Real Analysis| Three limits of sequences by the definition.

We give three examples where we prove the value of a few limits of sequences using the precise epsilon-N defintion.

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[FGL_tv]

Just wanted to say these videos are incredibly helpful. If I need a refresher on something I can pop in for 20 minutes and get a cogent, informative lecture. Really appreciate your work!

[yuyaogawa]

I am self-studying real analysis using a book, Understanding Analysis, and there are some points here and there that sort of lack clear explanations in the book which really confuses me sometimes. But your are making things super clear and explain things in the terms that are easily comprehensible for someone like me who is not used to the language of mathematics in general, so I cannot appreciate enough Michael! Thank you so much!!

[wtt]

Great professor.Your explanation is so clear!

[SamuelAdu-Gyamfi-jp]

I am Sam, always watching you from Ghana .. I am huge fan of your real analysis lectures ... it is really preparing me for my exams 😊

[saicharanritwikchinni]

YEE!! The best explanation I have seen till now.

[gabriellaaileenmendrofa]

i don't get it, 2-n/(n²+n)<ε and 2-n/n² is bigger than 2-n/(n²+n) but what gruarantees 2-n/n² < ε? i mean, it sounds like this 5<7, 8>5 so 8<7

[caladbolg]

If you took a strict inequality in the definition of the limit, "for all n>N", you wouldn't need to constantly deal with this N=ceiling+1 thing, no?

[rogerlie]

In the second example I would just choose N = ⌈15/4𝛆⌉. You only need to find one such N and not necessarily the best one.

[jhd]

I appriciate your effort and work really much. You do a difference because you are SO great at explaining things. :-)

[phyjob]

Thanks for your videos. Speaking of limits of sequences, would it be possible that, in a future video, you address the limit of S(x)=\sum_{n=0}^\inf (-1)^n x^{2^n}, when x tends to 1 from below?

[backyard]

18:10 you don't even have to worry about the maximum, just pretend your sequence doesn't have the first term, I.e. it starts at n=2, because for any convergent sequence, the new sequence achieved by eliminating the first k elements will have the same limit since the rest of the infinitely many terms behave the same way anyway. More precisely lim of a(n) = lim of a(n+k) for any natural number k, which can easily be proven

[viniciusteixeira]

Great work, Mic!
Which book you recommend to use to study this topic?

[ACTION]

Why add 1 to equality? I didn't get that.

[Will-nfgf]

So am I wrong or does this definition say "if after some point (N) in a sequence (a_n within real numbers) the value of the sequence is always contained within some 'arbitrarily small' region (epsilon) that continuously approaches a value (L), then the limit as a point in the sequence approaches infinity is equal to the value being approached by the epsilon region" and the whole n >= N and N = ceil(something containing epsilon) + 1 business is to 'seed out' this point N? Also, why does it matter that N is a natural number? I hope I am at least on the right track in terms of a starting point

[MounirGalati]

Thank you so much for your explanation it was amazing.
I would improve my skill in epsilon N definition could you please give me some advice or where I could find some exercise with solution if it is possible

[leroycheng]

Why do we need to add a 1 next to the (1 over epsilon)? Why do we need to make sure it is big enough? Thanks in advance!

[davidmoss]

I'm a little confused why you need to add + 1 in order to guarantee N sufficiently big. Shouldn't the ceiling function suffice?

[ayoubnouni]

6:15 i think the floor( x) <= x not the inverse !!!!
So, to fixe the issue, i think n > N should be enough to say that n> 1/epsilon^2

[pavybez]

In one of the proofs you choose N = ceil(15/(4.epsilon) - 5/2) + 1. But that doesn't guarantee that ceil(15/(4.epsilon) - 5/2) is positive. I think you should add +3 instead of +1?

[jshook]

15:20 can someone help me understand how making the abs(An - L) term larger will preserve the fact that it is less than epsilon?