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!
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 ;)
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.
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!!
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.
Hi. As we know that we can control |x-c|, we are trying to obtain an inequality of that form. Once we have established that, we start by saying, let us take x to be in a neighborhood of 1 to 2, i.e. x belongs to (2-1,2+1), which helps us put an upper bound on the value of |x+2|, further helping us in developing the inequality required. Now, one thing you need to make note of is, by saying x is in a neighborhood of 1 to 2, we have assumed delta to be less than or equal to 1 (less than or equal to, because the neighborhood could be smaller as well, the value 1 is just an arbitrary number we have set to our convenience, so as to establish a strict upper bound). Upon putting this, we get an inequality, |x-2| less than epsilon/5. Now, as per our previous assumptions, we also know that |x-2| must be less than or equal to 1, as x belongs to (1,3). So, we cleverly have chosen delta to be the minimum of these two (1 and epsilon/5). This might be a little confusing as to why we did that, but the explanation to it lies in the last slide. In order to prove that |f(x)-L| is less than epsilon, we need one term to generate a 5, and and the other term to generate an epsilon/5 so that the 5 gets cancelled and we get the necessary inequality. In order to do that, we need to make sure that x is less than 3, which we make sure by taking delta as we have. Take 2 cases, one with delta less than 5 and one with it being more than 5, which will give you a better understanding. And this also ensures that |x-2| is less than epsilon/5. Hope this helps!
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...
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?
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.
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?
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!
Do you mean why is |x-c|>0? Like, why do we require x to be distinct from c? This is because x needs to be in the domain of f but c does not. We can define the limit of a function at a point not in its domain, such as the limit of (x^2-9)/(x-3) as x approaches 3.
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!
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 ;)
Thanks so much! Glad it helped!
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.
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!!
Very thorough explanation. Well done, sir! 😊
Thank you Ezra!
You reached 85k by the time I was watching this video.. So I guess some congratulations is in order!!! 🎉
Very nice explanation. Well done.
Thank you!
Excellent video! Thank you.
My pleasure - thanks for watching!
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.
Hi. As we know that we can control |x-c|, we are trying to obtain an inequality of that form. Once we have established that, we start by saying, let us take x to be in a neighborhood of 1 to 2, i.e. x belongs to (2-1,2+1), which helps us put an upper bound on the value of |x+2|, further helping us in developing the inequality required. Now, one thing you need to make note of is, by saying x is in a neighborhood of 1 to 2, we have assumed delta to be less than or equal to 1 (less than or equal to, because the neighborhood could be smaller as well, the value 1 is just an arbitrary number we have set to our convenience, so as to establish a strict upper bound). Upon putting this, we get an inequality, |x-2| less than epsilon/5. Now, as per our previous assumptions, we also know that |x-2| must be less than or equal to 1, as x belongs to (1,3). So, we cleverly have chosen delta to be the minimum of these two (1 and epsilon/5). This might be a little confusing as to why we did that, but the explanation to it lies in the last slide. In order to prove that |f(x)-L| is less than epsilon, we need one term to generate a 5, and and the other term to generate an epsilon/5 so that the 5 gets cancelled and we get the necessary inequality. In order to do that, we need to make sure that x is less than 3, which we make sure by taking delta as we have. Take 2 cases, one with delta less than 5 and one with it being more than 5, which will give you a better understanding. And this also ensures that |x-2| is less than epsilon/5. Hope this helps!
@@saiaditya8594you are a lifesaver
At 2:47 shouldn't we say that f(x) itself converges to L but not the limit, since the limit is L itself?
Can you make a video explaining the definition of definite integral in depth pls
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...
I'd ask your professor. Personally, I don't remember what my class was like in that regard.
THANKS
You're welcome!
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?
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.
@@WrathofMath i will try that, thank you!!!
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?
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!
@@WrathofMaththank you and yes I did mean to say |x-A|
What is the reason to exclude c(limit point) when we write the defination
Do you mean why is |x-c|>0? Like, why do we require x to be distinct from c? This is because x needs to be in the domain of f but c does not. We can define the limit of a function at a point not in its domain, such as the limit of (x^2-9)/(x-3) as x approaches 3.
@@WrathofMath ok get it
Super sir, please class join
Thank you!
Dude. Your content...Fuck. THANK YOU !!
Is approaches or closer and closer the right term ?
I'm not sure what you mean, could you rephrase?
@@WrathofMath functions like Lorenz Attractor and Chaos functions... How will we use those terms ?