsilon=0.0001 and epsilon=0.00000000001 to make it really clear....
I have done εδ many times. You can see the description for the most detailed explanation I have. 😃
There is a theorem that is quite easy to prove which states that power functions are continuous in every point of their domain. If a function is continuous in a certain point by the definition of continuity of a function the limit can be evaluated by evaluating the function in that point. Who knows the proof of this theorem and uses simple substitution should than be considered as part of the “adult calculus” set.
Oooh, Ima take a stab at it. If we're trying to prove the limit of x^n at x=a, then:
| x^n - a^n | < epsilon
| (x - a) * (x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1)) | < epsilon
|x - a| * | x^(n - 1) + x^(n - 2)*a + x^(n - 3)*a^2 ... + a^(n-1) | < epsilon
So from there, set an arbitrary limit on delta, figure out what the maximum value of the second absolute value is over that set of x-values (let's call it "M"), and we're left with |x - a| * M < epsilon.
@@kingbeauregard Well done! But instead to take the maximum it is better to take a majorant of that set, because to apply Weierstrass theorem f needs to be continuous in the closed interval [a-h;a+h], where h is a real number. The proof would than be circular. You can also prove it, considering Lim |x^(n+h)-x^n| for h->0.
It becomes than Lim|x^n||x^h-1|=0 because x^h-1∼(x-1)h for h->0.
@@paolo_benda I had never heard of a "majorant" until now. I see your point though, the majorant would be better.
BPRP’s proof is like the Calc 1, week 1 homework assignment. Your version is like what the tutor tells the student to turn in because it’s more efficient and fits the topic though the student may not directly understand it, haha.
@@stephenbeck7222 After doing a LOT of thinking about epsilon-delta (and I'm not saying it's high-quality thought, just the best I'm capable of), I feel like the practical approach to arriving at a delta is, start with | f(x) - f(a) |, find some way to peel off an |x - a| term, and remove any x's from whatever's left. It just so happens it's real easy to pull an |x - a| out of | x^n - a^n |. I'll take it!
If there is no uncertainty at the point, then it makes no sense to fence the garden.
This seems unnescessrily complicated both in definition and in proof.
The way I was taught the definition of limit didnt have the delta part it just had that there exist a such that |f(a)-L|
What grade are you in brother?
no we do not have to prove it using the epsilon delta definition as the definition already proves the existence of limits. this is a finite limit, so by the proof of that we already have the proof in place and we can simply turn the mathematical handle.
Youre using a circular argument. What prevents anyone from saying that the left limit is 1/3 and not 1/2?
It may seem trivial to you, but he showed the more rigorous way of doing it because later on if you continue down the math route, you will need to do epsilon-delta proofs a lot
@@nestorv7627 computer programmer so I am only interested in turning the mathematical handle. I am not saying that the epsilon delta proof is invalid or unnecessary. I am saying that the limit of finite series is already been proven. It is a valid argument to use already existing proofs without having to keep proving the existing proofs. Prove that 1+1=2. You see we do not always prove everything in a proof.
Eid Mubarak for all mathematicians ❤️❤️
The x =\= a in your explanation of the limit seems wrong for continuity, as it classifies 1-point indicator functions as continous
I don’t know what a 1 point indicator function is, but he wasn’t defining continuity at any point. He was only defining the limit. For continuity you would remove the x=/= a condition.
My Point is about That his definition fails for non contionous functions.
Oliver Schmidt no it doesn’t. If he didn’t have “0 < …” on his “suppose” line then it would fail for functions with a removable discontinuity.
You are right that in the usual definition of continuity, there is no requirement that x != a. That doesn't make his definition wrong! He's defining a different, but related concept. You can actually give an equivalent definition of continuity in terms of limits as "f is continuous at a if lim_{x -> a} f(x) = f(a)", which you may check does indeed exclude indicators of singletons.
The reason you can't have x = a in the definition of a limit is that you need to be able to evaluate limits like (sin x)/x at 0, where the function you are taking a limit of might not even be defined. The most famous example is the difference quotient in the definition of the derivative.
"Adult Calculus" (Baby Real Analysis)
😂
Isn't this easier than Baby Rudin? I would have thought it's Baby Baby Real Analysis.
Genius ❤️🔥 is always genius
Good job 👌
This was great
Feel left out bc I don't think there's a classification for 13 year old calculus
there you just graph it and eyeball it
0:27 because the intellectual who recreate it said that it was correct
God i wish i saw this before my calc exam
Slightly disappointing for me since I would never discover the trick of choosing a delta that is the min of a two element set that includes 1(or some other convenient constant)
Delta is a function of Epsilon as your limit proof showed.
Thanks for the mild criticism of the term arbitrary in conjunction with phrase “arbitrary epsilon” Epsilon > 0 is all that is necessary 😀
I love mathematics
Happy birthday
It would be nice if I could understand half of what he says.
It seems like adult calculus is baby calculus with more steps.
Okay
17 gotta be your favourite number huh
Guess I don't wanna be an adult mathematician 😂
let's hope that it's only asked as a mcq and not a long answer your question
Me, who writes prf for proof:
老师您是台湾人吗？是否认识齐震宇老师？
That looks like 1/2-+0.0
Ah man, a proof? Do I have to?
How real men solves equations:
“If we come here,”
Super glue eating livestream when? How is this real.
i don't understand this method
I imagined that could an adult doing calculus
As a beginner highschooler
Why is density there ?😂
Wow, you can do class of children, I can’t with myself xd
That thing of: ‘’now we are adults’’ is like very angry, please don’t do that
So all you have to do is prove that x-a is smaller than something and f(x)-L is also smaller than something?? How does that prove anything. Where does the epsilon delta definition come from
The epsilon delta definition can seem obtuse since he didn't really explain what epsilon and delta actually were in this video. I'm pretty sure he has other videos on it, but let me try to explain:
The reason we use absolute values is because both delta and epsilon are _distances_ from some point. Delta is a distance in the input space (usually the x-axis), and epsilon is a distance in the output space (usually the y-axis). So when we say that |x - a| < d, what that means is we can pick a point x in the input space which is within a distance d of the point a. Then, putting this point into the function we want that f(x) will be within the distance e of the point L in the output space. And make sure this holds for any epsilon!
So, what this means is if we are given some small area around a point in the output space, we should be able to pick a small area in the input space which gets mapped entirely within the area in the output space if the function is continuous. If we can't do that, the function is discontinuous and our limit won't necessarily converge.
i dont want to do calculus anymore :(
noo you abandoned the pokeball