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Math Help - Epsilon Limit Question

  1. #1
    Super Member craig's Avatar
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    Epsilon Limit Question

    Q: Use the formal definition of convergence to prove that the sequence \frac{\pi n + \sin{n}}{n} tends to a limit.

    A: The formal definition is:

    A sequence x_n converges to a limit if, given \epsilon > 0, there is a natural number N such that |x_n - x| < \epsilon.

    Just not too sure on where to start with this one. I'm guessing that splitting up the fraction into \pi + \frac{\sin{n}}{n} and then proving that the latter part tends to a limit would be the way to go?

    Thanks in advance for your help

    Craig
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    TheEmptySet's Avatar
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    Yes then just plug into your def, but first you need to "guess" what the limit L is

    \displaystyle \bigg|\pi-\frac{\sin(n)}{n}-L \bigg|

    Then show you can make the above as small as you wish or less than epsilon
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  3. #3
    Super Member craig's Avatar
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    Thanks for the reply.

    Well my educated guess will be that the limit is \pi, giving me \displaystyle \bigg|\frac{\sin(n)}{n}\bigg|



    Now \displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n} which is obvious.

    In the solutions they've got this:

    Let \epsilon > 0, take N > \frac{1}{\epsilon}, then for all n>N:

    \displaystyle \bigg|\frac{\sin(n)}{n}\bigg| \leq \frac{1}{n} < \frac{1}{N} < \epsilon , so the sequence tends to \pi.

    Just wondering how they took N > \frac{1}{\epsilon} in the first place?

    Sorry for the confusion, really not getting this whole epsilon limit business.

    Thanks again

    Craig
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    This may help, think of epsilon as a challenge. Someone gives you maximal error estimate (epsilon) you need to find an N such that
    |x_n-x|< \epsilon
    is always true.

    As you noted

    \displaystyle |\frac{\sin(n)}{n}| \le \frac{1}{n}

    It you want

    \frac{1}{n} \le \epsilon \implies  \frac{1}{\epsilon} \le n

    Does this help
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  5. #5
    Super Member craig's Avatar
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    Thankyou that does actually help, not thought about setting it as a "challange" before.
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