Results 1 to 5 of 5

Thread: differentiability of Sin(1/x)...

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    9

    differentiability of Sin(1/x)...

    consider: f: R $\displaystyle \rightarrow$ R defined as:
    f(x) = $\displaystyle sin(\frac{1}{x})$ if $\displaystyle x \neq 0,$
    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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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 $\displaystyle \sin(1/x)=1$, for instance consoder the points:

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

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Here is what CaptainBlank did more formally.

    Assume that $\displaystyle \sin \frac{1}{x}$ is continous then for any sequence $\displaystyle (x_n)\in \mathbb{R} - \{ 0 \}$ so that $\displaystyle \lim x_n=0$ we must have that $\displaystyle \lim f(x_n) = f(0)=0$. But using the sequence that was shown above we see that fails.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2007
    Posts
    9
    Quote Originally Posted by ThePerfectHacker View Post
    Here is what CaptainBlank did more formally.

    Assume that $\displaystyle \sin \frac{1}{x}$ is continous then for any sequence $\displaystyle (x_n)\in \mathbb{R} - \{ 0 \}$ so that $\displaystyle \lim x_n=0$ we must have that $\displaystyle \lim f(x_n) = f(0)=0$. 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 $\displaystyle x_0 = 0$ if for any sequence $\displaystyle x_n$ $\displaystyle \epsilon$ R, such that $\displaystyle \lim_{n \to \infty}(x_n) = x_0$, we have $\displaystyle \lim_{n \to \infty}f(x_n) = f(x_0)=0$

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

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

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

    but, if we consider, $\displaystyle \lim_{n \to \infty}f(b_n)$ we have:
    $\displaystyle \lim_{n \to \infty}f(b_n)$ = $\displaystyle \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}}$ = $\displaystyle \lim_{n \to \infty}sin((2n + 1/2)\pi)$ = 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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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 $\displaystyle x_0 = 0$ if for any sequence $\displaystyle x_n$ $\displaystyle \epsilon$ R, such that $\displaystyle \lim_{n \to \infty}(x_n) = x_0$, we have $\displaystyle \lim_{n \to \infty}f(x_n) = f(x_0)=0$

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

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

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

    but, if we consider, $\displaystyle \lim_{n \to \infty}f(b_n)$ we have:
    $\displaystyle \lim_{n \to \infty}f(b_n)$ = $\displaystyle \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}}$ = $\displaystyle \lim_{n \to \infty}sin((2n + 1/2)\pi)$ = 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 $\displaystyle x_n$ is probably redundant as we have $\displaystyle f(0)=0$, so we need only display a sequence who's image under $\displaystyle f$ does not go to $\displaystyle 0$ to prove that $\displaystyle f(x)$ is discontinuous at $\displaystyle x=0$.

    RonL

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. differentiability
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 16th 2012, 06:31 AM
  2. Differentiability
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 13th 2010, 11:45 AM
  3. Differentiability
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 21st 2010, 07:16 PM
  4. Differentiability
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 20th 2009, 11:14 AM
  5. Differentiability
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 30th 2008, 01:08 PM

Search Tags


/mathhelpforum @mathhelpforum