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Xcos(1/x) limit at 0

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Define f: (0,1)---->R by . Does f have a limit at 0?



    2. Relevant equations
    0<absvalue(x-x_0)<delta
    absvalue(f(x)-L)<epsilon


    3. The attempt at a solution
    epsilon is arbitrary.
    0<absvalue(x-0)<delta. Then absvalue(xcos1/x-0)<epsilon.
    0<absvalue(x)<delta
    absvalue(xcos1/x)<epsilon
    absvalue(x)absvalue(cos1/x)<epsilon

    This is where I get stuck? Do I divide by absvalue(cos1/x) and let delta= sec(1/x)?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2008 #2

    Mark44

    Staff: Mentor

    I think you are confused between the ideas of finding the limit of a function and proving (with epsilons and deltas) that a certain number is the limit of that function.

    The question is "Does f have a limit at 0?" The answer should be either "yes" or "no." From your work it would appear that you believe the answer is yes. You can justify this belief with this inequality:
    [tex]0 \leq x cos(\frac{1}{x}) \leq x[/tex]

    As x approaches 0, what happens to each end of the inequality above?
     
  4. Nov 13, 2008 #3
    No, I am supposed to prove it. I'm just not sure about my proof.
     
  5. Nov 13, 2008 #4

    HallsofIvy

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    A proof does not necessarily mean an "epsilon- delta" proof. If you can use the "sandwiching" theorem, that would work nicely.
     
  6. Nov 13, 2008 #5

    D H

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    Science Advisor

    Replace that nasty cos(1/x) term with something much simpler. If you can find some f(x) such that |cos(1/x)| < |f(x)| you can use this function in lieu of cos(1/x).

    Note: You should prove this or find the theorem in your text that proves this for you.
     
  7. Nov 13, 2008 #6

    Mark44

    Staff: Mentor

    You didn't include that in your problem statement.

    Little tip: instead of typing absvalue over and over, use the | key. It'll save you a lot of typing.

    So, given an epsilon, you need to find a delta so that x < delta ==> |x cos(1/x)| < epsilon. Something that might be helpful is that |x cos(1/x)| <= |x| for all x != 0. Maybe you can use this idea.
     
  8. Nov 13, 2008 #7
    So, I have lxcos(1/x)-0l < epsilon
    lxcos(1/x)l < epsilon
    Then lxl lcos(1/x)l < epsilon
    But lxcos(1/x)l < lxl

    I'm not sure where to go with this idea....
     
  9. Nov 13, 2008 #8
    So, could I then say since lxl<epsilom, that delta=1/lxl?
     
  10. Nov 13, 2008 #9

    HallsofIvy

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    Well, that says that if |x|< epsilon, then |x cos(1/x)|, which is even smaller, will be less than epsilon. How do you choose delta to be sure |x|< epsilon?
     
  11. Nov 13, 2008 #10
    Let delta =epsilon?
     
  12. Nov 13, 2008 #11
    Let delta=epsilon because if epsilon =delta then lxcos(1/x)l < delta?
     
  13. Nov 13, 2008 #12

    Mark44

    Staff: Mentor

    Yes, that's where Halls and I were heading.
     
  14. Nov 13, 2008 #13
    Ok, so it's plain delta = epsilon? Just that simple? I wasn't sure since I did another problem where I had delta=epsilon/2
     
  15. Nov 14, 2008 #14

    HallsofIvy

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    You can't expect all problems to have the same answer!
     
  16. Nov 14, 2008 #15
    Ok, thanks. i just thought that seemed a little too easy, I guess, haha.
     
  17. Nov 14, 2008 #16
    If you have to prove a limit which involves trig functions, an epsilon-delta argument usually works nicely. The absolute values give a nice bound on sin and cos functions. The squeeze theorem is useful as well, but you should try deriving it on your own using the epsilon-delta.
     
  18. Nov 19, 2008 #17
    Ok, I'm unsure on the step where I say [tex]\left|x\right|<\epsilon[/tex]. How do I get this?
     
  19. Nov 20, 2008 #18
    Anybody have any hints on where this comes from? like I know lxl < delta.
     
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