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Show that f(x) = sin(x^2) is continuous for all a [-sqrt(pi), sqrt(pi).

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Given f : [-[tex]\sqrt{\pi}[/tex], [tex]\sqrt{\pi}[/tex] ] [tex]\rightarrow[/tex] [-1, 1]

    f(x) = sin(x[tex]^{2}[/tex])

    a)
    Show that f is continuous for all a [tex]\in[/tex] [-[tex]\sqrt{\pi}[/tex], [tex]\sqrt{\pi}[/tex] ]

    b)
    Find a [tex]\delta[/tex] so that |x - y| [tex]\leq[/tex] [tex]\delta[/tex] implies that
    |f(x) - f(y)| [tex]\leq[/tex] 0.1 for all x and y in [-[tex]\sqrt{\pi}[/tex], [tex]\sqrt{\pi}[/tex] ]

    2. Relevant equations



    3. The attempt at a solution

    I honestly don't know where to begin. But would be very pleased for any help. A full solution would be the optimal though. But anything helps.
     
  2. jcsd
  3. Sep 20, 2010 #2

    vela

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    If you want to go back to the definition of continuity, you need to show that for every point a in the interval,

    [tex]\lim_{x \to a} f(x) = f(a)[/tex]

    (with the appropriate one-sided limit at the endpoints). If you can assume other facts, then you can argue f(x) is continuous in a less tedious way. What do you think you're supposed to do?
     
  4. Sep 20, 2010 #3
    I'm supposed to use the definition of continuity.

    I know how to show continuity at a point, but have no clue of how to do it for an interval.
     
  5. Sep 20, 2010 #4

    vela

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    You just have to show f is continuous at every point in the interval. It essentially amounts to saying "Let a∈[-√π,√π]" and showing f is continuous at x=a like usual.
     
  6. Sep 20, 2010 #5
    Lol, that simple, thx. I kept thinking I had to use the interval in the proof somehow and not just say a = the interval. OK, that I can do. How about b?
     
  7. Sep 21, 2010 #6
    I think I solved it, I will write it down when I get home so someone can tell if its correct or not, but have to go now.
     
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