When math teacher struggles with this proof: kzfaq.info/get/bejne/irllptGbnJm3YZs.html

Q: Prove that f(x) = x² is continuous. A: Come on, it is obviously continuous!

This is easy: 1) f(x) = x is continuous 2) The product of continuous functions is continuous

The new part starts at 5:14. Thanks to Victor for a much better way!

Imagine being his student and giving your own solution to his questions. He's sure to give you an A+ and not mark you incorrect (given that you did it right).

This one makes much more sense. I really like that you care for your subscribers and improve yourself if you see something new. Kudos to you and congratulations to your students!❤️

I really appreciate your educational videos on mathematics man, and I'm sure a lot of other students do too. You explain well.

Behind every successful mathematician is a Pokemon stuffy.

I appreciate the passion for teaching 🙏 The education system needs teachers like you now more than ever.

I usually watch this channel once or several times a day to see what you are up to. I enjoy your effortless execution as you masterfully destroy one math problem after another. Don't get me wrong, im no math major, I've failed calculus before, but I still enjoy trying the odd math problem from my university days, seeing how much of the calculus part of my brain has deteriorated in my retail environment. Some days I'm feeling mighty confident in my abilities, neither days I am quickly humbled. You are indeed a genius. But that doesn't stop me from failing with grace

I've taken 2 years of various calc classes (calc 1 through diff eq) and not once have a had a teacher proove this so rigorously, thank you so much!!

Great job on Dr Tom Crawford's channel yesterday. That was crazy to watch.

Can you do videos that teach us about math proving methods? (example : direct method or maybe by induction etc...) Please show us the beauty of proofing!!

This is the best tutorial on this topic, I have seen so far. Thanks for this.

Great video :) congrats on 100mil total views!

Love these analysis videos. Do you think you're going to do more linear algebra or abstract algebra?

Appreciate your honesty and humor regarding the difficulty of this proof! Never understood why delta was assumed to be min of (1, 1/(2abs(a)+1). Triangle inequality did the trick and thank you again for modeling the difficulty faced by people who would be trying to discover that choice of delta the first time they attempted it. Also nothing sacrosanct about 1. Could be any positive constant.

This is the coolest video on KZfaq

Wow. You made this so clear. Brilliant.

Beautiful!

This is sooo satisfying to watch now, after seeing the a=-1/2 clip Yay!

Holy your bread is so long O-O. I've been watching a splattering of your videos never thought you can grow that

If f(X) is continuous function at a point like take 'a' then these must follow this argument: lim f(x). lim f(x) x-> a^- = x->a^+ = f(a)

I saw the Pikachu in his hand, and now I have a huge amount of respect for him. 🛐

I love your pikachu! That should be an element of the channel carrying everything while solving math from the death star to pikachu!!!

Sehr interessant und sehr informativ Very interesting and much informations

You sir, are awesome 😎

Much easier proof: A continuous function is one such that f^(-1)(U) is open for every open set U. Taking preimages commutes with taking unions, so we only need to show this is the case for intervals (a,b). Take 0

For the record, some people are saying that you can instead just use the fact that x |-> x^2 : R -> R+ is differentiable everywhere, to then conclude it is continuous everywhere. The problem with this is that you would need to prove the function is differentiable, and prove that differentiability implies continuity, and both proofs require using the definition of a limit, which is the same definition you use to prove continuity directly instead. As such, proving continuity directly is actually just literally simpler. If you want to take the differentiability for granted, then you may as well also take continuity for granted. We cannot be cherry-picking about which statements to prove and which to not prove.

Thanks

This reminds me of Russell's Principia Mathemathica proving that 1+1=2 with logical axioms and dozens of pages proving it and then there goes Gödel to say that every axiomatic system is incomplete. Bruh moment.

i see why the minimum is used, but it's possible without it. the only difference is that we get δ(δ+2|a|)=ε, which is a quadratic equation so we choose δ equal to the positive solution, and that solution is sqrt(|a|^2+ε)-|a| to me, that seems easier than using a minimum, but i guess that's subjective

Hey man... May you please try integration of (2x-sinx)^1/2

0:33 that laugh😂😂😂

Wasn’t this uploaded a few hours ago?

Hi Blackpenredpen!

When you're doing the proof right but you still included an OFC ... of course.

Of course we can put "of course" on the board. We might not get the best grades, but life goes on.

Pleas show how can we prove discontinuty of function by delta epsilon

Where did that 2a come from in the triangle part

My proof would have been similar, except that, instead of saying |x + a| < 1 + 2·|a|, I would have said |x + a| < δ + 2·|a|, which is actually true for any δ > 0, and so I would not have "pre-chosen" δ before hand. The idea is that |x - a| < δ is given, and the triangle inequality implies |x + a| = |(x - a) + 2·a| = < |x - a| + |2·a| = |x - a| + 2·|a| < δ + 2·|a|. The reason I find this to be even simpler and more intuitive is that no further assumptions are being used. With |x - a| < δ and |x + a| < δ + 2·|a|, the consequence is that |x - a|·|x + a| = |x^2 - a^2| < δ·(δ + 2·|a|). Thus the δ to then "choose" would be the largest solution to the equation δ·(δ + 2·|a|) = ε, which always exists for every ε > 0 and always satisfies δ > 0 for every such ε. This does become obvious if you simply solve for δ explicitly: by completing the square, you can prove that δ = sqrt(ε + a^2) - |a|, which for every ε > 0, satisfies δ > 0. I also like this because it avoids using the minimum function.

"And always choose 1. Why 1? Because 1 is eazzyy." 😂😂

f(x) tends to f(a) as x tend to a if and only if f(x) - f(a) tends to 0 as x tends to a if and only if f(a + h) - f(a) tend to 0 as h tends to 0 But f(a+h) - f(a) = (a+h)^2-a^2 = 2ah + h^2 which clearly tends to 0 as h tends to 0.

Can you show a case where we use the epsilon delta formula to prove a function is not continuous?

may i know how to prove this function is on uniformly continuos on the domain?

I feel like Nintendo is gonna start charging this guy rent any day now

1) f: x -> x^2 is admitting a derivate in any point given (use the def and we even find that f'(x)=2x) 2) as f admit a derivative in any point given f is continus in any point given (the proof is trivial using a limited development)

Beware , Pikachus can be really dangerous ,it could give you a really powerful thunderbolt lol .

do the defimition of continuity at a point a involve 0

Oooh that's why it was unlisted

Since I saw the video of buff pikachu I cant stop staring that pikachu 🤣🤣🤣

Thanks for a good video .DrRahul Rohtak Haryana India

Of course it is! QED

Bit late to comment, but I think that maybe even letting δ be |a| instead of 1 could be helpful, because if you follow the same logic with triangle inequality you just get |x+a|

Pikaaachu⚡🌩⚡

Almost easier to show that f is differentiable which then implies continuity 😂

Of course

你好! Bprp. I am a bit different than other because when I saw the start of the video, I was sure "this epsilon-delta".

do you know how to solve integral -pi/2 to +pi/2 cos^2x/1+3^x

Just use the tangent and the convexity? Both can easily established from basic algebra. Together they give you a funnel forcing continuity.

Awesome! I wish I can be a genius in mathematics as you guys are😥

Derivatives of √cos(x^2 +1)^2 by first principle

Can you also give a delta without using the min? So just one delta edit: i found one, you have to use quadratic formula. then its delta= sqrt( |a|^2+epsilon )-|a| in total this comes to delta (delta + 2|a|) = (sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )-|a|+2|a|) =(sqrt( |a|^2+epsilon )-|a|) (sqrt( |a|^2+epsilon )+|a|) =|a|^2+epsilon-|a|^2 =epsilon with 3rd binomial formula

I love math Anyway, is there anyone who is a Foreign worker/works abroad here who can answer this questionnaire? I need this for online class, thanks ❤️❤️ - How long you have stayed abroad? - What are the purposes of your stay there? - What are your most unforgettable or memorable experiences there? - How will you compare your country with other countries in terms of culture, religion, economy, politics, law, social practices, etc? - Do you want to go back abroad or to other countries in the future? Why or why not?

Where can I get that derivatives for you board behind you?

How does the concept of continuity deal with transcendental numbers? For example, wouldn't the function be discontinuous at the square root of e since e cannot be a solution to y = x^2?

g(x) = x is continuous (taking delta = epsilon shows this in definition). The product of two continuous functions is continuous. Hence f(x) = x*x = x^2 is continuous.

I feel like we need to see some videos where epsilon-delta makes things easier. Like, proving f(x) = x^2 / [(x^2 + 1)(x^2 + 2)(x^2 + 3)] is continuous. That looks like a nightmarish function, but thanks to the magic of the epsilon-delta process, that denominator can be replaced by the number 6, so all that complexity reduces to proving that x^2 / 6 is continuous.

Why do we need delta do be min{1, E/(1+2|a|)} and not just E/(1+2|a|) ? Is the min{1 part just there so that we can say that |x-a| < 1 , since delta can now be no greater than 1?

Hey sir i need your help , here is my question integral_0 to π _ X/(a²cos²x + b²sin²x)dx without using P6 property ❤️l🎉❤️🙏

cant we say that bc f(x) is differentiable for all values (-inf, inf), it is therefore continuous by default?

How do I contact you? Need urgent help

Q: Prove f(x) = x^2 is continuous Other maths branch : Obviously it is continuous Real Analysis : Here is the proof

it is eassy to prove . FOR INDIANS

Please make more videos on bprp asmr

I don't get it. Why can the triangle inequality theorem be applied?

backpenredpenyellowpokemonmic! Yay! :-D

我只知道神奇寶貝的可愛是無窮無盡的連續到永遠

The easiest things are either the easiest to prove or the hardest to prove. No in between.

The proof? Trivial and left as an exercise to the reader. Wait, what do you mean I can’t say that in an exam?

We can do this in one sentence using the the sequential criterion for functional limits

Also, you can indeed prove a function is discontinuous using the definition. However, be careful. For example, the function f : R\{0} -> R\{0}, f(x) = 1/x is not actually discontinuous anywhere. On the other hand, the function g : R -> R, f(x) = 1/x if 0 < |x|, f(0) = 0 is discontinuous somewhere: namely, at 0. This is an important distinction. The domain of f is not connected, but f is continuous everywhere in the domain. So you cannot demand for f to be proven discontinuous with the definition, because it is not discontinuous. You can, however, demand for g to be proven discontinuous at 0, since 0 actually is in the domain of g. You can indeed use the definition to do this. How so? Because |x - 0| = |x| < δ with δ > 0, and |1/x - f(0)| = |1/x| = 1/|x| < ε, with ε > 0, and 1/|x| < ε is equivalent to 0 < 1/ε < |x| for every ε > 0. |x| < δ does not imply, for every δ > 0, that 1/ε < |x|, because 1/ε < |x| and |x| < δ implies 1/ε < δ, which is equivalent to δ = 1/ε + γ for some γ > 0. Therefore, |x| < δ = 1/ε + γ implies 1/(1/ε + γ) = ε/(1 + ε·γ) < 1/|x| not 1/|x| < ε.

Why not check the limit of its left hand side and right hand side is equal check?

When did u decided to do analysis ? It's so funny

product of real-valued continuous function

Nice pikachu ⚡⚡

Q: Prove f(x)=x^2 is continuous for all real x A: The proof is trivial and left as an exercise for the grader

I still don't really get why we chose 1 as one of the values for δ.

woah where did you get the pikachu mic

f is differentiable. hence f is continuous. QED

Solve this:(integral(1, 2, (x^2 (e)^x)/root(2, x^2 - 1), x)

Why not just use the limit laws? lim_(x -> a) x = a, so lim_(x -> a) x * lim_(x -> a) x = a * a = a^2, but because lim_(x -> a) x * lim_(x -> a) x = lim_(x -> a) x * x = lim_(x -> a) x^2, it holds that lim_(x -> a) x^2 = a^2.

Hello sir I have a question related to integrals. Can we multipy and divide some function within integration? Yes or not plz give some explanation

I've seen about 200 different explanations of the epsilon delta limit definition over the years, along with hundreds of applications of it. And I *still* don't understand how to actually prove anything with it. It always seems to come down to some sort of magic somewhere that isn't intuitively obvious. And it isn't the same magic every time, either. I'm pretty sure it has to do with how little attention there was to algebra on inequalities back in various levels of education. Take that and mix in the mind-meltiness that limits and the like can get into and it's a perfect storm of confusion, I think.

I would have just shown that it is differentiable, which is much simpler, and if it is differentiable then it has to be continuous

How come delta= epsilon /1+2mod(a)

Q: prove that X² is continuous: A: Every polynomial function is continuous

Its continuos because It is differentiable. Fully and with no doubt proved! Give me my Field Medal.

But what if there was a hole in the function? The limit might exist but the function is not continuous.

I think epsilon delta is so daunting because the concept isn't clear. I'm pretty sure this is the explanation that would click with me, and might click with other students. Suppose you have a function f, and you want to prove the limit of the function equals L at x=a. So, imagine drawing a rectangle centered on the point (a, L) tall enough that the function never touches the top or bottom of the rectangle. Now, can you make smaller and smaller rectangles -- smaller height and also smaller width -- such that the function never touches the top or bottom of the rectangles? If you can do that, then that means you have a limit. So, the job is to prove the existence of a simple relationship between the height and width, that lets you show that, for any rectangle height, it's possible to set a rectangle width such that the function never touches the top or bottom of the rectangle. And, as the height goes to zero, so does the width. The trick is actually doing the math, which usually involves proving that there is a simpler function that is always further away from the limit than the original function is. Then, if you can prove that the simpler function can be contained in rectangles as described above, it follows that the original function will likewise be contained in those rectangles. You may have to do a few iterations of this, so that eventually you're dealing with a function pretty far removed from the original one. If you can come up with a straight line function, that is the simplest situation. Remember, the point is that, in any rectangle you draw around (a, L), you can be confident that the function never touches the top or bottom edges of the rectangle. A straight line doesn't have any wiggles that could threaten to touch the top or bottom edge, no matter how small your rectangle is. So it's safe to say that, if you can get to the point of a straight line function, you're pretty much done. Pretty often, you will want/need to set a maximum width for the rectangles, so that, below that maximum width, you can introduce a straight line that is always further away from the limit than the original function is. That way, in the "linear" region, you have your straight line function. So the height of our rectangles is 2*epsilon: that's a distance of epsilon above the limit, and a distance of epsilon below the limit. And the width of our rectangles is 2* delta: that's a distance of delta to the left of the point and a distance of delta to the right of the point. If it is possible to mathematically prove a delta that is defined in terms of epsilon, and delta goes to 0 as epsilon goes to 0, that proves our limit. All of that is just the concept. Once you understand that, I think, the math will make more sense.