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Math Help - differentiability of Sin(1/x)...

  1. #1
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    differentiability of Sin(1/x)...

    consider: f: R \rightarrow R defined as:
    f(x) = sin(\frac{1}{x}) if x \neq 0,
    f(0) = 0.
    prove f is not differentiable at 0.

    ------------------------------------------------------------------------
    section A

    going by the definition of differentiability, i understand it has to be shown that

    \lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{\sin \frac{1}{x} - sin(0)}{x - 0} = \lim_{x\to 0} \frac{\sin \frac{1}{x}}{x} .

    DOES NOT EXIST

    • i attempted computing the left and right hand limits but i dont think that is the right method?
    ------------------------------------------------------------------------
    section B

    i also know if a function is differentiable at 0 then it has to be continuous at 0, which sin(1/x) is not. but i don't have a good idea on how to prove that sin(1/x) is not continuous at 0
    ??

    also, would it be sufficient to just show that sin(1/x) is discontinuous at 0 to prove it is not differentiable at 0?

    -------------------------------------------------------------------------
    please provide assistance greatly appreciated. once again, i'm assuming this must be simple and straight forward for an advanced mathematician
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by heyo12 View Post
    also, would it be sufficient to just show that sin(1/x) is discontinuous at 0 to prove it is not differentiable at 0?
    Yes.

    To show that its not continuous at 0 we need only observe that there are points arbitarily close to 0 where \sin(1/x)=1, for instance consoder the points:

    x=\frac{1}{(2n+1/2)\pi}\ \ n=1, 2, 3, ..

    RonL
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  3. #3
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    Here is what CaptainBlank did more formally.

    Assume that \sin \frac{1}{x} is continous then for any sequence (x_n)\in \mathbb{R} - \{ 0 \} so that \lim x_n=0 we must have that \lim f(x_n) = f(0)=0. But using the sequence that was shown above we see that fails.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Here is what CaptainBlank did more formally.

    Assume that \sin \frac{1}{x} is continous then for any sequence (x_n)\in \mathbb{R} - \{ 0 \} so that \lim x_n=0 we must have that \lim f(x_n) = f(0)=0. But using the sequence that was shown above we see that fails.
    thnx captain and perfecthacker.

    ok this is what i understood from what you said perfecthacker. here's my proof from both your help..

    you used the sequential definition of continuity? which implies that:
    continuity at 0 = limit at 0 exists and equals f(0)?

    PROOF
    sin(1/x) is continuous at x_0 = 0 if for any sequence x_n \epsilon R, such that \lim_{n \to \infty}(x_n) = x_0, we have \lim_{n \to \infty}f(x_n) = f(x_0)=0

    now let x_n = \frac{1}{(2n)\pi} and b_n = \frac{1}{(2n + \frac{1}{2})\pi}

    so \lim_{n \to \infty}(x_n) = \lim_{n \to \infty}(b_n) = 0

    if we consider, \lim_{n \to \infty}f(x_n) we have:
    \lim_{n \to \infty}f(x_n) = \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n)\pi}} = \lim_{n \to \infty}sin(2\pi n) = 0

    but, if we consider, \lim_{n \to \infty}f(b_n) we have:
    \lim_{n \to \infty}f(b_n) = \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}} = \lim_{n \to \infty}sin((2n + 1/2)\pi) = 1

    we see there are points arbitrarily close to 0 where sin(1/x) = 1.
    so this means the limit at 0 doesn't exist
    as the limit doesnt exist. the function is not continuous. this implies that it is not differentiable at 0. Q.E.D


    is this completely correct? and is there anything else i need to add onto this proof? thanks
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by heyo12 View Post
    thnx captain and perfecthacker.

    ok this is what i understood from what you said perfecthacker. here's my proof from both your help..

    you used the sequential definition of continuity? which implies that:
    continuity at 0 = limit at 0 exists and equals f(0)?

    PROOF
    sin(1/x) is continuous at x_0 = 0 if for any sequence x_n \epsilon R, such that \lim_{n \to \infty}(x_n) = x_0, we have \lim_{n \to \infty}f(x_n) = f(x_0)=0

    now let x_n = \frac{1}{(2n)\pi} and b_n = \frac{1}{(2n + \frac{1}{2})\pi}

    so \lim_{n \to \infty}(x_n) = \lim_{n \to \infty}(b_n) = 0

    if we consider, \lim_{n \to \infty}f(x_n) we have:
    \lim_{n \to \infty}f(x_n) = \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n)\pi}} = \lim_{n \to \infty}sin(2\pi n) = 0

    but, if we consider, \lim_{n \to \infty}f(b_n) we have:
    \lim_{n \to \infty}f(b_n) = \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}} = \lim_{n \to \infty}sin((2n + 1/2)\pi) = 1

    we see there are points arbitrarily close to 0 where sin(1/x) = 1.
    so this means the limit at 0 doesn't exist
    as the limit doesnt exist. the function is not continuous. this implies that it is not differentiable at 0. Q.E.D


    is this completely correct? and is there anything else i need to add onto this proof? thanks
    That looks pretty good to me. Though on second thoughts the sequence x_n is probably redundant as we have f(0)=0, so we need only display a sequence who's image under f does not go to 0 to prove that f(x) is discontinuous at x=0.

    RonL

    RonL
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