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Evaluate this simple limit

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data
    lim(x->infinity) sin(x)






    3. The attempt at a solution
    undefined.
     
  2. jcsd
  3. Jul 26, 2007 #2

    CompuChip

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    Why do you think that?
    Please give an argument.
    And please say what knowledge you have of limits (basis calculus, real analysis, etc)
     
  4. Jul 26, 2007 #3
    Can you give a proof why this limit doesn't exist?

    The sequential criterion for limits is handy, if you know this criterion.

    The sequential criterion for limits is:

    [itex]\lim_{x \to a}f(x)=b[/itex] iff for every sequence x_n in the extended real numbers (abstractly, the domain of f) converging to a has the property that f(x_n) converges to b.

    Can you think of a sequence converging to infinity such that sin(x) converges? Can you think of another sequence converging to infinity such that sin(x) converges to a different value?
     
    Last edited: Jul 26, 2007
  5. Jul 26, 2007 #4
    I don't think you should just say that the limit is something or other. If you want to take limits in simple non-series functions, then you should look at the outer rims of the function.

    For example sin(x) is defined as -1 < x < 1. Now you must work out if the limit is negative or positive infinity. And you should look at the definition that ZioX posted. It comes in handy.
     
  6. Jul 26, 2007 #5

    CompuChip

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    Or doesn't exist :smile:
    Which is what he said (just didn't prove yet).
     
  7. Jul 26, 2007 #6

    HallsofIvy

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    In other words, everyone is saying "yes, the limit doesn't exist, but you must tell why it doesn't exist"!
     
  8. Aug 11, 2007 #7
    Actually I reviewed this part of calculus a couple of days ago. So it is my bad om that one ;)

    And the simplest proof of seeing that the limit doesn't exist is graphical.
     
  9. Aug 11, 2007 #8
    Ziox showed it doesn't exist.

    x=2n*(pi)
    x=(2n+1/2)*(pi)
    when n->infinite, then got two different values.
     
    Last edited: Aug 11, 2007
  10. Aug 11, 2007 #9

    HallsofIvy

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    This is non-sense. x can be any real number. I think what you meant was that if y= sin(x) then [itex]-1\le y\le 1[/itex]. And you don't need to "work out if the limit is negative or positive infinity"- it's neither one, it just doesnt' exist.
     
  11. Aug 12, 2007 #10

    Gib Z

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    The point is that the limit as x approaches infinity of finite period functions does not exist, as it will oscillate between the range of the function.
     
  12. Aug 13, 2007 #11

    CompuChip

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    Depends what you call a proof. And how rigorous a proof you want.
     
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