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Continuity Proof

  1. Aug 8, 2009 #1

    jgens

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    Gold Member

    1. The problem statement, all variables and given/known data

    Prove that the sine function is continuous on its domain.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I think that I've gotten this right but I would appreciate it if someone checked my solution . . .

    Let [tex]\epsilon > 0[/tex]. We define [tex]\delta[/tex] such that,

    [tex]0 < |x - a| < \delta = \epsilon[/tex]

    Now, by the mean-value theorem of differential calculus, we have that,

    [tex]1 \geq \frac{sin(x) - sin(a)}{x - a} \geq -1[/tex]

    or similarly,

    [tex]1 \geq \left |\frac{sin(x) - sin(a)}{x - a} \right |[/tex]

    To complete the proof, we utilize this inequality such that

    [tex]0 < |sin(x) - sin(a)| \leq |x-a| < \epsilon[/tex]

    As desired. This proves that the sine function is continuous on its domain.
     
  2. jcsd
  3. Aug 8, 2009 #2

    nicksauce

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    I don't think I agree with your proof. You're trying to show that sine is continuous and you invoke the mean value theorem. But the mean value theorem requires that the function it's applied to is continuous. In other words, you're assuming your conclusion.
     
  4. Aug 8, 2009 #3

    jgens

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    Darn! Forgot about that, you're right. :(
     
  5. Aug 9, 2009 #4
    A suggestion:

    First show sin(x) is contunuous at x = 0 by exploiting the fact that for [itex]-\frac{\pi}{2}\leq x \leq \frac{\pi}{2}[/itex] you have [itex]|\sin (x)| \leq |x|[/itex].

    Then using the Pythagorean identity for cosine in terms of sine (near x = 0) you can show that cosine is also continuous at x = 0.

    Then using the identity that [itex]\lim_{x\rightarrow a}{f(x)} = \lim_{h\rightarrow 0}{f(a+h)}[/itex], and the angle sum identity for sine you can show that sin(x) is continuous at any arbitrary number a.

    I hope this helps.

    --Elucidus
     
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