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Babynini

  • one year ago

Proving limits with delta..

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    Gaah sorry it is sideways. I got stuck at the end there, not sure where to go. The answer, I think, should be some min{ ?}

  3. Babynini
    • one year ago
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    @jim_thompson5910 you free? :)

  4. zepdrix
    • one year ago
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    Let delta be sufficiently small, say \(\large\rm \delta<1\). Then, \(\large\rm |x-2|<1\)\[\large\rm -1< x-2<1\]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.\[\large\rm |x-2||x+4|<|x-2|\cdot 7<\epsilon\]

  5. zepdrix
    • one year ago
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    \[\large\rm |x-2|<\frac{\epsilon}{7}\]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.\[\large\rm \delta=\min\left\{1,~~\frac{\epsilon}{7}\right\}\]

  6. zepdrix
    • one year ago
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    Yes, we were trying to find a way to "deal with" the |x+4| :)

  7. zepdrix
    • one year ago
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    I 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

  8. Babynini
    • one year ago
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    hrmm what you did looks pretty close to what we did in class with other examples.

  9. jim_thompson5910
    • one year ago
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    @zepdrix I think as long as a > 0 and b > 0, then \[\Large a < x+4 < b \implies |a| < |x+4| < |b|\]

  10. Babynini
    • one year ago
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    So then it comes out like: ...?

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  11. zepdrix
    • one year ago
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    For part \(\large\rm i)\) at the last line I think you want to write:\[\large\rm |x-2||x+4|<\epsilon\]I don't think we can relate this to delta just yet at that point. hmm.

  12. zepdrix
    • one year ago
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    And then in part \(\large\rm ii)\) I would maybe make a connection here:\[\large\rm 5<x+4<7\qquad\implies\qquad |x+4|<7\]

  13. Babynini
    • one year ago
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    ahh right right.

  14. zepdrix
    • one year ago
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    And 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 "="

  15. zepdrix
    • one year ago
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    Oh 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."

  16. zepdrix
    • one year ago
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    Assume \(\large\rm \delta<1\). Maybe that sounds a little nicer >.< I dunno

  17. Babynini
    • one year ago
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    Dude I was just looking at that too. So redundant haha

  18. zepdrix
    • one year ago
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    These 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.

  19. Babynini
    • one year ago
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    Oh goshh, really? So you're pro then!

  20. Babynini
    • one year ago
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    hahaha best description of these ever. But I like English, so it's ok xP

  21. Babynini
    • one year ago
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    Is it all looking good now? :)

  22. zepdrix
    • one year ago
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    Nooo I'm not a pro :O noob. learning. Looking good? I think soooo.

  23. zepdrix
    • one year ago
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    We 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<|x-2|<delta implies |f(x)-L|<epsilon or some nonsense like that, ya? :D

  24. zepdrix
    • one year ago
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    Maybe I'm being a little fussy :) But these words mean things

  25. Babynini
    • one year ago
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    Ah 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!

  26. zepdrix
    • one year ago
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    Are 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, mean-value theorem, lots of cool normal math stuff.

  27. Babynini
    • one year ago
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    mhm! I am taking Calc 1 this quarter.

  28. Babynini
    • one year ago
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    Okay! one last time!! xD

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  29. Babynini
    • one year ago
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    oh goood, I hope it gets more fun haha

  30. zepdrix
    • one year ago
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    I 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?

  31. Babynini
    • one year ago
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    oh, 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

  32. Babynini
    • one year ago
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    I guess I could just write delta > 0 there though. Since I go back at the end and do that fancy "Therefore ..."

  33. zepdrix
    • one year ago
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    Hmm 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.

  34. Babynini
    • one year ago
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    Would it be redundant/dumb to do both? o.o

  35. zepdrix
    • one year ago
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    You 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?

  36. Babynini
    • one year ago
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    Ah ok. Yep, it just helps me from getting lost haha

  37. zepdrix
    • one year ago
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    woah woah woah we're forgetting something very crucial

  38. zepdrix
    • one year ago
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    wow im glad i caught this before it was too late

  39. zepdrix
    • one year ago
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    \[\Huge \blacksquare\]

  40. zepdrix
    • one year ago
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    buhahaha

  41. Babynini
    • one year ago
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    hahaha I will put it in right now!!

  42. Babynini
    • one year ago
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    thanks!

  43. zepdrix
    • one year ago
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    yay team

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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