There is a small mistake in the video at 9:04. I gave as the reason for using the "less than or equals" sign (as opposed to the stricter "less than" sign) that |x - y| might be zero. What I should have said is that |x + y| might be zero. The proof is still correct; just the stated justification was incorrect.
seriously I was so frustrated because I didn't understand the procedure to show continuity, after your video a lot got so much clearer thank you so much!
Thank you for the feedback. Don't lose hope. Keep this thought in mind: there is always a reason why things are defined the way they are. Unfortunately, you will not always be told. Just hold out hope that it actually DOES make sense and try to find out how. My own such search is what leads me to make these videos.
Excelent video. Thank you so much. I needed to remember how to do these proofs and you explained them flawlessly. Never really understood why we could select a "first delta", but now I know.
Hello, i understand the whole video, but something bothers me. How are we allowed to pick a value for delta ? I mean, i thought this is what we were trying to find ? Thanks.
The idea behind the proof is that for every epsilon you could think of, there exists some delta so that the criterium holds. So we say "let epsilon be what it may". Then we use that epsilon to create a delta that makes the criterium hold, thereby proving that such a delta must exist (since we just gave one example of one such delta). Hope that helps.
Excellent video, thanks so much! At 10:20 you have that delta( |y - x| + |2x| ) = delta(delta + |2x|). Could you tell me how you got that to be so? thanks again!
You're welcome! Glad you found the video helpful. At 10:20 we have delta( |y - x| + |2x| ) < delta(delta + |2x|) because our assumption (written right above the chain of inequalities) is that |y-x| < delta. So we are basically "subbing in" delta for |y-x|. Hope that helps!
I don't really get why we chose delta to be the minimum of (1, Epsilon/(1+2|x|)). I know that we need our delta to be less or equal to 1, but if 1 < (Epsilon / (1+2|x|)) Delta will then be equal to one, and our proof won't be valid anymore..?
hi, on your first proof why epsilon becomes equal to |f(y)-f(x)| instead of epsilon being less than |f(y)-f(x)|? it's the last part written in green on 5:34
hi, thanks for your question. i think the reason for the confusion is this: in the proof you refer to, there are several = signs followed by a < sign, followed again by a = sign. taken together, this means that < holds. hope that helps!
indeed, thanks for your video!, i wasnt sure you would answer so i asked on MO but that was indeed the confusion math.stackexchange.com/questions/2185681/in-this-epsilon-delta-continuity-proof-why-do-inequalities-change-to-equalities
Thanks for the comment. If you look closely, in the first example, delta only depends on epsilon, and not on x or y at all: delta := epsilon/3. But proofs like this can be confusing at first.
Great video! However, your voice is too gentle lol makes me want to sleep while studying midnight watching your video haha Anyway, keep up the good work !!!
