Man this is incredible , I wish i had you back in my college days. Till now i found this as the most effective and literally understandable video on yt. keep bringing up this type of videos...thanks a ton!
Finally, an epsilon-delta video I like! And you're even wearing my favorite hat! I really like your observation that there's always an |x-a| hiding in the |f(x) - f(a)|, and you have to get good at factoring it. This is where your videos on the definition of derivatives have trained us well, because we need to use the same skills: the ability to cope with "f(x+h) - f(x)" is pretty much what you need to cope with "f(x) - f(a)". In my experience, the main problem people have with epsilon-delta is conceptual: it's hard to see why we're doing this "f(x) - f(a)" business. So I think of it like this: imagine a rectangle centered at (a, L) that is tall enough, and narrow enough, that the function f(x) never touches the top or bottom edges of the rectangle. Now, can you also shrink that rectangle down to any size, all the way down to nothing, and f(x) still never touches the top or bottom edges? If you can work out the geometry of a rectangle that makes that possible, then it means that "x" reliably gets closer to "a" as f(x) gets closer to "L". And if you can say that, then the limit exists. The height of those rectangles is 2*epsilon (i.e. it goes from L - epsilon to L + epsilon), and the width of those rectangles is 2*delta (i.e. it goes from a - delta to a + delta). So that is the game, counterintuitively enough: if you can work out dimensions for the shrinking rectangle, then the limit exists. From there, it is as you described: you factor out an |x-a| and then deal with the rest. To clean up that mess, there are two tricks you can use: you can limit your delta to a narrow region around "a" (in this case you used a value of "1"), and then you replace the entire mess with a value that you are confident will be larger than the mess in the narrow region. At that point, you're actually shifting to doing epsilon-delta on a different and simpler function, and then counting on squeeze proof logic: if your simpler function satisfies epsilon-delta, so must the original function. We probably need to say that delta = min {1, epsilon / 5} so that we don't forget that we've restricted our deltas to that narrow region.
Could u do more epsilon delta proofs for functions with polynomial in numerator and denominator, many more slightly harder proof for square root, reciprocal functions, trigonometric functions. Thank you so much, u educated with so much patience, love and a happy face, and explain every step.
2:59 1. Write general Definition of a limit 4:06 2. Apply the if and then condition to the actual problem at hand from the general definition Thank u 4:20 3. Scratch work - look for delta as a function of epsilon 13:00 4. Check the delta chosen as a function of epsilon to prove the limit. |F(x) -L|< epsilon QED In short For scratch work |F(x)-L| leads to delta as a function of epsilon For proving the limit Delta leads to epsilon using |F(x)-L|?
Wait a minute. Your answer is not 100% correct. Since you restricted |x-1|< 5, then your answer should be that delta = d = min{1, epsilon/5}. That is, if epsilon is greater than 5, then you choose d =1. if epsilon
6:56 We can also divide (x to the power 3 - 2x +1) by x-1 and get Thanks x to the power 2 +x-1 or Or by equating the coefficients of same powers of x. Thanks
Regarding August 28, 2023 "Epsilon-Delta" proof of cubic function. At 9:20 you presented an analogy/metaphor with "income and bills." Wonderful! Would it be possible with your busy schedule to continue the metaphor to the conclusion?
What a clear explanation. Thank you so much ! But i have one question: Within the range of delta = 1 there is between 0 and 2 a value of x that makes a zero of |x^2+x+1| at x = 0.618. Am i thinking the wrong way?
Because |F(x) -L| gets a zero in the denominator between - 1.6 & 0.6. Does it have any impact on how big x-1 can it be or it’s still 5 times. Should this restriction be a consideration, or it does not have any impact on delta?
Could u explain the following exercise? Prove Lim x tends to infinity 4x2-3x+2/8x2-6x+1 =1/2 Tried to understand it with the help of someone else, it’s difficult to clearly understand. Thank you
Some parentheses to separate the terms and maybe a couple of carets (^) to indicate where 2 is a power might help make things more clear: (4x^2-3x+2)/(8x^2-6x+1) would be the way I would write what I think the formula is supposed to be. If you have access to special characters you could replace ^2 with a ² if you prefer. Here's a non-rigorous way to look at it: As x goes to infinity, the x^2 terms are going to dominate the x terms and the constants. This means that the numerator and denominator will more and more closely resemble 4x^2 and 8x^2 as x becomes large. Feel free to work out a few evaluations at, say, x = 1, x = 10, x = 100 and so on to see that this happens. Don't divide numerator and denominator for this. Do them separately. This means that the ratio of the numerator and denominator will look more and more like (4x^2)/(8x^2), which is easily seen to cancel to be 4/8, which simplifies to 1/2.
Hey prime , instead of going through all those steps to find the value of X, can't u just get a value that when one is subtracted (x-1) gives 1 since value of a is 1
well the point is we want to pick a bigger number so that if x was in the interval from 0 to 2 that x we chose will be smaller than x in that interval, so we can say that for every x in that interval will smaller than the number we let x to be, which is 2 you can also pick 2,1 it doesnt matter actually if we let x=1,5 then there will be value of x that greater than 1,5 in the interval that we chose before
![int{(logx)(log(logx)]x} dx](http://i.ytimg.com/vi/B1OXPV1_RP8/mqdefault.jpg)
![int{(logx)(log(logx)]x} dx](/img/tr.png)
I really appreciate the way factor the cubic function.
Your prodigy in making hard things become completely livable is really unmatched.
You are a genius Sir. The explanations are simple and understandable
This was always such a hard topic for my students to understand. You did an excellent job.
AAAAA FINALLYY SOMEONE MADE VIDEO ABOUT THE CUBIC AND THE EXPLANATION WAS SOO PERFECTT LURVE LURVE LURVE!!!😭❤️❤️❤️
this is very good. You are an excellent teacher
Man this is incredible , I wish i had you back in my college days. Till now i found this as the most effective and literally understandable video on yt. keep bringing up this type of videos...thanks a ton!
Finally, an epsilon-delta video I like! And you're even wearing my favorite hat!
I really like your observation that there's always an |x-a| hiding in the |f(x) - f(a)|, and you have to get good at factoring it. This is where your videos on the definition of derivatives have trained us well, because we need to use the same skills: the ability to cope with "f(x+h) - f(x)" is pretty much what you need to cope with "f(x) - f(a)".
In my experience, the main problem people have with epsilon-delta is conceptual: it's hard to see why we're doing this "f(x) - f(a)" business. So I think of it like this: imagine a rectangle centered at (a, L) that is tall enough, and narrow enough, that the function f(x) never touches the top or bottom edges of the rectangle. Now, can you also shrink that rectangle down to any size, all the way down to nothing, and f(x) still never touches the top or bottom edges? If you can work out the geometry of a rectangle that makes that possible, then it means that "x" reliably gets closer to "a" as f(x) gets closer to "L". And if you can say that, then the limit exists.
The height of those rectangles is 2*epsilon (i.e. it goes from L - epsilon to L + epsilon), and the width of those rectangles is 2*delta (i.e. it goes from a - delta to a + delta). So that is the game, counterintuitively enough: if you can work out dimensions for the shrinking rectangle, then the limit exists.
From there, it is as you described: you factor out an |x-a| and then deal with the rest. To clean up that mess, there are two tricks you can use: you can limit your delta to a narrow region around "a" (in this case you used a value of "1"), and then you replace the entire mess with a value that you are confident will be larger than the mess in the narrow region. At that point, you're actually shifting to doing epsilon-delta on a different and simpler function, and then counting on squeeze proof logic: if your simpler function satisfies epsilon-delta, so must the original function.
We probably need to say that delta = min {1, epsilon / 5} so that we don't forget that we've restricted our deltas to that narrow region.
أستاذ شرحك رائع أشكرك على كل شيء تقدمه
ارجو المزيد من الفيديوهات في هدا الموضوع
أتابعك من ليبيا 🇱🇾
Grazie per la spiegazzione così chiara e semplice, sei molto didattico per spiegare questi concetti astratti,
I love this. I just love it. Thanks for such a lucid explanation of the Epsilon-Delta limit.
Could u do more epsilon delta proofs for functions with polynomial in numerator and denominator, many more slightly harder proof for square root, reciprocal functions, trigonometric functions.
Thank you so much, u educated with so much patience, love and a happy face, and explain every step.
This guy is wonderful.
2:59
1. Write general Definition of a limit
4:06
2. Apply the if and then condition to the actual problem at hand from the general definition
Thank u
4:20
3. Scratch work - look for delta as a function of epsilon
13:00
4. Check the delta chosen as a function of epsilon to prove the limit.
|F(x) -L|< epsilon
QED
In short
For scratch work
|F(x)-L|
leads to delta as a function of epsilon
For proving the limit
Delta leads to epsilon using
|F(x)-L|?
Awesome video. Epsilon delta proofs were my favorite topic in advanced calculus
Je trouve votre style d'enseignement très sympathique et bien sûr efficace
(I find your teaching style very nice and of course effective).
I’m loving your videos sir ,🔥🔥
you are a amzaing teacher wow!. Keep posting!
Wait a minute. Your answer is not 100% correct. Since you restricted |x-1|< 5, then your answer should be that delta = d = min{1, epsilon/5}. That is, if epsilon is greater than 5, then you choose d =1. if epsilon
Wonderful exposition! Thank you!
OMG I LOVE THIS ONE !! REALLY II HELPFUL THANKYOU SO MUCH ❤❤❤❤
tnx bro you really made it clear for me
Stellar explanation! Thanks for the help
I am so greatful for this
Helped a lot Sir
You're a genius!
I can't thank you enough
6:56
We can also divide (x to the power 3 - 2x +1) by x-1 and get
Thanks x to the power 2 +x-1 or
Or by equating the coefficients of same powers of x.
Thanks
Thanks for your video... master
Regarding August 28, 2023 "Epsilon-Delta" proof of cubic function.
At 9:20 you presented an analogy/metaphor with "income and bills." Wonderful! Would it be possible with your busy schedule to continue the metaphor to the conclusion?
Lovely and satisfying
I got you on a minor error. Let d represent delta. As you say very carefully, d>0 so (as I am sure that you know) you need to say that 0
You are correct. There's a great chance I'd have to redo this video
@@PrimeNewtons Maybe we can do the video together?
Limitless math knowledge comes from Prime Newtons! 🎉😊
Long live my blissful sir
Lovely and loveable
What a clear explanation. Thank you so much ! But i have one question: Within the range of delta = 1 there is between 0 and 2 a value of x that makes a zero of |x^2+x+1| at x = 0.618. Am i thinking the wrong way?
if it is zero , it will be less than epsilon for all epsilon. What's your problem?
Thank you so much!
WELL SPOKEN
Because |F(x) -L| gets a zero in the denominator between
- 1.6 & 0.6. Does it have any impact on how big x-1 can it be or it’s still 5 times. Should this restriction be a consideration, or it does not have any impact on delta?
❤❤❤
Please make a video in proofing a trigonometric function in epsilon delta definition
Please email me an example
чел, ты крут!
thank you
another easy way is to simply use synthetic division (remainder theorem) and find the factors.
To be fully rigorous we must take delta= min(epsilon/5, 1)
Could u explain the following exercise?
Prove
Lim x tends to infinity
4x2-3x+2/8x2-6x+1
=1/2
Tried to understand it with the help of someone else, it’s difficult to clearly understand.
Thank you
Some parentheses to separate the terms and maybe a couple of carets (^) to indicate where 2 is a power might help make things more clear:
(4x^2-3x+2)/(8x^2-6x+1) would be the way I would write what I think the formula is supposed to be. If you have access to special characters you could replace ^2 with a ² if you prefer.
Here's a non-rigorous way to look at it: As x goes to infinity, the x^2 terms are going to dominate the x terms and the constants. This means that the numerator and denominator will more and more closely resemble 4x^2 and 8x^2 as x becomes large. Feel free to work out a few evaluations at, say, x = 1, x = 10, x = 100 and so on to see that this happens. Don't divide numerator and denominator for this. Do them separately.
This means that the ratio of the numerator and denominator will look more and more like (4x^2)/(8x^2), which is easily seen to cancel to be 4/8, which simplifies to 1/2.
Hey prime , instead of going through all those steps to find the value of X, can't u just get a value that when one is subtracted (x-1) gives 1 since value of a is 1
Why take a=1?
I didn't choose a=1.
The only hard part was finding the trick to manipulate inequality 💀
You let 0 < x < 2, then let x = 2, why not let x = 1.5?
well the point is we want to pick a bigger number so that if x was in the interval from 0 to 2 that x we chose will be smaller than x in that interval, so we can say that for every x in that interval will smaller than the number we let x to be, which is 2 you can also pick 2,1 it doesnt matter actually
if we let x=1,5
then there will be value of x that greater than 1,5 in the interval that we chose before