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Limit of Arcsin Approaching Infinity

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    limit approaching infinity: (arcsin(x))/(x)

    = 0

    Question is: Why? The 'Sandwich Theorem' 0=[(arcsinx)/x]=0 gives this
    solution, but looking at the graph of (arcsinx)/x , this appears

    2. Relevant equations

    lim x->OO [arcsin(x)] - {DNE)
    lim x->OO [(arcsinx)/x] - {DNE/OO}

    lim x->OO [-Pi/2x] = lim x->OO [(arcsinx)/x] = lim x->OO [Pi/2x]
    => 0 = [(arcsinx)/x] = 0

    => lim x->OO [(arcsinx)/x] = 0

    HOWEVER: (-Pi/2)<[arcsinx]<(Pi/2) ... SO ...
    ... (x) never reaches (OO) for (Pi/2x) to reach the limit x->OO (Pi/OO)
    = 0.

    3. The attempt at a solution

    The sandwich theorem gives and answer of "0". Maple 11 gives the same answer, so does my teacher. This just doesn't seem possible.
  2. jcsd
  3. Feb 28, 2008 #2
    The only way you can get "0" is if you're using complex limits. I'm pretty sure. Let me check up on that.
    Last edited: Feb 28, 2008
  4. Feb 29, 2008 #3
    If you use, L'Hospitals Rule, you get a [tex]\frac{1}{\sqrt{1-x^2}}[/tex], for x tending to infinity. Thats a zero, but a complex one, dont know if its even applicable here...
  5. Feb 29, 2008 #4
    arcsin(x) isn't defined for x outside of [-1, 1] considering only the reals. So unless x is complex, I don't think this limit exists.
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