He is so positive. When I am sad that I do not understand math I watch him )))
@ChaineYTXF
3 жыл бұрын
It's a passion and it shows. Best way to go about it🙂 very communicative
@Kernviter
4 жыл бұрын
I did not catch the "delta cannot depend on x" part in my lecture so this was a great help
@gianlucabrandi9675
3 жыл бұрын
i still don't get it
@xer_t3661
3 жыл бұрын
@@gianlucabrandi9675 for all epsilon > 0 there is delta > 0 such that, for all x, if |x-3|< delta then |x^2 - 9| < epsilon delta is defined for all x therefore it can not depend on the x value
@PrincessCupncake
3 жыл бұрын
Why couldn’t my professor just say that? That totally makes sense why it can’t depend on X when you put it like that
@beatadongwi5804
2 жыл бұрын
@@xer_t3661 I noted your wonderful explanation in my book.
@SlingerDomb
4 жыл бұрын
It took me 2 years to really appreciate this video !. As I’m taking Real Analysis right now, this video is amazing! Thank you Dr.Peyam. Your video are always great. Even though I didn’t understand it at first
@BootesVoidPointer
4 жыл бұрын
I'm in a similar position. I encountered the epsilon-delta definition of limits first when I was brushing up my calculus skills about two years ago, but I couldn't understand it thoroughly. Now I am preparing for real analysis and after watching these videos it finally starts to really click.
@josuearreaga7142
4 жыл бұрын
You did get me to smile when the results canceled into epsilon. pretty satisfying thank you
@keyyyla
6 жыл бұрын
Nice! More ε-δ-proofs with more complicated limits would be fun :)
@cajintexas7751
4 жыл бұрын
I've been looking at videos and reading explanations of this proof for days and could never understand why the step involving the minimum of 1 and epsilon even matters, and this video finally cleared it up. Well done and thanks.
@carlosvargas2907
6 жыл бұрын
Recuerdo haber estudiado esto solo y me gusta saber que hay quien lo entienda y muestre sin reparo. Muchas gracias, Dr Peyam!
@charlottemanier4997
3 жыл бұрын
Thank you so much for being such a kind voice of reason. This really calmed me down and helped me understand this concept when I was freaking out.
@mekkiferiel7832
4 жыл бұрын
thank you for the hard work ! i have finally understood the epsilon delta limits you are a great positive and kind teacher
@princesrivastava6153
4 жыл бұрын
I'm very glad to see ur video sir You are always be a inspiration for many poor children Thanks a lot for this video sir
@theprocessss
11 ай бұрын
This dude is the definition of a TANK!!! Thanks for the help brodie
@sjoerdmeijer3116
5 жыл бұрын
"It's like finding nemo but for delta" that made me laugh hard.
@xtv_
5 жыл бұрын
This professor is so passionate, and his teaching is so clear. Went with USC over UCI and the other UCs, but if I was at UCI right now, I would no doubt enroll in this professor's classes. Thank you so much!
@drpeyam
5 жыл бұрын
Awwwww, thanks ❤️
@wvyabhi
Жыл бұрын
dang im now at usc and that's why im on this video
@youssefaly7067
3 жыл бұрын
Thank you epsilon limit made no sense at first so I decided to watch both you and black pen red pen :).
@juancanekortegasanchez7961
3 жыл бұрын
Such a wonder, to see someone that explains it with so much positivity :)
@karanesh5249
4 жыл бұрын
Just loved your videos on epsilon-delta definition.......i was facing a hard time understanding proofs from my book,and your videos helped me a lot.
@n_x1891
6 жыл бұрын
Good explanation, when I first learned this I got lost in the notation but the idea is simple and clear!
@txikitofandango
4 жыл бұрын
In the past, I have wondered why it's called the (epsilon,delta) definition and not (delta,epsilon), when delta is associated with x and epsilon is associated with y. But now I see that it's because delta can depend on epsilon, and we put the independent variable first. Maybe?
@ugwuokechinedu8393
3 ай бұрын
Thank you sir😀 I'm really happy about delta-epsilon Now I understand what it is All about all due not whole process. But I'm happy thank you once again AND God bless you 😀
@binocentsenopile3008
5 жыл бұрын
so my lecture just made assumptions outta nowhere, we were all lost,thanks man now it makes sense.
@katiaarellano919
3 жыл бұрын
thank you Dr. Peyam!!!!!! One of the best worked out proofs on youtube
@baltimoredude1
4 жыл бұрын
This is one of the best explanations of the ε-δ proof. You and Prof. Rob Bob did an excellent job of this definition. I had to watch your video and bookmark it. Stewart's book did a horrible explanation of this subject. Thanks a million, Peyam.
@drpeyam
4 жыл бұрын
Thank you! Really appreciate it :)
@thundercraft0496
9 ай бұрын
thanks this is helpful i was planning on self studying analysis next year as my goal that i set for my self
@neerajbhatt700
3 жыл бұрын
Thanku so much sir , love from India ..way of your teaching is Amazing
@milajerwodckk6189
5 жыл бұрын
Great explanation!! I understand this so much now, Thanks!!
@chill4r585
3 ай бұрын
Hi !Dr Peyam, I love your videos and the way you explain things as intuitively as possible. I'd like to know whether you are missing at 9:52, since we know that delta has to be > 0 and epsilon also has to > 0, shouldn't we write delta > 0 somewhere? I mean, we've already established that epsilon > 0, that's good, but we haven't established that delta > 0. Is it missing or am I actually overthinking xD And sir, we know that we have to make delta = to min{1 , epsilon/7}. But why do we use epsilon / 7 at the end even though we don't know which one is smaller? Many thanks!
@GermanSnipe14
6 жыл бұрын
Can you do an example where the limit does not exist, or goes to infinity? I remember seeing an example but I don’t think it was explained well enough. Great video Dr. Peyam :)
@drpeyam
6 жыл бұрын
It’s one of my first videos, it has the same title but it’s Example 3 or 4
@GermanSnipe14
6 жыл бұрын
Ahh ok, thanks!
@Asli_bnk
4 жыл бұрын
the only one who makes it easy for me ! thank you so much .
@caridadekali1021
3 жыл бұрын
Thank you Dr peyam for making this concept very clear.
@shreyanshpandey4022
5 жыл бұрын
समझदार आदमी😂😂i mean very intelligent man ,,,like from india
@callmelilly1376
3 жыл бұрын
All i needed to get this topic, thankss!!
@atomf3550
3 жыл бұрын
hello sir, Can we find limit (not proving) by Epsilon delta definition ? please reply
@mohamedabouch3267
2 жыл бұрын
Thank you for all the valuable parameters, may Allah reward you. I have a question that I don't understand yet. Why did you choose the minimum value to take delta. (1,eppssilon/7) min and why not take the maximum value for delta?
@joaobellato4668
3 жыл бұрын
Thank you Dr. Peyam for your help, i didn't understad this in my classes.
@JavierTJL99
4 жыл бұрын
Hi, I'm still not very sure about the last part of the proof, why did you use |x-3| < ε/7 instead of |x-3| < 1, is it because you want to introduce ε into the inequality or is there some other reason?
@drpeyam
4 жыл бұрын
You want epsilon at the end :)
@kelvinmasasire6737
2 жыл бұрын
You really inspire me Doc 👍
@kingkong-gx9kf
6 жыл бұрын
This was a great calm lesson, thank you Very much
@mujiismail531
4 жыл бұрын
U'RE THE BEST MATH TEACHER EVERR!!!!
@m7mdabohashem
3 жыл бұрын
oh maaaan i love you , thank you very much you helped me a lot . And your style of explanation is unique and very good I love it. Keep the great work up🥰🥰😘
@BrainGainzOfficial
5 жыл бұрын
I love these proofs! Thanks for this video!
@drpeyam
5 жыл бұрын
Thanks so much!!!
@ivaniaareas6095
5 жыл бұрын
This video helped a lot, thank you Dr. Peyam
@dominicellis1867
4 жыл бұрын
I’ve been stuck on brilliants problem of this for weeks thank you sooooo much I crave to understand this
@drpeyam
4 жыл бұрын
:)
@MrBillonaire99
4 жыл бұрын
Great video! I found this example on James Stewart book and theres even the demonstration but I found it difficult to understand it, tough is very similar that the one you just show but I understand it very easily, for that you got one new subscriber
@drpeyam
4 жыл бұрын
Thank you!
@PrincessCupncake
3 жыл бұрын
THANK YOU!!!!!!! You just gave me my ah-ha moment and I’ve been trying to figure this out for a whole day!
@fromblonmenchaves6161
2 жыл бұрын
What happens if we choose (epsilon/5) instead of epsilon/7 ?
@SimchaWaldman
6 жыл бұрын
Till this day I still do not like the proof for this limit. Are there any other ways?
@everettmeekins1582
6 жыл бұрын
יהודה שמחה ולדמן since it is a quadratic, instead of choosing a number like 1, you could just do it more generally with epsilon. You would then have to use the quadratic equation to get your delta in terms of epsilon, and from there you could do the proof as normal
@SimchaWaldman
6 жыл бұрын
Example?
@everettmeekins1582
6 жыл бұрын
l x-3 l < d -> -d < x-3 < d -> 6-d < x+3 < 6+d -> l x-3 l * l x+3 l < d * (6+d). Now we can let d^2+6d = e, and then find d in terms of e.
@SimchaWaldman
5 жыл бұрын
Thanks. I saw this solution just a few weeks ago and I like it.
@carlosvargas2907
6 жыл бұрын
Sería interesante ver un ejemplo con funciones circulares
@streamerbox3172
3 жыл бұрын
Thank you dr 🇩🇿🇩🇿
@RUTHLESSS19
3 жыл бұрын
Damn it haha. As soon as you were like "ahha!" with regards to delta = epsilon / |x+3| I knew I found the right video because I tried using that as an answer and it was wrong haha.
@nahomtadesse7045
4 жыл бұрын
What a great approach ! Respect
@joaovictorandrade5720
5 жыл бұрын
Great explaination, hugs from brazil
@karabopaledi1491
3 жыл бұрын
Negative function video please ... My delta is negative on limit of (-3x+5) = -7 as x is approaching 4
@user-jc9kj7wt6j
2 жыл бұрын
If I start with |x-3|
@drpeyam
2 жыл бұрын
Yes
@user-jc9kj7wt6j
2 жыл бұрын
@@drpeyam Thank you!
@simonabodakpi3319
Жыл бұрын
So why did doc own start from min{€/7,1} instead of min{1,€/7}
@physicsandmaths7013
2 жыл бұрын
i like how you teach math its coool
@shandyverdyo7688
3 жыл бұрын
Nice explanation i like it.
@reniaschangadzo7257
3 жыл бұрын
Thanks a lot, finally i can now solve such problems
@ArtistRadhey
5 жыл бұрын
Sir Please solve this problem : Prove that simultaneous limit does not exist When f(x,y) = 1 , if (x, y) not equal to (0,0) And, f(x,y)= 0 ,if (x,y) = (0,0)
@thetruereality2
3 жыл бұрын
what do you do if you have combination of non linear functions such as 3x^2 + 4y +xz+1 =6 at (1,1,-1)
@albertomunoz4867
5 жыл бұрын
Great example. Thank you!
@mervegulec6954
3 жыл бұрын
Thank you teacher, i'm Turkish student :)
@drpeyam
3 жыл бұрын
Hos geldiniz
@mervegulec6954
3 жыл бұрын
@@drpeyam hos bulduk :)
@sorry4all
3 жыл бұрын
I like your English
@ivyzheng8681
4 жыл бұрын
Nice explanation! Thank you!
@helenallen7004
4 жыл бұрын
how do you know that the absolute value of x-3 is small and to assume that is it less than 1?
@charliedah
5 жыл бұрын
great video, thanks a lot!
@welfarewagonrepairs
4 жыл бұрын
can you explain why both epsilon/7 and epsilon/5 would be valid answers, in lecture we were told that only epsilon/7 was valid because its the smaller one
@gerardoelizondo9182
4 жыл бұрын
Sometimes they are equal but not always, so stay with the smaller one because if you choose the bigger you could exceed the other side of the interval.
@AdoptedPoo
2 жыл бұрын
how did u get the inequality -1
@devkumardenray2439
4 жыл бұрын
Great explaination!
@joeyasro8785
5 жыл бұрын
Does this definition rely upon inductive evidence? Is it true to say that the definition is saying that if one cannot find an epsilon for which there is NOT such a corresponding delta, then L is a limit? Or is there a rigorous method to prove that in a given case, no such epsilon exists?
@joeyasro8785
5 жыл бұрын
@@angelmendez-rivera351 Thank you for your reply. Given your answer, it seems to me, that such a definition relies upon induction. In other words, it doesn't prove that the limit DOES exist; it can only prove that the limit DOES NOT exist. For instance, just because I can't find an epsilon, doesn't mean that there isn't one. It's like science. Empirical evidence can't prove the theory, it can only disprove the theory. (Actually science can't even do that unless one presumes that empirical evidence is not subject to error.)
@physicstorque8084
4 жыл бұрын
But why exactly do we take delta to be the minimum of both?
@carinaut4371
3 жыл бұрын
Help me a lot!! Thank you so much!
@terefechali326
Жыл бұрын
Sorry ; why we can't say €/5?
@noway2831
4 жыл бұрын
How could this be used in computer science? If I wanted to find the derivative fo a function, instead of (f(x+0.0001)-f(x))/0.0001 we do something more rigorus?
@ishan4866
2 жыл бұрын
sir how to be as happy as you?
@kbolternorris2676
3 жыл бұрын
thanks friend that was useful
@sethchiasson5969
4 жыл бұрын
ur a legend man
@klementhajrullaj1222
Жыл бұрын
For me, it's more simple, because, when x->3 => |x+3|=6, so |x-3|
@aebramdylanoduca1021
4 жыл бұрын
You mentioned that since everything is positive, it's okay to say that 5< |x+3|
@aebramdylanoduca1021
4 жыл бұрын
I hope you will see this comment. I'm pretty desperate right now because it's our exam on Tuesday and I am still not confident with this proof.
@drpeyam
4 жыл бұрын
You set |x+4|
@aebramdylanoduca1021
4 жыл бұрын
@@drpeyam that's what I did, |x+4|
@drpeyam
4 жыл бұрын
|x| < 5, the one number among -3 and -5 with the highest absolute value
@aebramdylanoduca1021
4 жыл бұрын
@@drpeyam ohhhh so basically, I just need to find the absolute value of those negative numbers? Does that mean the answer would be epsilon/5 when the delta that I'm looking for is equal to epsilon/|x|? I really appreciate your help Dr. I feel like I'm starting to get it little by little.
@tomasbeltran04050
2 жыл бұрын
I loved this
@Dinghly
5 жыл бұрын
Thank you! Professor!
@gauravnimrani7021
4 жыл бұрын
what if i wrote that for each delta greater than zero there exists an epsilon
@Gabrieljesus-hi2ft
4 жыл бұрын
Thank u, helped me a lot!
@p.nandhukutty7331
4 жыл бұрын
Thank You Sir...
@tanmaysingh1131
6 жыл бұрын
It used to difficult for me to solve such questions but now it seems like a piece of cake. 👌
@drpeyam
6 жыл бұрын
So true! For me too :)
@ikhsandwiseptiansyah7326
3 жыл бұрын
many thanks sir
@planetofmathematics4099
3 жыл бұрын
Very helpful
@jimallysonnevado3973
6 жыл бұрын
But delta is just assumed to be like that why does it work if delta is just an assumption
@user-hy7um7cp5l
4 жыл бұрын
Please can you help my with this Lim1/x=2 when x approaches to 1/2 i really don’t know how i can solve it 🥺
@rachelwilliam9650
3 жыл бұрын
Thanks very much
@mohammedal-haddad2652
5 жыл бұрын
I think this is also a proof that every left-handed is a genius.
@bhataktiatma1080
5 жыл бұрын
it was helpful, thankyou
@unzi876
3 жыл бұрын
can you help me if lim x->2 (-x^2+5x-2)=4 then what happens if we look 0delta and |x-2||x-3|< epsilon then assume ; |x-2|
@drpeyam
3 жыл бұрын
|x-3| < 2
@unzi876
3 жыл бұрын
@@drpeyam oh ok, thank you very much sir❤️
@parsat5567
6 жыл бұрын
Awesome dude 👍
@aintrocketscience.1239
Жыл бұрын
Thank you sir
@timothystudies2753
6 жыл бұрын
you should do a series discussing the mathematical papers you have written
@drpeyam
6 жыл бұрын
Look at my 100th video special!
@adibchyy
4 жыл бұрын
How did you add the 3? Where did that come from?
@gerardoelizondo9182
4 жыл бұрын
perhaps you refered to this part -1
@adibchyy
4 жыл бұрын
@@gerardoelizondo9182 Thanks for the clarification my friend
@user-hy7um7cp5l
4 жыл бұрын
Thank you so much 💗💗💗😭
@pedrosso0
3 жыл бұрын
Part II just feels like you're doing part I again... I don't see the point of it
@dolevgo8535
6 жыл бұрын
so.. what is the purpose/usage of the epsilon-delta definition?
@XanderGouws
6 жыл бұрын
dolev goaz It gives us a rigorous way to prove limit identities/theorems.
@Koisheep
6 жыл бұрын
Imagine you have a function f(x,y) and you want to calculate its limit when x goes to 3 and y goes to 2. This is, lim [(x,y) -> (3,2)] f(x,y) (1) A theorem states that, if such limit exists (let's call the limit L), then we have L=lim(x -> 3) [lim (y -> 2) f(x,y)]=lim(y -> 2) [lim (x -> 3) f(x,y)] Nonetheless, the converse is not always true; the limit of the limit can exist despite (1) doesn't converge to any value. So what you have to do is to calculate the iterated limit AND THEN prove that For any epsilon > 0 exists a delta > 0 such as (x-3)²+(y-2)² < delta implies |f(x,y)-L| < epsilon
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