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What is sin x when x tends to infinity?

  1. Jun 4, 2005 #1
    This question popped up in my head. What is sin x when x tends to infinity? Since sine is a preiodic functions which repeats itself, is the answer 1, -1 or 0 or something else altogether?
     
  2. jcsd
  3. Jun 4, 2005 #2
    It is undefined; there is no limit! Sin[x] does not converge to any value as x increases - it maintains its periodic character. Limits only apply when something converges to something somewhere.

    A proof is readily constructed from the def. of limits at [tex]\mbox{$\infty$}[/tex]. Choose a small [tex]\mbox{$\epsilon$}[/tex] ("1"is small enough) and show that for any x0, no matter how large, there exists an x>x0 such that

    [tex]\|f(x)-f(x_0)\|\geq\epsilon[/tex].​

    Plug in [tex]\mbox{$\epsilon=1$}[/tex] and [tex]\mbox{$x=x_0+\frac{\pi}{2}$}[/tex].

    What you show is that is not convergent, thus there is no limit.

    Hey, does anyone know a better way to do inline LaTeX here?
     
    Last edited by a moderator: Jun 5, 2005
  4. Jun 5, 2005 #3

    dextercioby

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    It looks good to me until [itex]\geq[/itex].

    Daniel.
     
  5. Jun 5, 2005 #4
    What's wrong with [itex]\geq[/itex]? You want to show that [itex]\|f(x)-f(x_0)\|<\epsilon[/itex] does not hold for all [itex]x>x_0[/itex], so you find an [itex]x[/itex] where the relation is [itex]\geq[/itex] instead of [itex]<[/itex].
     
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