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Limits/Absolute_Value/Inequalities Proof

  1. Aug 31, 2011 #1
    Hi folks. We're working towards understanding limits in my math class and, as part of that, here's one of our exercises:

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

    Find a natural number N so that absValue( (2n+3)/(4n+5)-1/2 )<.01 for all n > N

    2. Relevant equations

    3. The attempt at a solution

    After some fiddling around with the absolute value, I have very good reasons to suspect that absValue( (2x+3)/(4x+5)-1/2 )<.01 for all real numbers x greater than or equal to 11.25. So, that would make N 11 assuming n is also a natural number.

    By substituting 12 (the next natural number greater than N) for n in the original expression absValue( (2n+3)/(4n+5)-1/2 ), it simplifies to ~.0094 which is obviously less than .01.

    It is fairly easy to show that the terms of a sequence given by (2n+3)/(4n+5) are strictly decreasing, so it makes sense then that if ( (2n+3)/(4n+5)-1/2 )<.01 then the same will be true for n+1 and n+2 and so on.

    We also know that the terms of the sequence given by (2n+3)/(4n+5) are always positive.

    So, at this point I know these three things:

    1. When n=12, absValue( (2n+3)/(4n+5)-1/2 )<.01
    2. The sequence where the nth term is given by (2n+3)/(4n+5) is strictly decreasing
    3. The sequence where the nth term is given by (2n+3)/(4n+5) is always positive

    But I don't think this is enough to prove that absValue( (2n+3)/(4n+5)-1/2 )<.01 for all n > 12

    I've been thinking about using induction to prove it, but I'm not sure how to go about doing so.
  2. jcsd
  3. Aug 31, 2011 #2


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    You could prove that if an > 0, an+1 ≤ an, and aN ≤ c for some positive integer N, then an ≤ c for all N ≥ n by induction. But that's getting pretty pedantic. If you are in a first year Calculus class learning limits I would guess you could just state it as obvious.
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