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Math Help - Limits approaching infinity

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    Limits approaching infinity

    If b>0 prove that limn->infinity 1/1+nb=0 using the limit definition.

    I can't seem to make any headway in choosing my delta to make progress in this exercise.
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    Re: Limits approaching infinity

    Quote Originally Posted by renolovexoxo View Post
    If b>0 prove that limn->infinity 1/1+nb=0 using the limit definition.
    Suppose \epsilon >0 now we want \left|\frac{1}{1+nb}-0\right|=\frac{1}{1+nb}< \epsilon.

    So make n\ge N>\frac{1-\epsilon}{b \epsilon}
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    Re: Limits approaching infinity

    what we need to do is show that for ANY ε > 0 (even, or perhaps especially the very tiny ones), we can find SOME positive integer N, such that:

    for ALL n > N:

    \left|\frac{1}{1+nb} - 0\right| < \epsilon

    note we're not using "delta" because n isn't tending to a certain definite real number (|x - ∞| < δ doesn't make any sense). instead, we're using the idea that "close to infinity" means "really big".

    we expect that if ε is very small, N should be quite large.

    now:

    \left|\frac{1}{1+nb} - 0\right| = \left|\frac{1}{1+nb}\right| = \frac{1}{1+nb} (since we can insist n > 0 (we are, after all, going "all the way positive to infinity") and we know that b > 0).

    if we want:

    \frac{1}{1+nb} < \epsilon, then we want:

    1 < (1+nb)\epsilon, so:

    \frac{1}{\epsilon} < 1 + nb \iff \frac{1}{\epsilon} - 1 < nb \iff \left(\frac{1}{b}\right)\left(\frac{1}{\epsilon} - 1\right) < n

    this suggests we pick N > \left(\frac{1}{b}\right)\left(\frac{1}{\epsilon} - 1\right), or N = 1 (just to keep N positive).

    so, if we choose:

    N = \max\left(1,\frac{\1-\epsilon}{b\epsilon}\right), we have for all n > N:

    \left|\frac{1}{1+nb}} - 0\right| = \frac{1}{1+nb} < \frac{1}{1+Nb} < \frac{1}{1 + \left(\frac{1-\epsilon}{b\epsilon}\right)b}

    =\frac{1}{\frac{\epsilon + 1 - \epsilon}{\epsilon}} = \frac{1}{\frac{1}{\epsilon}} = \epsilon, as desired.

    let's see how this works for a specific ε, and a specific b (we'll actually find the N that works).

    suppose b = 2, and ε = 0.1 or 1/10.

    according to what we did above, the integer N we want is:

    N > (1/2)(9/10)(10) = 9/2, so we should pick N = 5.

    note that 1/(1 + 2*4) = 1/9 > 1/10, so N = 4 doesn't work.

    now if n ≥ 5:

    1/(1 + 2n) < 1/11 < 1/10.
    Last edited by Deveno; October 2nd 2012 at 02:10 PM.
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