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Babynini
 one year ago
Proving limits with delta..
Babynini
 one year ago
Proving limits with delta..

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Gaah sorry it is sideways. I got stuck at the end there, not sure where to go. The answer, I think, should be some min{ ?}

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 you free? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Let delta be sufficiently small, say \(\large\rm \delta<1\). Then, \(\large\rm x2<1\)\[\large\rm 1< x2<1\]Adding 6,\[\large\rm 5<\color{royalblue}{x+4<7}\]And I think we can do something withhhhh this. Hmm. I'm just trying to figure out how we relate this to absolute value.\[\large\rm x2x+4<x2\cdot 7<\epsilon\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\large\rm x2<\frac{\epsilon}{7}\]And then you have to choose your delta carefully, depending on which is bigger, this epsilon/7 or the 1 restriction that we put on delta.\[\large\rm \delta=\min\left\{1,~~\frac{\epsilon}{7}\right\}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Yes, we were trying to find a way to "deal with" the x+4 :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I was able to get this relationship: \(\large\rm 5<\color{royalblue}{x+4<7}\) But not something like this: \(\large\rm a< x+4<b\) I hope that isn't going to matter.. I'm trying to think about it :\ hmm

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hrmm what you did looks pretty close to what we did in class with other examples.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2@zepdrix I think as long as a > 0 and b > 0, then \[\Large a < x+4 < b \implies a < x+4 < b\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0So then it comes out like: ...?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3For part \(\large\rm i)\) at the last line I think you want to write:\[\large\rm x2x+4<\epsilon\]I don't think we can relate this to delta just yet at that point. hmm.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3And then in part \(\large\rm ii)\) I would maybe make a connection here:\[\large\rm 5<x+4<7\qquad\implies\qquad x+4<7\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3And after that `"So" line` I would get rid of the `equals sign` on the next line. It reads like, the epsilon from the previous line is equivalent to the next stuff. Maybe just the word "Then," instead of "="

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Oh also, maybe too much words at the start of \(\large\rm ii)\) hehe Let \(\large\rm \delta\) be \(\large\rm \delta<1\). When I'm reading this in my head it reads like: "Let delta be delta is less than 1."

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Assume \(\large\rm \delta<1\). Maybe that sounds a little nicer >.< I dunno

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Dude I was just looking at that too. So redundant haha

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3These are tricky huh? :d it's like combing Math and English class! lol I'm taking a whole class on these types of problems :3 it's a pain sometimes.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Oh goshh, really? So you're pro then!

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hahaha best description of these ever. But I like English, so it's ok xP

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Is it all looking good now? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Nooo I'm not a pro :O noob. learning. Looking good? I think soooo.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3We don't want a therefore on the last line. We gotta pick a delta first! :D Use another word like ummm... `Let` delta=min{stuff}. Therefore, 0<x2<delta implies f(x)L<epsilon or some nonsense like that, ya? :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Maybe I'm being a little fussy :) But these words mean things

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ah well I think you're doing excellently then! xP oh gosh. What is all this fanciness haha. Nah I think it's good. Fussy is better than having it incomplete!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Are you near the beginning of a Calc 1 course? :o That's when I remember first going over these types of problems. Everything was super fun after this stuff. Related rates, max/min problems, meanvalue theorem, lots of cool normal math stuff.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0mhm! I am taking Calc 1 this quarter.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Okay! one last time!! xD

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0oh goood, I hope it gets more fun haha

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I don't understand the first line of your proof: \(\large\rm \delta=\qquad\qquad >0\) Do you go back later and fill in that gap after you have found a delta or something?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0oh, yes. That is how the prof does it. So now i'd go back and put in the delta = min{1, e/7} in there. I just forgot to do that yet xD

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0I guess I could just write delta > 0 there though. Since I go back at the end and do that fancy "Therefore ..."

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Hmm that's a clever way to do that :) I like that. Yah either way should be ok I think, since you're stating your delta that you found at the end.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Would it be redundant/dumb to do both? o.o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3You mean, would it be dumb to put \(\large\rm \delta=\left\{1,\frac{\epsilon}{7}\right\}>0\) at the top since you already have it at the bottom? Nahhh seems fine. Why are they numbered i) and ii)? Is that just a way to stay organized?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ah ok. Yep, it just helps me from getting lost haha

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3woah woah woah we're forgetting something very crucial

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3wow im glad i caught this before it was too late

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3\[\Huge \blacksquare\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0hahaha I will put it in right now!!
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