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Thread: 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 $\displaystyle {\epsilon}>0$ there exists a $\displaystyle {\sigma}>0$ such that $\displaystyle |x-0|=|x|<{\sigma} ->|sin(1/x)-L|<{\epsilon}$

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

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

    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 $\displaystyle \displaystyle \epsilon - \delta$ proof? Couldn't you just say that since the function oscillates, it is impossible to say that $\displaystyle \displaystyle \sin{X}$ approaches a particular value as $\displaystyle \displaystyle X \to \infty$?
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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 $\displaystyle \displaystyle L$ such that $\displaystyle \displaystyle \forall \epsilon > 0, \exists \delta$ such that $\displaystyle \displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L \right| < \epsilon $ if $\displaystyle \displaystyle |x - 0| = |x| < \delta$.



    On any interval $\displaystyle \displaystyle (-\epsilon + \bar{x}, \bar{x} + \epsilon)$ we have $\displaystyle \displaystyle x \in (-\delta, \delta)$ such that $\displaystyle \displaystyle f(x) = \pm 1$. Consequently for $\displaystyle \displaystyle \epsilon < 2$ we can't have $\displaystyle \displaystyle \left|\sin{\left(\frac{1}{x}\right)} - L\right| < \epsilon$ for any $\displaystyle \displaystyle L$ and any $\displaystyle \displaystyle \delta > 0$. Thus we can not satisfy the definition no matter what we choose for $\displaystyle \displaystyle L$.
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  5. #5
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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

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

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

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

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