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Math Help - Prove lim(sin(1/x)) as x tends to 0 does not exist

  1. #1
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    Prove lim(sin(1/x)) as x tends to 0 does not exist

    assume there is a limit, L. So we have that for all {\epsilon}>0 there exists a {\sigma}>0 such that |x-0|=|x|<{\sigma} ->|sin(1/x)-L|<{\epsilon}

    Consider when x>_0 and let {\epsilon}=1/2.

    Then |x|=x -> 0<x<{\sigma}-> |sin(1/x)-L|<1/2

    I need a contradiction.

    Thanks
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  2. #2
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    Do you need to use an \displaystyle \epsilon - \delta proof? Couldn't you just say that since the function oscillates, it is impossible to say that \displaystyle \sin{X} approaches a particular value as \displaystyle X \to \infty?
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  3. #3
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    no I need to use epsilon delta
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    We need an \displaystyle L such that \displaystyle \forall \epsilon > 0, \exists \delta such that \displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L \right| < \epsilon if \displaystyle |x - 0| = |x| < \delta.



    On any interval \displaystyle (-\epsilon + \bar{x}, \bar{x} + \epsilon) we have \displaystyle x \in (-\delta, \delta) such that \displaystyle f(x) = \pm 1. Consequently for \displaystyle \epsilon < 2 we can't have \displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L\right| < \epsilon for any \displaystyle L and any \displaystyle \delta > 0. Thus we can not satisfy the definition no matter what we choose for \displaystyle L.
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    Sorry, I don't quite understand. what x bar? Do you mean L. Then I don't know why f(x) must take the values + or - 1 in any (-delta,delta). Lastly you seem to imply L must be eith + or -1 meaning the distance between sin(1/x) and L will be 2 at some point. Why is this the case?
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    \displaystyle \bar{x} is some \displaystyle x close to \displaystyle 0.

    The proof basically states that for values infinitessimally close to \displaystyle x = 0, it can be said that \displaystyle f(x) is equal to both \displaystyle 1 and \displaystyle -1, which is enough to show that you can not "zoom in" on an \displaystyle L as you make \displaystyle \epsilon and \displaystyle \delta smaller.
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    Consider the sequence x_n = \frac 2 {n\pi}, for n = 1, 2, ....
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    In fact, consider the two sequences x_n= \frac{1}{2n\pi} and x_m= \frac{1}{n\pi}. You need to know that f(x) goes to L as x goes to a if and only if f(xn) goes to L for every sequence {xn} that goes to L.
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  9. #9
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    You need to know that f(x) goes to L as x goes to a if and only if f(xn) goes to L for every sequence {xn} that goes to L. Is that a general theorem?

    Don't know how you know 1/2npi goes to L but I will roll with it for now

    sin(2npi)=0 so need delta such that if 0<|1/(2npi)|<delta then |-L|=|L|< e. But we can make e arbitarily small so it can be made less than |L| which is a contradiction. So f((1/2npi))has no limit so f(x) has no limit. Am I correct?
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    I've done more working
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  11. #11
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    Re: Prove lim(sin(1/x)) as x tends to 0 does not exist

    Given any \delta> 0, there exist an integer n such that \frac{2}{n\pi}< \delta. That means that every number from -\frac{2}{n\pi} to +\frac{2}{n\pi} are less than \delta. For those values of x, sin(1/x) becomes -sin(n\pi/2)= -1 and sin(n\pi/2)= 1. That is, no matter how small \delta is, there will exist values of x less than \delta such that sin(1/x) takes on values of 1 and -1. You can't guarentee any \epsilon less than 2.
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