Results 1 to 7 of 7

Math Help - Continuity

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    25

    Continuity

    I have two homework problems that I was hoping someone could help me with:
    1. Prove that f(x)= sin(1/x), where x can't equal zero and f(0)=0 is not continuous at x=0 by finding an ε for which there is no reply.


    and

    2. At what values of x is
    f(x) = (piecewise defined) 0, if x is irrational or = sinx if x is rational


    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    For the first one see this thread

    For the second one, my guess would be that it's continous at all points of the form k\pi where k \in \mathbb{Z} since \sin(x)=0 for these points (and only for these).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by CoraGB View Post
    I have two homework problems that I was hoping someone could help me with:
    1. Prove that f(x)= sin(1/x), where x can't equal zero and f(0)=0 is not continuous at x=0 by finding an ε for which there is no reply.


    and

    2. At what values of x is
    f(x) = (piecewise defined) 0, if x is irrational or = sinx if x is rational


    Thank you
    To prove f(x) is continuous at x=k\pi for k\in\mathbb{Z}, we need to prove that \forall~\epsilon>0, \exists~\delta>0 s.t. |f(x)-f(k\pi)|=|f(x)|<\epsilon if |x-k\pi|<\delta. To do this, take two cases:

    1) x is irrational. This is trivial though, because |f(x)|=0<\epsilon.

    2.) x\in\mathbb{Q} and |x-k\pi|<\delta. Since \sin(x) is continuous, |f(x)|=|\sin(x)|=|\sin(x)-\sin(k\pi)|<\epsilon when |x-k\pi|<\delta.

    So f(x) is continuous at x=k\pi for k\in\mathbb{Z}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    And to prove it's continuous nowhere else, let x_0\in\mathbb{Q}. Since \forall~\delta>0, there exists an irrational number x\in(x_0-\delta,x_0+\delta), |f(x_0)-f(x)|=|f(x_0)|=|\sin(x_0)|>0. So choosing \epsilon=\frac{\sin(x_0)}{2} ensures that f is not continuous at x_0.

    A similar argument can be constructed if x_0 is irrational.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    151
    For what it is worth, another solution to 1.

    You know that \sin(1/x)=1 when \frac{1}{x}=\frac{\pi}{2} + 2k\pi where k\in\mathbb{Z},
    and \sin(1/x)=-1 when \frac{1}{x}=\frac{3\pi}{2}+2k\pi

    Lets define a sequence x_n=\frac{\pi}{2} +2n\pi where n\in\mathbb{N}.
    Also the define the sequence g_n=\frac{3\pi}{2} +2n\pi, where n\in\mathbb{N}.
    It is clear that \lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\inft  y}g_n=\lim\limits_{x\to0^+}\frac{1}{x}=+\infty
    So we will substitude the sequences for \frac{1}{x}

    Observe that
    \lim\limits_{n\to\infty}\sin(x_n)=1
    And
    \lim\limits_{n\to\infty}\sin(g_n)=-1

    Now we have two sequences x_n and g_n with the same limit as 1/x such that \sin(x_n) and \sin(g_n) have two different limits.
    Hence \lim\limits_{x\to0^+}\sin\left(\frac{1}{x}\right) does not exist, so \lim\limits_{x\to0}\sin\left(\frac{1}{x}\right) does not exist either.

    Hjörtur
    Last edited by hjortur; October 22nd 2009 at 02:17 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by hjortur View Post
    For what it is worth, another solution to 1.

    You know that \sin(1/x)=1 when \frac{1}{x}=k\frac{\pi}{2} where k\in\mathbb{Z},
    and \sin(1/x)=-1 when \frac{1}{x}=k\frac{3\pi}{2}
    Actually,

    \sin(1/x)=1 iff \frac{1}{x}=\frac{k\pi}{2} and k\equiv1\mod4

    \sin(1/x)=-1 iff \frac{1}{x}=\frac{k\pi}{2} and k\equiv3\mod4
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2009
    Posts
    151
    Of course you are correct, I wonder why I didnt notice the error?
    Any way corrected version above.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continuity
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 20th 2011, 02:36 PM
  2. y=x^3 (continuity)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 16th 2010, 11:50 PM
  3. continuity
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 9th 2010, 05:23 PM
  4. Continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 10:10 PM
  5. Continuity, Uniform Continuity
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 1st 2009, 09:36 PM

Search Tags


/mathhelpforum @mathhelpforum