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anonymous

  • one year ago

I only have two questions left in my homework that's due tonight before midnight and I need help with them. 1) Prove the statement using the ε, δ definition of a limit. lim x^2 = 0 x→0 Given ε > 0, we need δ > 0 such that if 0 < |x − 0| <  δ, then |x^2 − 0| <  ε ⇔ (blank) <  ε ⇔ |x|< (blank). Take δ = (blank). Then 0 < |x − 0| <  δ right double arrow implies |x^2 − 0| < ε. Thus, lim x^2 = 0 by the definition of a limit. x→0 2) Use the given graph of f to find a number δ such that if |x − 1| < δ then |f(x) − 1| < 0.2 δ = (Blank) given graph: http://www.webassign.net/scalcet7/2-4-001.gif

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  1. beginnersmind
    • one year ago
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    For the second one they are asking how far can you go from x = 1 in the x direction so that the values of y are within 0.8 and 1.2 .

  2. beginnersmind
    • one year ago
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    E.g, if you go 0.2 to the right, to x = 1.2 is the value of the function still within 0.2 of 1? That is is it in the range [0.8, 1.2]? What if you only move 0.01? What's the most you can move horizontally to stay in the strip ]0.8, 1.2[ vertically?

  3. zepdrix
    • one year ago
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    So are the (blank)s all the things that you have to fill in?

  4. anonymous
    • one year ago
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    Yes.

  5. anonymous
    • one year ago
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    @zepdrix

  6. anonymous
    • one year ago
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    The notes that my professor gave my class today for these questions don't even come close to them at all.

  7. zepdrix
    • one year ago
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    haha that's weird :D

  8. zepdrix
    • one year ago
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    Oh there is a graph! I didn't even see that part.

  9. anonymous
    • one year ago
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    I asked him to do the first one I have listed up on the board and he completely does a different question.

  10. anonymous
    • one year ago
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    did*

  11. zepdrix
    • one year ago
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    What a silly billy -_-

  12. zepdrix
    • one year ago
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    Oh I guess I'm a little confused then. This looks like two different problems mashed together...

  13. anonymous
    • one year ago
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    Sorry about that...

  14. zepdrix
    • one year ago
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    Or that's just a second problem? :D Ok ok ok my bad

  15. zepdrix
    • one year ago
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    Thinkinggg +_+

  16. anonymous
    • one year ago
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    There, before anyone else joins and gets confused as well.

  17. zepdrix
    • one year ago
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    I think maybe this is what they're looking for... If x is within delta of zero, \(\large\rm 0\lt |x-0|\lt\delta\) Then x^2 is within epsilon of the limit value, zero, \(\large\rm |x^2-0|\lt\epsilon\) Therefore \(\large\rm |x^2|\lt\epsilon\qquad\implies\qquad |x|\cdot|x|\lt\epsilon\) Since \(\large\rm |x|\lt\delta\) this implies that \(\large\rm |x|\cdot|x|\lt\delta\cdot\delta\) Choose \(\large\rm \delta\) such that \(\large\rm \delta^2\le\epsilon\) So that \(\large\rm 0\lt|x-0|\lt\delta\le\sqrt{\epsilon}\) Which shows that \(\large\rm |x|\lt\sqrt{\epsilon}\)

  18. zepdrix
    • one year ago
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    I've never been very good at these epsilon-delta thingies :\ Lemme try to suggest what they want for the blanks here....

  19. anonymous
    • one year ago
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    I got the first blank! x^2

  20. zepdrix
    • one year ago
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    So \(\large\rm |x^2-0|\lt\epsilon\) if and only if \(\large\rm x^2\lt\epsilon\) Ah ok they just simplified :o Hmm

  21. zepdrix
    • one year ago
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    Do you get multiple guesses or no? :3

  22. anonymous
    • one year ago
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    I have 100 tries.

  23. zepdrix
    • one year ago
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    lol

  24. zepdrix
    • one year ago
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    I'm thinking they want: \(\large\rm |x|\lt\underline{\quad\sqrt{\epsilon}\quad}\) For the next blank... maybe +_+

  25. anonymous
    • one year ago
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    Webassign took it.

  26. zepdrix
    • one year ago
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    Accepted it? :O Noiceeee

  27. zepdrix
    • one year ago
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    Maybe what your teacher wanted to do, to get from that step to the next one,\[\large\rm x^2\lt \epsilon\]Is take the square root of both sides of this inequality.\[\large\rm \sqrt{x^2}\lt \sqrt{\epsilon}\] Recall that \(\large\rm |x|:=\sqrt{x^2}\) That's what will happen on the left side. And we shouldn't have any problems with the inequality or square root since epsilon is positive.

  28. zepdrix
    • one year ago
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    Take delta = _______ Hmmmmmmmmmm

  29. zepdrix
    • one year ago
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    Normally we would make a restriction on delta, force it to be small, like delta <= 1, and then use that to form our epsilon and find a connection between the delta and epsilon. Then we would let delta be equal to the smaller of those two restrictions. Like \(\large\rm \delta=min\{\sqrt{\epsilon},1\}\) that or something... Maybe try it? D: I dunno.. grrr

  30. anonymous
    • one year ago
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    It can't understand it.

  31. zepdrix
    • one year ago
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    Then maybe just \(\large\rm \delta=\sqrt{\epsilon}\) :d

  32. anonymous
    • one year ago
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    Yes!

  33. zepdrix
    • one year ago
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    Is that all of them for number 1? :OO I'm sorry if that doesn't help you further understand the subject ;c I know how crazy these can be. At least we got them filled in though

  34. anonymous
    • one year ago
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    Yes, indeed! I just want to get this done before 11:59 because that's when my homework is due. Now I just have that one regarding the graph.

  35. zepdrix
    • one year ago
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    Did you read what Beginnersmind wrote about that question? :)

  36. anonymous
    • one year ago
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    Yes but it confused me further. I'll feel like an idiot if those were the answer.

  37. zepdrix
    • one year ago
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    |dw:1442015519645:dw|

  38. zepdrix
    • one year ago
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    I dunno if you can read my sloppy handwriting there lol When epsilon puts us 0.2 away from y, how far away from x does that put our delta?

  39. anonymous
    • one year ago
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    It's better than my chicken scratch. Um... 0.2...?

  40. anonymous
    • one year ago
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    Nope wait! That was a brain fart!

  41. zepdrix
    • one year ago
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    So that epsilon puts our line up at 1.2. That distance between 1.2 and 1 is 0.2, our epsilon. That line at 1.2 corresponds to an x value of 0.7. So how far is your 0.7 from 1?

  42. anonymous
    • one year ago
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    Yeah... I have nothing... This is going to be a long semester with this professor.

  43. zepdrix
    • one year ago
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    I'm asking you how far it is from 0.7 to 1. To calculate distance between numbers, you can either count up, or do subtraction. The distance \(\large\rm \delta\) is going to be 1-0.7.

  44. zepdrix
    • one year ago
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    That's not our answer though, so don't put that in the box.

  45. anonymous
    • one year ago
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    I did long before I came here for help.

  46. zepdrix
    • one year ago
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    ??

  47. zepdrix
    • one year ago
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    I'm just asking a simple question... what is the distance between 0.7 and 1? Hello? :o

  48. anonymous
    • one year ago
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    0.3

  49. anonymous
    • one year ago
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    I don't know.

  50. zepdrix
    • one year ago
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    |dw:1442016501759:dw|Ok good, let's call that delta_1, the distance from x on the left side :)

  51. zepdrix
    • one year ago
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    |dw:1442016540368:dw|Let's check the other side. If we TAKE AWAY delta from our y, that shows us the corresponding x value of 1.1. So what is our delta_2? :)

  52. anonymous
    • one year ago
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    0.1?

  53. zepdrix
    • one year ago
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    Good! So we've determined that when we are 0.2 away from our y value, that corresponds to being 0.3 and 0.1 away from our x value, \(\large\rm \delta_1=0.3\) and \(\large\rm \delta_2=0.1\) And \(\large\rm \delta=min\{\delta_1,\delta_2\}\) So for your delta, choose whichever one is smaller! :)

  54. anonymous
    • one year ago
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    ...This shouldn't have taken me this long...

  55. zepdrix
    • one year ago
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    lol :)

  56. anonymous
    • one year ago
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    Thank you for the help!

  57. zepdrix
    • one year ago
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    yay team \c:/

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