A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

By the method of delta-epsilon prove limx->5 ((1/2)x+6)= 8 and a1/2

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ugh.... epsilons and deltas.....

  2. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x)-L < e |x-c|<d

  3. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    at least its a line and not a curve

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know all of that sorry let me tell you where am having a problem

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know all of that sorry let me tell you where am having a problem

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i simplified it to get 1/2 (x-2) < e so how do i find my delta from there?

  7. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |x-5| < d right? id have to work it out meself to recall how to do it lol

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its |x-5|<d

  11. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a1/2 ?? whats that mean? -e < (1/2)x+6) - 8 < e

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    8 and a half

  13. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -e < (1/2)x-2 < e 2 -e < (1/2)x < 2+e 2(2 -e) < x < 2(2+e)

  14. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    0 < x-5 <d 5 < x < 5+d

  15. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    5+d < 2(2+e) i got know idea if this is working lol

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    arent we suppose to be converting x-2 to x-5?..no?

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    usually you work these problems backwards.

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this one is tricky...

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    start with | \[|.5x+6-8.5|<\epsilon\]

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[-\epsilon < .5x-2.5< \epsilon\]

  22. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohh....8' 1/2 ....

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[2.5-\epsilon<.5x<2.5+\epsilon\]

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[5-2\epsilon<x<5+2\epsilon\]

  25. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    looks familiar so far lol

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can u do it in terms of epsilon and delta?

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can u do it in terms of epsilon and delta?

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how am i doing so far?

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    good

  30. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, other than having the right numbers; looks the same as mine lol

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh my delta will depend on epsilon. i am just trying to figure out what the delta needs tgo be

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kk

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the game is played like this: i say the limit is L. you say ok, make it within epsilon of L. i say ok use this delta which obviously depends on epsilon. that is delta will be written in terms of epsilon

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok but in class we normally would deleta write where u wrote your epsilon but its problem a different example

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how'd you get the .5 to 5?

  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i am working the problem backwards. then the proof is to work it forwards. you say for example pick delta such that \[5-2\epsilon < \delta< 5+2\epsilon\]

  38. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the 5?

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we had \[2.5- \epsilon<.5x<2.5+\epsilon\]

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to get x by itself i doubled everything

  41. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so now we have \[5-2\epsilon <x < 5+2\epsilon\]

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this says the same thing as \[|x-5|<2\epsilon\]

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so the delta i pick is \[2\epsilon\]

  44. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok sorry i had to move from the pc a sec

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because that is what we wanted. we wanted to say that we could find a delta such that if \[|x-5|<\delta\] then \[|.5x-6+8.5|<\epsilon\]

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    don't forget what the goal is. you want to show that \[lim_{x->5} (.5x+6)=8\]

  47. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    8 1/2

  48. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    meaning that given any epsilon > 0 y ou can find a delta such that if \[|x-5|<\delta\] then \[|.5x+6-8.5|<\epsilon\]

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok truthfully i dont understand what u did

  50. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i say pick \[\delta = 2 \epsilon\] i win by running the algebra backwards

  51. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the way you have to do these problem is work backwards. you have to find the delta, which depends on epsilon. i.e. \[\delta = delta(\epsilon)\]

  52. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    delta will be an expression in epsilon. so you start with the thing that you want, namely \[|.5x+6-8.5|<\epsilon\] do some algebra, and see what delta you get.

  53. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got \[\delta = 2\epsilon\]

  54. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  55. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you!

  56. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    run the algebra forward and see that it works. just take the steps i wrote and write them backwards. in other words start with \[|x-5|<2\delta\] and see that you get \[|2.x+6-8.5|<\epsilon\]

  57. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    typo

  58. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i meant start with \[|x-5|<2\epsilon\]

  59. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i will write it if you like

  60. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont undestand what your asking me to do

  61. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me write it and maybe it will make sense. hold on and i will write it all out.

  62. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okat

  63. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You want to show that given any \[\epsilon>0\] you can find a \[\delta\] such that if \[|x-5|<\delta\] then \[|.5x+6-8.5|<\epsilon\] let \[\delta=2\epsilon\] then \[|x-5|<2\epsilon\] means \[-2\epsilon<x-5<\epsilon\] giving \[5-2\epsilon < x < 5+2\epsilon\] gives \[2.5-\epsilon < .5x < 2.5 +\epsilon\] so \[-\epsilon < .5x-2.5<\epsilon\] or \[|.5x-2.5|<\epsilon\]

  64. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry it took so long to write. but i never would have found delta = 2 epsilon without working the problem backwards

  65. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hope this helps

  66. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow THANK YOU SO MUCH

  67. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    welcome.

  68. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok like thank you so much i get it but i never would have figured that out on my own thanks a billion seriously

  69. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.