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Jamierox4ev3r

  • one year ago

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  1. Jamierox4ev3r
    • one year ago
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    Given that \[\lim_{x \rightarrow 2}(5x-5) =5\], find values of \(\delta\) that correspond to \(\epsilon\)=0.1, \(\epsilon\)=0.05, and \(\epsilon\)=0.01

  2. ganeshie8
    • one year ago
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    finally you have started reviewing epsilon delta!

  3. Jamierox4ev3r
    • one year ago
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    Looks like it :P

  4. Jamierox4ev3r
    • one year ago
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    I would think you would start off by making this equation: -0.1<5x-5<0.1

  5. Jamierox4ev3r
    • one year ago
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    then solving for x -0.1<5x-5<0.1 +5 +5 +5 ------------- 4.9<5x<5.1 /5 /5 /5 --------- 0.98<x<1.02

  6. Jamierox4ev3r
    • one year ago
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    I'm just kind of confused as to what you would do here... Since a=2, wouldn't you subtract 0.98 and 1.02 from 2?

  7. ganeshie8
    • one year ago
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    0.98<x<1.02 thats same as : |x| < 0.02 yes ?

  8. Jamierox4ev3r
    • one year ago
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    right

  9. ganeshie8
    • one year ago
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    so \(\delta = 0.02\) works

  10. Jamierox4ev3r
    • one year ago
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    so you don't need to subtract from whatever the value of a is? (which in this problem, it's 2)

  11. ganeshie8
    • one year ago
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    what that means is if you're with in \(0.02\) around \(x=2\), the value of function stays within \(0.1\) around \(5\)

  12. ganeshie8
    • one year ago
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    a graph might help you want to spend some time and see whats going on

  13. ganeshie8
    • one year ago
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    or you may continue with rest of the problems and hope things become clear after doing few problems

  14. Jamierox4ev3r
    • one year ago
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    It's funny, I've already done a lot of problems. And things seemed pretty straightforward until i got to this one.

  15. Jamierox4ev3r
    • one year ago
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    For example, I had a problem that read, "Find a number \(\delta\) such that if \[\left| x-2 \right|=\delta\], then \(\left| 5x-10\right|=\epsilon\), where \(\epsilon =0.1\)

  16. ganeshie8
    • one year ago
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    ikr! often times the concepts look so easy when you read them but you only get to really learn them only after actually solving the problems

  17. Jamierox4ev3r
    • one year ago
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    ^ agreed. But this problem that preceded...I was able to get it with no problem. my process looked like this: -0.1<5x-10<0.1 +10 +10 +10 ------------------ 9.9<5x<10.1 /5 /5 /5 ---------- 1.98<x<2.002

  18. Jamierox4ev3r
    • one year ago
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    from there, I subtracted 2 from 2.02, and 1.98 from two, which both gave me 0.02. So I concluded that \(\delta =0.02\), which was a correct answer according to the computer software I'm using.

  19. Jamierox4ev3r
    • one year ago
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    So I assumed from then that I knew what I was doing XD I suppose I was just confused about the fact that the functions are different, but they wield the same answers

  20. ganeshie8
    • one year ago
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    let me ask you a question there

  21. Jamierox4ev3r
    • one year ago
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    Sure

  22. ganeshie8
    • one year ago
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    you have ``` 1.98<x<2.002 ``` then you say : ``` from there, I subtracted 2 from 2.02, and 1.98 from two, which both gave me 0.02.`

  23. Jamierox4ev3r
    • one year ago
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    yes

  24. ganeshie8
    • one year ago
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    that is wrong how can both give you 0.02 ? double check

  25. Jamierox4ev3r
    • one year ago
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    wait...

  26. Jamierox4ev3r
    • one year ago
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    2.02-2=0.02 2-1.98=0.02 ^^ they both do give 0.02 o-o

  27. ganeshie8
    • one year ago
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    Oops sry, you're right!

  28. ganeshie8
    • one year ago
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    so it seems the same \(\delta\) works in both cases

  29. Jamierox4ev3r
    • one year ago
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    yeah :P weird how the solutions are exactly the same, regardless of the functions being different

  30. Jamierox4ev3r
    • one year ago
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    Why is that? do you know?

  31. ganeshie8
    • one year ago
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    they both are kinda same functions, they only differ by a constant

  32. ganeshie8
    • one year ago
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    f(x) = 5x-5 g(x) = 5x-10

  33. Jamierox4ev3r
    • one year ago
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    that's true...I suppose that makes sense, thy only vary by where in the graph they start

  34. Jamierox4ev3r
    • one year ago
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    Another thing i don't get is that with the 5x-10, you use 2 in order to find \(\delta\) but with 5x-5, you use 1 in order to find \(\delta\) I don't know why I get hung up on these tiny details XD

  35. ganeshie8
    • one year ago
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    teamviewer ?

  36. Jamierox4ev3r
    • one year ago
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    o-o pardon?

  37. ganeshie8
    • one year ago
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    you don't have teamviewer/skype?

  38. Jamierox4ev3r
    • one year ago
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    nope

  39. Jamierox4ev3r
    • one year ago
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    I could screenshot you what I'm looking at o-o

  40. ganeshie8
    • one year ago
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    not necessary, i have a khan academy video that explains exactly this one sec..

  41. ganeshie8
    • one year ago
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    here is it is https://www.khanacademy.org/math/differential-calculus/limits_topic/epsilon_delta/v/epsilon-delta-limit-definition-1

  42. Jamierox4ev3r
    • one year ago
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    omg @ganeshie8 that's actually a super clear explanation, I got it! Thank you u.u

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