Thanks dr πam, When I was thaught this at uni I thought: what are they doing if we need to demonstrate de epsilon part, but you clarified some doubts I had up until now. I think I saw these kind of demonstrations more clearly in vector calculus idk why hahaha
@sevarussnape6797
2 жыл бұрын
한국 학생입니다. 한국에서는 3차함수에 대한 엡실론-델타 논법 증명자료가 인터넷에 별로 없었는데, 우연히 이 채널을 발견하게 되어 과제에 도움이 되었습니다. 감사합니다.
@yasho772
Ай бұрын
Great video! I was so stuck on the fact on how can we assume δ to be 1, great clarification. Thank you so much for your efforts!
@sanjeevkumardhiman6783
4 жыл бұрын
Old thumbnail still in this video that’s absolutely amazing I wanna see more epsilon delta coz it would clear 99.99999% of my doubts thank u dr Peyam Love from India happy 48 K subs may ur subs diverge like the harmonic series
@aryamankejriwal5959
4 жыл бұрын
But.... doesn’t that mean... his channel would grow really slowly? 😂
@Kdd160
4 жыл бұрын
@@aryamankejriwal5959 slow but steady
@MrCigarro50
4 жыл бұрын
Very good example, just as the others. Thank you Dr. Peyam.
@sadiakhan6500
3 ай бұрын
this guy is so damn funny, I already love math, but he makes these difficult proofs so fun. I saw his other video on limits as x-> infinity and he brought up Nemo again. Idk why but delta being Nemo will always make me chuckle. Great videos, thank you so much :)
@foreachepsilon
4 жыл бұрын
This is a good way to explain the motivation for the use of min{*}.
@user-di9xh8xy2t
2 жыл бұрын
Thanku brOther...from Pakistan❤🤗
@MathemaTeach
4 жыл бұрын
Well explained Dr Peyam. Thank you. 👍
@geraldnsunga
2 жыл бұрын
Dr Peyam. Thank you a thousand times. I wouldn't have known all this but because of you I am able to understand and solve. Thank you very Much. Allah bless your life. Vielen Danke. Grüss Got. 🤝🤝🤝.
@artwitha.g.8955
4 жыл бұрын
God bless you and your videos you truly are saving my semester :D
@zainabsidiq9403
2 жыл бұрын
I just legit asked this on a previous epsilon video and saw this!!!🥺Thank you Dr.
@CelalCankutAldemir
3 жыл бұрын
Wow! I really like ur way to tell, you are the best instructor I've ever seen !
@gabrielemarchioni567
4 жыл бұрын
4:17 HUUUUUGE! However, fantastic video AS ALWAYS :D !
@plaustrarius
4 жыл бұрын
I have been waiting for this one oh heck yes!!
@user-dx9gi8ql2c
8 ай бұрын
Another way to do this could be by setting z = x-2. This lets you get x^3 - 8 = (z+2)^3 - 8 = z^3 + 6z^2 + 12z. From there you can try and derive the following proof: Let e > 0. Let d = min{e/19, 1}. If 0 < |x-2| < d, then |x^3 - 8| = |(x-2)^3 + 6(x-2)^2 + 12(x-2)| ≤ |x-2|^3 + 6|x-2|^2 + 12|x-2| < d^3 + 6d^2 + 12d = d(d^2 + 6d + 12). If e ≥ 19, then d = 1, so |x^3 - 8| < d(d^2 + 6d + 12) = 19 ≤ e. If e < 19, then d = e/19 < 1, so |x^3 - 8| < d(d^2 + 6d + 12) < d(19) = 19d = e.
@MathswithMuneer
4 жыл бұрын
An amazing teacher, I wonder if we could work together
@historybuff0393
4 жыл бұрын
You said that Ix-2I is less than the min {1, epsilon/19}, but in the end you chose only epsilon/19 instead of 1. On your second graph you showed that 1 could be greater than epsilon/19. Why did you choose only epsilon/19 in the end as being less than Ix-2I and not 1?
@drpeyam
4 жыл бұрын
I also used delta < 1 because we needed |x-2| < 1 to get the 19 part
@tomasbeltran04050
2 жыл бұрын
Lovely
@ajiwibowo8736
4 жыл бұрын
That line makes me cant stop laughing. "x can be HUUUUUUUGE"
@nathanisbored
4 жыл бұрын
is there a problem with defining the limit only in cases where evaluation doesnt work? like, lim x->2 = f(2) if f(2) exists, and use the epsilon delta definition otherwise
@Kdd160
4 жыл бұрын
Yeah we can show that a limit does not exist using epsilon delta; he did that in a video ----. Lim as n->inf of (-1)^n doesnt exist
@thedoublehelix5661
4 жыл бұрын
No we can't do that because not all functions are continous. Take the function that is 1 everywhere except for x=0 where it jumps to 2. The lim f(x) as x goes to 0 should obviously be 1 but its 2 under your definition.
@nathanisbored
4 жыл бұрын
@@thedoublehelix5661 why do we need this property? "should obviously be 1" for example
@dstigant
4 жыл бұрын
The problem with your suggested approach is that continuity is defined in terms of limits. F(x) is continuous at x=c iff lim x->c- f(x) = lim x->c+ f(x) = f(c). You’re using the fact that the function is continuous at x= 2. But how would you prove that fact? By showing that the limit as x-> 2 is the same as f(2). So you have to have established that the limit exists first.
@nathanisbored
4 жыл бұрын
@@dstigant i never said anything about continuity. i just said define the limit such that: lim x->a = f(a) if f(a) exists, and use epsilon-delta def otherwise
@sanjeevkumardhiman6783
4 жыл бұрын
In my book abs(x^3-8)
@pandabearguy1
4 жыл бұрын
Know how I know its true? I have wolfram alpha and impeccable eye sight
@pchaitanya4230
4 жыл бұрын
Does y=X and y=2x meet at +/- infinity???
@AryssaRiyasat
Жыл бұрын
No.
@randomstuff9960
3 жыл бұрын
This is so cool!
@cecildesist
Жыл бұрын
But what is epsilon as a decimal number.
@Absilicon
4 жыл бұрын
Bruh you're looking spiff 👌
@klementhajrullaj1222
Жыл бұрын
For me it's wrong again, because, when x->2 => |x^2+2x+4|=12, so |x-2|
@jayassharma659
4 жыл бұрын
Thank you for the help!!!!!
@tamajongmichaelnkeh1978
4 жыл бұрын
I recently got stuck on the proof for x->0 , (cos(x)-1)/x -> -1/2
@drpeyam
4 жыл бұрын
Check out my proof of sin(x)/x
@tamajongmichaelnkeh1978
4 жыл бұрын
Im sorry that function should be (cos(x)-1)/x² i will check your proof thanks
@tamajongmichaelnkeh1978
4 жыл бұрын
@@drpeyam please can you formulate an epsilon delta proof for functions like these?
@ruffifuffler8711
4 жыл бұрын
Dominant field frying electric soldiers, never miss a beat, They dance in the absence of the known, 'till they find a seat. After that, it's a bow to their feet. Kinked Daddy Long Leggs.
@liminf5606
4 жыл бұрын
Nimo=δ…
@anjummuneer5081
4 жыл бұрын
I think you have caught cold.
@drpeyam
4 жыл бұрын
No, just allergies
@anjummuneer5081
4 жыл бұрын
@@drpeyam Ok, do stay safe at home during lockdown.
@iliashamdad6859
3 жыл бұрын
mnin tchri kaka mn ouedkniss
@gobberman09
4 жыл бұрын
I less than 3 your jokes Dr.
@iabervon
4 жыл бұрын
In this section, I epsilon greater than his jokes instead.
@gobberman09
4 жыл бұрын
Ha ! I didn't think out that one!
@TSR1942
8 ай бұрын
Gentleman please do not obstruct what you are writing.
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