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Homework Help: Continuity of sin(1/x) on (0,1)

  1. Dec 20, 2007 #1
    1. The problem statement, all variables and given/known data
    how do you show that sin(1/x) is continuous on (0,1)? (i know it's also continous on (0, infinite)).


    2. Relevant equations



    3. The attempt at a solution

    |f(x)-f(xo)| = |sin(1/x)- sin(1/xo)|= |2sin((xo-x)\2)cos((xo+x)/2)|

    =< 2|sin((xo-x)/(2xox))|=< |(xo-x)/(xox)|. is this inequality true? a similiar one is used in a different example. if it is, why? is it because sin(x)=<x ? when x is positive? now since x<1 choosing [tex]\delta[/tex]=xo[tex]\epsilon[/tex] then if

    |xo-x|<[tex]\delta[/tex] then |f(xo)-f(x)|<[tex]\epsilon[/tex]

    is this answer correct? what about the endpoints?
     
  2. jcsd
  3. Dec 20, 2007 #2
    i guess this is true for all xo and x different from 0 so it is true for (0, infinite) right?
     
  4. Dec 20, 2007 #3
    A cheap way would be to use the theorem that differentiability implies continuity
    The derivative of [tex]\sin(\frac{1}{x})[/tex] is [tex]\frac{-1}{x^2}\cos(\frac{1}{x})[/tex].
    The only place where the derivative is undefined is at x=0 which is not in the given interval so the function is continuous on the interval.

    I don't really like that way because it seems out of order (differentiability comes after continuity)

    Here's another.
    Definition of Continuity: The function is defined at every point on the interval and
    [tex]\lim_{x\to c}f(x) = f(c)[/tex]

    We know that the function is defined at every point except 0, so it is defined in the interval.
    The limit as x goes to c of the function is [tex]\sin(\frac{1}{c})[/tex] which is defined and equal to f(c) at every point except c=0, which isn't in the interval
    Therefore, the function is continuous on (0,1)
     
  5. Dec 20, 2007 #4
    sinx is continuous, and 1/x is continuous on (0,1). The composition of continuous functions is continuous. So sin(1/x) is continouous on (0,1)

    Are you required to prove that sinx itself is continuous?
     
  6. Dec 20, 2007 #5

    HallsofIvy

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    You say "(i know it's also continous on (0, infinite))". If a function is continuous on a set, A, it is continuous on any subset of A.
     
  7. Dec 20, 2007 #6
    ANY subset of A? What if the subset has an empty interior? How can f inverse of an open subset of f(A) (in the subspace topology of R) be open in A?
     
    Last edited: Dec 20, 2007
  8. Dec 20, 2007 #7
    What's the epsilon-delta proof for the continuity of sinx itself?
     
  9. Dec 20, 2007 #8

    morphism

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    One can prove that |sin(x) - sin(y)| <= |x - y|.
     
  10. Dec 20, 2007 #9

    morphism

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    The empty set and A itself are still open subsets of A.
     
  11. Dec 20, 2007 #10
    Ok, I got it. If A is given the subspace topology of R, then any continuous function restricted to A is still continuous.

    But an epsilon-delta proof of this fact will not work if A is a set with an empty interior.
     
    Last edited: Dec 20, 2007
  12. Dec 25, 2007 #11
    i see you have all posted great replies, but can anyone tell me if my proof actually works. i want to understand the proper use of the epsilon-delta proof. the teacher will surely ask us to do such a proof. it seems okay.
     
  13. Dec 25, 2007 #12
    There's no need to use the epsilon-delta proof, and I don't think your prof expects it.
     
  14. Dec 25, 2007 #13
    I don't think so either, the exercise does not seem to encourage you to take a look at the epsilon-delta kind of proofs.
     
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