The Limit of a Sequence is Unique Proof

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The Limit of a Sequence is Unique Proof. I hope this video is helpful.

[@louismensah4700]

Why e/2 ?

[@aaryadhurwey6282]

Why we had written there (an-a)less than epsilon by 2??

[@covariance5446]

If Xn converges to some value (call it a) then by definition you can say that when you have a certain "roadmarker" (call it N or N-nought or N-sub-1, the name doesn't matter) in the element of the naturals, it will be the case that the absolute value of the difference between Xn and a should be able to be smaller than any number you choose. Traditionally, the choice is to call that value Epislon, but it's just name, and you can call it whatever you want. If you're just trying to say that the abs.value of the difference between xn and a can be made arbitrarily small, you can also say that as n progresses past your roadmarker, then the diff between xn and a will also be smaller than epislon/two. The nice thing about doing this is so that when he (Math Sorc) sets up the inequality, you're adding two terms for which each term is an absolute difference between xn and the ostensible limit. If each one was said to be less than epislon over two, then the sum of both can also be said to be less than 2*epislon/2. Just like how if 1 < 5 and 2 < 5, then 1+2 is less than 5 + 5. So, now you have the sum of |a-xn| + |xn-b| being less than epsilon. This allows us to explicate the contradiction (of |a-b| being less than epislon, which means |a-b| being less than |a-b| which is an obvious contradiction).

[@TheMathSorcerer]

[@successorof68]

Its been 7 years since you uploaded, no way you are going to reply but worth a shot. how did you get eps/2+eps/2 from |a-an|+|an-b|? really hope you see this

[@TheMathSorcerer]

Each of those is less than eps/2 so i replaced each with eps/2

[@Hosseinshafii]

why epsilon/2 ?

[@joaohax52]

well, since you can set every epsilon within, epsilon/2 is one of the possibilities. that's just a step to show that there is an epsilon that doesnt follow these conditions (you can show that there are more than just one, but it is a different way..

[@londynocchipinti6607]

truly do not understand why professors make this so complicated. the way you explain it is perfect! thank you!

[@TheMathSorcerer]

You're very welcome!

[@enkixloki]

pls someone explain to me what does episilon,N,n....represent...im still confused

[@debendragurung3033]

Its been a month I am scratching my head around this proof I have on a lecture note. And it's been 4 years since this video been uploaded. I am glad I found it finally and I can actually make a good sense of it now. You are a saviour mate. Thanks and keep them coming.

[@TheMathSorcerer]

So glad it helped this stuff is hard😄

[@enkixloki]

pls someone explain to me what does episilon,N,n....represent

[@debendragurung3033]

@@enkixloki You have to understand from the very beginning and also understand why, why is the bigger question. First thing is we are using Natural number tools to understand real numbers. That's the reason why we see use of Natural numbers everywhere throughout this lecture. We have to assume we know very little about real numbers. That's even though we have been pretaught many things about real numbers before college and we have been using them without evidence of concrete proof. The best place to start as any lecturers teach is the existence of numbers that are not rational. First of course in the liat is sqrt(2). Earlier mathematician did believe that any numbers between two integers must be some fractional number . You can check an easy proof that sqrt(2) cannot be a fraction. Yet simple logic tells u it must exist and it is lesser than 2 but greater than 1. So what other numbers are there in the interval [1,2] that are not rational. Hence the irrational. Now what does sequence have to do with any of this. To define sequence we use NATURAL NUMBERS. It's the tool we are all born with. We can discretely count them 1,2,3.... And we can without any proof say 3 is greater than 2 and 2 is greater than 1. We are simply using this tool to reconstruct understanding of real numbers. We can of course never write out all real numbers using this fixed set of symbols in natural numbers. But best we can do is get very close to it. For instance π =3.14...... with endless digits of natural numbers. What we doing here is simply getting it's approximate value. 3

[@enkixloki]

@@debendragurung3033 i appreciate your explanation but im still stuck with it i will try to comprehend the topic more....i would like it if u helped me more it seems like u have understand it well and i wanna reach your level just show me z way

[@soumyaranjanbiswal1827]

Why r u taking€/2

[@elisha2358]

Because we assume (an) converges to 2 different values, namely, a and b.

[@smoosq9501]

my god!!!!, i was really confuse when the book only said the |a-b|

[@SequinBrain]

I don't get these, but I'll keep trying anyway.

[@TheMathSorcerer]

Yeah keep trying man,I should and will make more of these!!

[@jmitchell8082]

Thank you for this, so much clearer than my lecture notes

[@MrBABYCOLZ]

Why is it epsilon/2?

[@doodledeedah]

Probably a little late for your exam, but you may think of ε as (a-b)/2 it's the same thing and clearer. i.e Suppose a != b. W.l.o.g. we may assume that a < b. We choose ε = (|a-b|)/2 > 0 (as a is bigger than b). Our epsilon is half the distance between a and b shown below. a a+ε = b - ε b ^ ^ ^ Here a is one limit, b is the other. Since (an)n∈N converges to a, there is some N1 ∈ N such that a − ε < an < a + ε for all n ≥ N1 - this is the interval our limit for an lies in. Since (an)n∈N converges to a, there is some N2 ∈ N such that b − ε < an < b + ε for all n ≥ N2 - this is the interval our other supposed limit for an ALSO lies in. hence using n ≥ N1 and n ≥ N2 take the lower bound of the limit b's interval and the upper bound of limit a's interval we get b - ε < an < a - ε < ---- this is a contradiction since a+ε = b - ε must hold for our initial assumption a != b to be true. Therefore our initial assumption is false and limit a == b (limit is unique QED)

[@kendedetar]

@@doodledeedah You stated that a < b, then said "(as a is bigger than b)". So you should either write |b-a|/2 = ε, or just assume that a > b. But correct me if I'm wrong :)

[@joaohax52]

@@kendedetar |a-b| = |b-a| since it's always positive

[@lemyul]

it doesnt, but it will, later, hopefully

[@mohmd252]

oh sorry , your proof is correct, please forgive me

[@TheMathSorcerer]

+mohmd252 np:)

[@twebazegilbert6871]

You are actually complicating math😢

[@ROCKaholic]

What about delta (δ)? What if there is some positive delta that lives inside the Reals that exists in the definition of a limit? My book defines a limit as such: (Given a point of interest, a possible limit value, a defined function f: S -> R* where R* means the set of all real numbers and S ⊆ R* contains a deleted neighborhood of a) The limit of f(x) as x approaches a = L is defined as: IF [∀ (positive) ε ∈ R*, , ∃ (positive) δ ∈ R* such that ∀ x ∈ S; 0 < |x - a| < δ], then | f(x) - L | < ε. You didn't seem to include this type of definition, which is fair because every class and every book teaches it differently and your method is both true and helpful, but I learned it a little differently, so with what I gave, how could this proof be changed to pander to this formal definition of a limit?

[@joaohax52]

we need to realise that an and bn already converge to a and b, respectively. it doesnt need to be mentioned, since we just need to find an epsilon such that doesnt follow these condirions and then shows that is false (for every case)

[@joaohax52]

and if a = b, |a-b| < epsilon => 0 < epsilon, it returns into a true, that always epsilon is positive

[@shirleymoon9934]

Very helpful 👍🏻

[@ethanperry9544]

great proof and all but you should use Natural number instead of Z+😠

[@alexandertownsend3291]

Unique proof of what? Lol

[@tracyjannalzaro5099]

He meant, "The Limit of a Sequence is Unique, Proof." hhe. :)

[@Sureshiru]

Why u take Max{N1,N2} value

[@GagandeepKaur-mw4ih]

pls tell me why is it epsilon by 2

[@doodledeedah]

Probably a little late for your exam, but you may think of ε as (a-b)/2 it's the same thing and clearer. i.e Suppose a != b. W.l.o.g. we may assume that a < b. We choose ε = (|a-b|)/2 > 0 (as a is bigger than b). Our epsilon is half the distance between a and b shown below. a a+ε = b - ε b ^ ^ ^ Here a is one limit, b is the other. Since (an)n∈N converges to a, there is some N1 ∈ N such that a − ε < an < a + ε for all n ≥ N1 - this is the interval our limit for an lies in. Since (an)n∈N converges to a, there is some N2 ∈ N such that b − ε < an < b + ε for all n ≥ N2 - this is the interval our other supposed limit for an ALSO lies in. hence using n ≥ N1 and n ≥ N2 take the lower bound of the limit b's interval and the upper bound of limit a's interval we get b - ε < an < a - ε < ---- this is a contradiction since a+ε = b - ε must hold for our initial assumption a != b to be true. Therefore our initial assumption is false and limit a == b (limit is unique QED)

[@timechanges1670]

Bcz v hv to prove tat epsilon so dat v r taking epsilon/2

[@msk.nfazal9757]

Thanks :-)

[@khadijaeldesouky4586]

i read all the comments about why is it epsilon divided by 2 and why is epsilon equal to abs(a-b) but i still can't understand.

[@listen-to-korin]

I hope you got it but in case you didn't and for anyone else stuck, read the section 3 "common questions" part in my Real Analysis progress diary blog: rakugakistudy.com/tutorials/62708b581b716f001650bf0b Hopefully it helps :)

[@kalaivani.v9833]

Tq

[@enkixloki]

pls someone explain to me what does episilon,N,n....represent

[@TechToppers]

Look up the definition of convergence of a sequence.

[@azhinrafiq4676]

Thanks 👍🏻

[@TheMathSorcerer]

You are welcome!

[@suchetasinha2205]

Sir why we r taking the max value of n1 and n2 ....????

[@TheMathSorcerer]

well we need n > N_1 and n > N_2, so if we let M be the maximum, whenever n > M, it's also bigger than N_1 and N_2, so we have what we want(this way both inequalities hold)

[@shreyashraj6906]

Thanku🙂

[@TheMathSorcerer]

👍

[@nmtran]

Thank you so mucchh

[@lijinnan9647]

great!

[@jisungyu2]

How did you come up with epsilon = abs (a-b)? what if epsilon is not equal to abs(a-b)?

[@DOtman2222]

In the definition of the convergence of the sequence: "For ANY epsilon >0,.." this means whatever positive number you pick for epsilon the other conditions should work(so that this sequence is convergent). Now this allows us to pick any number we want for epsilon(and the conditions will remain true) therefore we pick |a-b| which will help in attaining contradiction.hope that helps

[@Codepmz.]

@@DOtman2222 well replied