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Math Help - epsilon delta proof

  1. #1
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    epsilon delta proof

    abs[sin(n)/n] <epsilon

    How do I solve for n so that I can find a delta to use in my proof that the limit of sin(n)/n is 0?

    Can I just say that since sin has max and min values between 1 and -1 that 1/n < epsilon? and then use 1/epsilon as my delta?
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  2. #2
    Senior Member roninpro's Avatar
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    I'm not very clear on what you are asking. Are you trying to show

    \displaystyle \lim_{n\to \infty} \frac{\sin n}{n}=0

    using the \varepsilon definition?
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  3. #3
    MHF Contributor
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    Use \displaystyle \left|\frac{\sin{n}}{n}\right| \leq \left|\frac{1}{n}\right|
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