Epsilon - Delta Proof (constant function)

[@Kwinnbujik]

This Man is saving my life very soon for the exams ahead.

[@Misteribel]

I wished they taught us the 'simple' stuff at secondary school as well. My teacher was more like "it's like that because it's defined like that". Thanks for fixing the gaps in the trains of my old teacher!

[@amlalashhab1719]

شكرا على ما تقدمه اقدم لك تحياتي من ليبيا 🇱🇾

[@JourneyThroughMath]

Ive been telling my students lately "Sometimes the easiest examples are the hardest examples". Which sounds weird, but I show them problems at a certain difficulty and they become accustom to that difficulty. Anything harder OR easier, it seems harder for them.

[@maarinahope]

at least you are being honest

[@goymath]

Great! Please do a similar video on quadratic functions

[@anonakkor9503]

3:25 Bro just this little part helped me so much😭😭😭 was confused about something but nothing I was doing this all the time…..

[@MariinaNuuyoma]

You are definitely the goat

[@mkhumbie5484]

Great work thanks, my contribution will be that I would rather have proven the other way around. What do I mean? The "IF statement" is what's given so by 0

[@glorrin]

This video is great I would like to see more for example a video where the proof doesnt work because there is no limit Can we use it to prove there is no limit ? also i was wondering if this was a proof of the limit or a proof that a function is continuous at this point... and then I realised it is the same thing.

[@josephlorizzo8997]

it's not the same thing, to be continuous in a certain point the function must be equal to its limit in that point, but It can even not be defined In that single point for the limit to exist

[@davicamponogara4005]

You are amazing man, thank you so much

[@PrimeNewtons]

Happy to help!

[@nbdb-u5b]

first i wanna say thank you what u r doing for us. i wanna ask you you said if you divided the episilon by zero youll get different options, does any number can be divided by zero i think the answer is undefined yeah

[@alexander8971]

How would you show this to be true when f(x)=c and x is approaching a?

[@shmuelzehavi4940]

Take this presentation, replace "4" with "c" and "3" with "a" and then you'll get the answer.

[@bizoitz86]

|f(x) - L| = |c-c| = 0 < |x-a| < delta = espilon

[@odalesaylor]

When you need a delta for this, could you just assign delta to be epsilon or maybe epsilon divided by 2.

[@shmuelzehavi4940]

Or to assign an arbitrary value (δ = 1) and show that the "ε" requirement is satisfied for any x which satisfies: 2 < x < 4 .

[@supe-ns3gl]

Yeah I never saw you divide a number by 0😂 I'm trying hard not to believe that

[@MekbibChaka]

Amazing

[@kalwenyaNdiholovanhu-zy3td]

Very helpful

[@Heybem05]

You are special ever but didn't get what you deserve, Hmmm, here is how the world treats ........ GREAT!!!

[@nakyejwesarah4040]

Thnx bro

[@WhateverIwannaupload]

because delta cannot be found in relation to epsilon it means the value of delta does not matter? and since all that is left is 0 |0| < epsilon that is because epsilon deals with the y value and the delta deals with the x value.

[@TegenawFenta]

It is best but you can do another trip

[@KurstesTebqegn]

How to prove by epsilon -Delta limit of X power of n =Y power of n. X approchs to Y

[@kingbeauregard]

I would cheat and rely on squeeze proof logic, and swap out "f(x) = 4" with "f(x) = x + 1". I am a big stinky cheater.

[@PrimeNewtons]

😀😀😀😀