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Does lim x →0 sin 1/x exist? also .

  1. May 24, 2009 #1
    Does lim x →0 sin 1/x exist? also.....

    1. The problem statement, all variables and given/known data

    does lim x →0 sin(1/x) exist?

    3. The attempt at a solution

    I am very new to calculus. From what i have understood till now, limit of a function at a point exists only if the left and right hand limits are equal. What would the graph of sin(1/x) look like? Would the left and right hand limits exist? Do they exist in all cases, for all functions?
     
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  3. May 24, 2009 #2

    matt grime

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    Re: Does lim x →0 sin 1/x exist? also.....

    try drawing it and find out.

    It would be a bit dull if left and right limits always exist. Remember that limits are unique. So, you need to think about sequences of real numbers x_n tending to zero.

    Alternatively we can think about

    lim sin(y)

    as y tends to infinity.
     
  4. May 24, 2009 #3
    Re: Does lim x →0 sin 1/x exist? also.....

    Btw, the existence of left and right limits of a function at a certain point is necessary but not sufficient for a limit to exist at that point. Left and right limits should also be equal for a limit to exist at the same point.
     
  5. May 24, 2009 #4

    HallsofIvy

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    Re: Does lim x →0 sin 1/x exist? also.....

    In particular, think about the sequences [itex]x_n= 1/(n\pi)[/itex], [itex]x_n= 2/((4n+1)\pi)[/itex], and [itex]x_n= 2/((4n-1)\pi)[/itex] a n goes to infinity.
     
  6. May 24, 2009 #5
    Re: Does lim x →0 sin 1/x exist? also.....

    In a real analysis class you would do something like the following:

    Let [tex] f(x) = sin ( \frac {1}{x} ) [/tex]

    Let [tex] s_{n} [/tex] be the sequence [tex] \frac {2}{n \pi} [/tex]

    [tex] \lim_{n \to \infty}s_{n} = 0 [/tex]

    but [tex] (f(s_{n}) )[/tex] is the sequence 1,0,-1,0,1,0,-1,0... which obviously doesn't converge.

    Therefore, [tex] \lim_{x \to 0} sin( \frac {1}{x}) [/tex] does not exist.
     
  7. May 24, 2009 #6

    tiny-tim

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    Hi hermy! :wink:

    matt's :smile: is the best way for general questions like this …

    it shows you what the difficulty is! :wink:
     
  8. May 25, 2009 #7
    Re: Does lim x →0 sin 1/x exist? also.....

    Could you give an example of a function where the left/right hand limit doesn't exist at a point (where the function is defined on the set of all real no.s)?
     
  9. May 25, 2009 #8

    D H

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    Re: Does lim x →0 sin 1/x exist? also.....

    f(x) = 0 if x is rational, 1 if it is irrational.
     
  10. May 25, 2009 #9

    Mark44

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    Re: Does lim x →0 sin 1/x exist? also.....

    Another example is f(x) = 1/x, x [itex]\neq [/itex] 0; f(0) = 0.
    This function is defined for all real numbers. The left hand limit, as x approaches zero is negative infinity, while the right hand limit, as x approaches zero, is positive infinity.
     
  11. May 25, 2009 #10
    Re: Does lim x →0 sin 1/x exist? also.....

    In calculus you need to use the idea of limits. Essentially the limit of sin x/x does equal 1 but you have to show it from both sides.

    If we consider a left hand limit if sin x/x then we can verify the limit with a table of calculations (just grab a calculator and calculate f(x) sin x / x for values approaching x)

    We can also consider the right hand limit also. For the right hand limit we can do the same thing by letting f(x) approach sin x/x.

    Now the limit is only valid if and only if the right hand limit equals the left hand limit. So
    essentially if this is the case and you find a value for it, then you can prove what that limit
    is.
     
  12. May 25, 2009 #11

    D H

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    Re: Does lim x →0 sin 1/x exist? also.....

    The original poster is looking at f(x)=sin(1/x), not g(x)=sin(x)/x.
     
  13. May 25, 2009 #12

    matt grime

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    Re: Does lim x →0 sin 1/x exist? also.....

    I don't want to give the game away but.... sin(1/x) as x tends to 0 perhaps? I can trivially extend this to a function on the whole of the real line if I want to, if you're worried about it being undefined at zero:

    f(x)=sin(1/x) x=/=0
    f(0)=0 (or anything else you want)
     
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