Thank you so much! I'm self-studying Calculus after going through Algebra and Trig, and this came up relatively early on. Needless to say, I had a hard time understanding it at first as the only things I had ever "proved" up until that point were Trig identities. After watching your video (at least a couple times, I admit), I feel like I understand it much better. Also, happy to say I completed the practice problem on my own too before checking my proof against yours. (Look at me, Mom! I did it *all by myself*!) Thank you!

I find it's super helpful to think of it like this. Suppose you're trying to prove the limit of f(c) = L. So, imagine a rectangle centered at the point (c, L) that is tall enough that the function never touches the top or bottom edges of the rectangle. Can you shrink the rectangle down to nothing -- no height, no width -- without the function ever touching the top or bottom edges at any size? If you can do that, it shows that, the closer you get to y = L, the closer you also get to x = c. And, that proves our limit. All the math is just a matter of mathematically representing that rectangle, whose height is 2*epsilon (that's L plus or minus epsilon), and whose width is 2*delta (that's c plus or minus delta). So it's all about showing that you can establish a relationship between epsilon and delta, such that you get rectangles with the right geometry. Turns out most functions are boringly continuous, unless there's an obvious divide-by-zero or if the function is defined with a discontinuity. But, someone's got to actually prove that functions are continuous, and epsilon-delta's the way to do it.

Oh my gosh your explanation was amazing. I enjoyed every bit of the video, plus finally understood the proof that my professor did :) kudos for this amaazing series on real analysis, u've won a sub ;)

@WrathofMath

11 ай бұрын

Thanks so much! Glad it helped!

Hi there. I just want to say your videos are so fantastic and we are very lucky to have someone that can explain these topics so clearly and eloquently. Thank you so much for making them and please don't stop. I have watched this video a number of times - so good! But I have to ask - at 14.51 is the background noise from a fiend or a friend lol? Very curious to know when I get to that point in the video! Thank you again. I studied these topics years ago and it was always tricky because our professors were fantastic mathematicians but not always natural teachers (plus no you tube or rewind button in those days). You make it all so clear and understandable!!

You reached 85k by the time I was watching this video.. So I guess some congratulations is in order!!! 🎉

Very thorough explanation. Well done, sir! 😊

@WrathofMath

Жыл бұрын

Thank you Ezra!

Excellent video! Thank you.

@WrathofMath

Жыл бұрын

My pleasure - thanks for watching!

Very nice explanation. Well done.

@WrathofMath

Ай бұрын

Thank you!

THANKS

@WrathofMath

2 ай бұрын

You're welcome!

Can you make a video explaining the definition of definite integral in depth pls

At 2:47 shouldn't we say that f(x) itself converges to L but not the limit, since the limit is L itself?

Hi, I am having trouble understanding how to place an upper bound on the Epsilon-Delta Proof 2. Can someone please help me? I have asked several friends and my professor and I still do not get it. I believe the confusion for me starts at 15:00 of this video.

@cjjk9142

2 ай бұрын

just fix delta as 1 and solve the abs problem, |x-c| -1+c. So xE(1+c,-1+c). Now your final solution should have delta= min{ 1, some function of epslon}

Dude. Your content...Fuck. THANK YOU !!

Do i need to show the scratch work on my exams, or is it enough to just write down the proof? Cause then i dont show how i found what delta was gonna be, but it works...

@WrathofMath

10 ай бұрын

I'd ask your professor. Personally, I don't remember what my class was like in that regard.

So this is my informal paraphrase of the delta epsilon definition: "if the limit L exists at x = A, no matter how small epsilon is, there will exist a set of x values, which satisfies the condition |x = A| ≤ delta (where delta is a value we have to find but we know it exists) so that |f(x) - L| ≤ epsilon"....is that correct or close to correct?

@WrathofMath

Ай бұрын

Yes that is correct assuming you meant to write |x - A| and not |x = A|, except I would place the existence emphasis on delta, not the "set of x values". The existence of the limit at x=A guarantees we will be able to find a delta (which is not unique) so that all the other stuff is true. Thanks for watching!

@minamishi

Ай бұрын

@@WrathofMaththank you and yes I did mean to say |x-A|

I'm having trouble in showing that a limit doesn't exists, suppose that f(x) → 4, as x → 3, where f(x) = (2x - 1). How can we show that this limit doesn't exists?

@WrathofMath

11 ай бұрын

Since the limit of f(x)=2x-1 as x approaches 3 is 5, the most intuitive way to show the limit is not 4 would probably be to show it IS 5, combined with the result that functional limits are unique. Alternatively, pick epsilon=0.5. Then, no matter what delta we pick, no matter how close we require x to be to 3, some algebra will show you that f(x) is not within 0.5 of 4 at all points in this interval.

@siriuss_

11 ай бұрын

@@WrathofMath i will try that, thank you!!!

Super sir, please class join

@WrathofMath

7 ай бұрын

Thank you!

Is approaches or closer and closer the right term ?

@WrathofMath

2 ай бұрын

I'm not sure what you mean, could you rephrase?

@lifeforever1665

2 ай бұрын

@@WrathofMath functions like Lorenz Attractor and Chaos functions... How will we use those terms ?