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Help with a limit

  1. Mar 23, 2007 #1
    I have to prove that lim sin(1/x) when x->0 does not exist. i used Haines definition of the limit to prove this. I found two sequences x=1/n*pi, so when n-> infinity x->0, and the other sequence that also converges to zero that i used in this case is x=2/(4n+1)*pi, this also when n->infinity then x->0. now when we take take the corresponding sequences we get:

    lim sin n*pi, n-> infinity, and lim sin (4n+1)*pi/2, as n-> infinity,

    the problem is here, i am not sure at this case if i can take lim sin n*pi, n-> infinity =0 and lim sin (4n+1)*pi/2, as n-> infinity= 1, i think i should do like this. But at this case what i am courious to know is why should i take these results??

    any help would do.
     
  2. jcsd
  3. Mar 23, 2007 #2

    JasonRox

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    Assume it does converge to a limit as x->0.

    Now, let the epsilon be 0.1, then we should be able to find a gamma that satisfies this condition by the definition of the limit.

    Can you see a problem that occurs when e=0.1?
     
  4. Mar 23, 2007 #3

    HallsofIvy

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    What exactly is "Haines definition of the limit"?
     
  5. Mar 23, 2007 #4

    JasonRox

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    His professors name? :confused:
     
  6. Mar 23, 2007 #5

    matt grime

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    You have a sequence x_n = sin npi, so that x_n=0 for all n.
    You have a sequence y_n = sin (4n+1)pi/2, so that y_n=1 for all n.

    They obviosuly converge, since they are constant. That x_n=sin npi is neither here nor there.
     
  7. Mar 24, 2007 #6
    yeah matt grime i do understand that x_n = sin npi, so that x_n=0 for all n, and also y_n = sin (4n+1)pi/2, so that y_n=1 for all n. But my question is :

    If it is still safe to jump to these conclusions when n->infinity?

    thnx
     
  8. Mar 24, 2007 #7

    matt grime

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    Of course it is. Why wouldn't it be? x_n is always zero. Irrespective of what n is. Do you think that at some point sin(npi) is going to stop being zero?

    Actualy scrub that. Let me put it this way. What has n going to infinity got to do with the value of sin(npi)? You appear to be thinking of things in the wrong order.

    taking the limit of sin(npi) as n tends to infinity does not affect the values sin(npi) at all. How can it. They're fixed (and always 0).
     
    Last edited: Mar 24, 2007
  9. Mar 25, 2007 #8
    ej matt grime thank you, this is what i was not quite sure about. Now it is all clear thnx for your help guys
     
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