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How to find lim as x -> inf. of sin(x) algebraically

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \to \infty } \sin x[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.
     
  2. jcsd
  3. Sep 2, 2012 #2

    Bacle2

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    The usual definition of limit as x→∞ is:

    [tex]\lim_{x \to \infty } \f(x)[/tex] =L

    if, given ε>0 , there is an M such that for all x>M , |f(x)-L|<ε

    So , algebraically, you can show the limit does not exist by choosing the right value

    of ε for which the def. is not satisfied.
     
  4. Sep 2, 2012 #3

    HallsofIvy

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    You can't "find the limit of sin(x) as x goes to infinity" because it does not have a limit. No matter how large x is, the will be larger values for which sin(x) has values of -1, 1, etc. sin(cx) does NOT get close to any one number as x goes to infinity.
     
  5. Sep 2, 2012 #4

    Bacle2

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    But PhizKind is not trying to find the limit,but instead to show , algebraically,

    that the limit does not exist.
     
  6. Sep 2, 2012 #5

    Ray Vickson

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    The function sin(x) is periodic, with period 2π. That means that sin(x+2kπ) = sin(x) for k = ±1, ±2, ±3, ... for any x. So, the graph of y = sin(x) for x between 10,000,000π and 10,000,002π is an exact copy of the graph of y = sin(x) for x between 0 and 2π.

    RGV
     
  7. Sep 2, 2012 #6

    Zondrina

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    You need to use a formal definition for this question. In particular :

    [itex]\forall[/itex]M>0, [itex]\exists[/itex]N | x>N [itex]\Rightarrow[/itex] f(x)>M

    Then start with f(x) > M and massage it until you can find your suitable N.
     
  8. Sep 2, 2012 #7

    Bacle2

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    I think you are referring here to the case where the limit is ∞, not the limit as x→∞.

    For f(x)>1 , you willfind no M with the property you described.
     
  9. Sep 2, 2012 #8

    Bacle2

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    Maybe you can do this by contradiction:

    If the limit existed, and it was equal to L, you would be able to say that, when you

    go far-enough (i.e., when x is largerthan M ), you can get to within any level of accuracy

    (i.e., within ε ) to the limit.

    This means, formally , that there is some M>0 with:

    |Sinx-L|<ε when x>M

    But this will not happen using , e.g., Ray Vickson's argument. You want:

    -ε< Sinx-L <ε

    But no matter what value of L you choose, L will oscillate.
     
  10. Sep 2, 2012 #9

    Zondrina

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    Ah yes I forgot about that, I meant to say |f(x)-L| < M, but using ε in this case would have been more appropriate than using M ( Even though it really doesn't matter just some formalism ).
     
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