Limits are unique

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Dr Peyam

Dr Peyam

Күн бұрын

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@FT029
@FT029 4 жыл бұрын
I like the contradiction end to the proof a lot more! Thanks for showing that.
@dr.peyamsfan7158
@dr.peyamsfan7158 4 жыл бұрын
Analysis at its finest!!!!
@tmlen845
@tmlen845 4 жыл бұрын
It seems the original statement lim ... = S already implies that this is true, because an = sign is used. And I general if a = b and a = c, it must be that b = c.
@srpenguinbr
@srpenguinbr 4 жыл бұрын
Not exactly. It could mean the limit can take 2 different values. For instance, consider x=arcsin(0) you know x can be 0 or pi, so does that mean pi=0? or consider roots in complex numbers, they can take many different values and none of them is more correct than the others
@orlandomoreno6168
@orlandomoreno6168 4 жыл бұрын
@@srpenguinbr But that's inconsistent notation. = is transitive. Sometimes notation is used informally like that when it means something else and it's inelegant.
@martinepstein9826
@martinepstein9826 3 жыл бұрын
Good observation. Technically, before you prove limits are unique you shouldn't use that notation. You should just say s is _a_ limit and t is _a_ limit. Every time one expression "equals" two different values simultaneously baby Jesus cries ;(
@sebmata135
@sebmata135 4 жыл бұрын
Thank you for the vid Dr. Peyam! I ran into the same question last night while I was attempting the chapter 9 homework.
@leprofeet9989
@leprofeet9989 3 жыл бұрын
this was very well explained
@jayjayf9699
@jayjayf9699 4 жыл бұрын
I’m glad you finally added a subscript to the N as in your other videos the notation is ambiguous
@eliyasne9695
@eliyasne9695 4 жыл бұрын
This is quite elegant!
@sebmata135
@sebmata135 4 жыл бұрын
Wow the argument at the end doesn’t rely on the Least Upper Bound axiom so it’s very powerful. You can use this theorem to prove Monotone Bounded Convergence => Least Upper Bound property
@drpeyam
@drpeyam 4 жыл бұрын
Indeed! Check out the playlist, I do precisely that :)
@heliocentric1756
@heliocentric1756 4 жыл бұрын
What if humans really don't have free will, how can we choose epsilon??
@drpeyam
@drpeyam 4 жыл бұрын
Maybe epsilon chooses you 😂
@jayasuryav8324
@jayasuryav8324 7 ай бұрын
what if w-w0 is episilon/2
@coreymonsta7505
@coreymonsta7505 4 жыл бұрын
I never thought of doing it like that! Could take epsilon as the average of s and t, and N as the max of the N1 and N2. Then past N, sn is always supposed to be in two disjoint intervals somehow. Even one instance of such a thing is crazy lol
@late7245
@late7245 4 жыл бұрын
Is there any limit teorem that limit value would be equal to (s-t)/2 or (s+t)/2 ? (s,t mentioned in this video)
@fullfungo
@fullfungo 4 жыл бұрын
Yes, the limit uniqueness, which is what this video is about. If S and T are as defined in the video, then (S+T)/2 is equal to S and is equal to T.
@coreymonsta7505
@coreymonsta7505 4 жыл бұрын
Could you clarify what your question is?
@coreymonsta7505
@coreymonsta7505 4 жыл бұрын
La Te my guess is that you’ve got the thumbnail picture in mind in some way?
@hach1koko
@hach1koko 4 жыл бұрын
In the usual sense, no. But you can define other types of limits for which that could be the case. For example, the sequence u(n)=(-1)^n has alternating values 1,-1,1,-1 and so on, so in the usual sense it does not have a limit, but its "cesaro limit" is 0 which is the average of 1 and -1. (the cesaro limit is the limit of (1/(n+1))*(sum u_k from k=0 to n), it's the limit of the average of the nth firsts u_k terms). One can prove that if a sequence has a limit L, then its cesaro limit is also L. So it's a more "general" definition of a limit.
@mikhailmikhailov8781
@mikhailmikhailov8781 4 жыл бұрын
R is Hausdorff. Done.
@mikhailmikhailov8781
@mikhailmikhailov8781 4 жыл бұрын
I never liked doing things in the style of real analysis, when there is a visual way to understand the result that works very generally. Limits and continuity are just that.
@late7245
@late7245 4 жыл бұрын
Please analyze lim n->inf (sin(arcsin(n)) in another episode?
@snejpu2508
@snejpu2508 4 жыл бұрын
Arcsin is inverse of sin, so sin(arcsin(n)) is just n.
@late7245
@late7245 4 жыл бұрын
@@snejpu2508 nope.graph( sin(arcsin(x))) != graph (x). For this reason limit must be analyzed.
@snejpu2508
@snejpu2508 4 жыл бұрын
@@late7245 Oh, right. It works only for x from -1 to 1. So actually the x cannot approach infinity, because it's not in the domain. It's like calculating the limit of x approaching minus infinity of sqrt(x). I'm not sure, if we can even do that.
@late7245
@late7245 4 жыл бұрын
Pardon. I meant arcsin(sin(n)) ?
@iamtrash288
@iamtrash288 4 жыл бұрын
arcsin takes only the values from -1 to 1 on the real number line though
@Domzies
@Domzies 4 жыл бұрын
I knew we should have used epsilon/2
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