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Limit of f(x) as it approaches from the positive side

  1. Dec 10, 2008 #1
    i. Write out the proper definition for
    lim f(x)
    x[tex]\rightarrow[/tex]a+

    ii. Let g(x)=x-1cos(x-1) Show that for every N and any b>0 there is a point t E (0,6) so that g(t)>N

    iii. Prove that for g as in (ii) we don not have
    lim g(x)
    x[tex]\rightarrow[/tex]0+


    I'm not sure what how to do any of these... like.. for the first one, i know that the definition of a limit is for every [tex]\epsilon[/tex]>0 there is some [tex]\delta[/tex] >0 such that, for all x if 0<|x-a|< [tex]\delta[/tex], then |f(x)-L| < [tex]\epsilon[/tex]
     
  2. jcsd
  3. Dec 10, 2008 #2

    Dick

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    Continue with the first one. All you have to do is drop the absolute value in the delta part. Do you see why that works?
     
  4. Dec 10, 2008 #3
    I can see why you drop the absolute value sign, it then becomes x>a meaning the x's approaching from the positive side.

    so what about the others?


    Thanks
     
  5. Dec 10, 2008 #4

    Dick

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    It's time for you to help. If t is in (0,6) then as t->0, 1/t becomes very large, right? What does cos(1/t) do?
     
  6. Dec 10, 2008 #5
    wouldn't it be either 1 or 0 or -1?
     
  7. Dec 10, 2008 #6

    Dick

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    It oscillates between 1 and -1, yes. So as 1/t becomes very large...?
     
  8. Dec 10, 2008 #7
    does it go to zero?
     
  9. Dec 10, 2008 #8

    Dick

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    No, (1/t) becoming very big times something oscillating between -1 and 1, will, I think oscillate between -infintity and +inifinity. You don't agree, right?
     
  10. Dec 10, 2008 #9
    i'm not following you. please elaborate.
     
  11. Dec 14, 2008 #10
    can someone please help me out with this problem.. the person that replied to this thread never fully answered my question
     
  12. Dec 15, 2008 #11

    Mark44

    Staff: Mentor

    As Dick said earlier, it's time for you to do some work. He was trying to explain what the function did without spoonfeeding you the answer. I think he lost patience.

    Go back and look at what he said. If you have specific questions, ask them and someone will help you out.
     
  13. Dec 15, 2008 #12

    Dick

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    I didn't lose patience (yet). But sketch a graph of the function. To help answer the given question think about the values of the function g(x) at x=1/(k*pi) where k is an integer.
     
  14. Dec 15, 2008 #13
    o i see, as it approaches zero, the graph starts to oscillate from negative infinity to positive infinity. But i'm not sure how to answer part ii?
     
  15. Dec 15, 2008 #14

    Dick

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    Try and answer the question I asked, what is g(1/(k*pi))?
     
  16. Dec 15, 2008 #15
    well, as k increases for g(1/k*pi)....
    it is negative infinity for k=1
    it is positive infinity for k=2
    and etc...
     
  17. Dec 15, 2008 #16

    Dick

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    Why would that be? If x=1/(k*pi) what is cos(1/x)? k is an integer. It's not some kind of infinity.
     
  18. Dec 15, 2008 #17
    well, i was referring to the actual function because the actual function is x^(-1)cos(1/x) soooo, as k increases... g(1/k*pi) is negative infinity for k=1 and positive infinity for k=2... no?
     
  19. Dec 15, 2008 #18

    Dick

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    Negative infinity? If k=1? How does that follow? I get pi*cos(pi). That's not infinity. Express the answer in terms of k.
     
  20. Dec 15, 2008 #19
    o i see, when k=1, pi*cos(pi)

    i'm sorry, i haven't gotten sleep for the passed few days...

    soooo, in terms of k... it would be....
    kpi*cos(kpi) right?
     
  21. Dec 15, 2008 #20

    Dick

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    Yes, cos(pi)=(-1), cos(2pi)=(+1) etc. You can also write that as kpi*(-1)^k. Does that help with question ii)? If that seems hard you are probably better off getting some sleep than trying to push on.
     
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