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Math Help - Delta-Epsilon Proof

  1. #1
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    Delta-Epsilon Proof

    This was a problem from my calculus final a year ago that I've wanted to see for a while:

    Use the delta-epsilon definition of a limit to prove that the limit as x approaches 0 of f(x) = sin(x)/(x^2 +1) is 0.

    I tried using the squeeze theorem in an effort to bound sin(x), because I really don't know how to deal with sin(x) in a delta epsilon proof. This problem has just been on my mind for a while. Thanks for the help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by amee View Post
    This was a problem from my calculus final a year ago that I've wanted to see for a while:

    Use the delta-epsilon definition of a limit to prove that the limit as x approaches 0 of f(x) = sin(x)/(x^2 +1) is 0.

    I tried using the squeeze theorem in an effort to bound sin(x), because I really don't know how to deal with sin(x) in a delta epsilon proof. This problem has just been on my mind for a while. Thanks for the help!
    For all x \in \mathbb{R} and x\ne 0: |\sin(x)|< |x|

    CB
    Last edited by CaptainBlack; August 6th 2009 at 05:58 AM. Reason: tidying up
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  3. #3
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    Can you show me how to do this proof? CB, what you wrote is true, but I'm not familiar with delta-epsilon proofs to know how to incorporate it into the rest of the proof.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    You want d such that |x- 0| < d then |f(x) - 0 | < e


    |f(x) - 0 | = | sin(x)/(x^2+1)| < |sin(x)|


    take d = arcsin(e)

    then if |x| < arcsin(e) then |sin(x)| < |sin(arcsin(e)| = e
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  5. #5
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    Quote Originally Posted by Calculus26 View Post
    You want d such that |x- 0| < d then |f(x) - 0 | < e


    |f(x) - 0 | = | sin(x)/(x^2+1)| < |sin(x)|


    take d = arcsin(e)

    then if |x| < arcsin(e) then |sin(x)| < |sin(arcsin(e)| = e
    But since |\, \sin (x) \,| < |\, x \,|\; \text{for}\; |\, x \, | > 0 then you can chose \delta = \varepsilon .
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