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Limits and infinite oscillations

  1. May 16, 2009 #1
    Hello everyone,

    I am having trouble understanding the concept of a limit not existing for functions like sin (1/x) when x tends to 0. The good book says that the function "does not settle on any value as we get closer to x" implying some infinite oscillation. I am having trouble visualizing it and why it should happen with this particular function.

    Any kind soul here willing to elaborate on this and help me understand this better? I would be extremely grateful.

    Many thanks,

    Luca
     
  2. jcsd
  3. May 16, 2009 #2

    dx

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    Think about sin(x) as x goes to infinity. Do you know what the graph of sin(x) looks like? Does it settle down to any particular value as x goes to infinity?
     
  4. May 16, 2009 #3
    Hi there,

    Thanks for the reply. The sine graph is oscillating between -1 and 1 every with a period of 2 pi. How would this change into infinite oscillations for sin(x) as x approaches infinity...

    Thanks,

    Luca
     
  5. May 16, 2009 #4
    Ok, I think I understand this better now as we approach infinity, the value will keep oscillating between -1 and 1 and there is no one value where the function will settle down to...fair enough :)

    Thanks,

    Luca
     
  6. May 16, 2009 #5

    dx

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    You said it yourself. It oscillates between 0 and 1. Just draw the graph of sin. As you go further and further in the positive direction of the x axis, does it settle down to some value? No, it keeps oscillating between 1 and 0.

    EDIT: Sorry, I don't know what I was thinking. I meant to say oscillating between 1 and -1.
     
    Last edited: May 16, 2009
  7. May 16, 2009 #6

    HallsofIvy

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    Another way to look at it:

    [itex]sin(n\pi)= 0[/itex] for all n. Let [itex]x_n= 1/(n\pi)[/itex]. Then as n goes to infinity, [itex]x_n[/itex] goes to 0 and [itex]sin(1/x_n)= 0[/itex] for all n.

    But [itex]sin(2n\pi+ \pi/2)= 1[/itex] for all n. Let [itex]x_n= 1/(2n\pi+ \pi/2)[/itex]. Then as n goes to infinity, [itex]x_n[/itex] goes to [itex]sin(1/x_n)= 1[/itex] for all n.

    If [itex]\lim_{x\rightarrow 0} sin(1/x)[/itex] existed, the limit of [itex]sin(x_n)[/itex] would have to be that value for any sequence [itex]x_n[/itex] converging to 0. Since the two sequences above have different limits, the limit of the function cannot exist.
     
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