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Proving Limit with Epsilon and Delta

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove the following sequence {an} converges to L=1/2
    an = n2/(2n2+n-1)

    3. The attempt at a solution
    Given ε>0 we can determine an N∈N so that |an - L|<ε for n≥N. We have:
    |an-L|=|(n2/(2n2+n-1)-(1/2)|
    = |(-n+1)/(2(2n-1)(n+1))|

    I'm not sure what to do once I get to this point. Any help would be great!
     
  2. jcsd
  3. Mar 27, 2014 #2

    LCKurtz

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    If you change the numerator to (+n+1) you can get that fraction to be less or equal to$$
    \frac 1 {2(2n-1)}$$What happens if you drop the ##-1## too?

    [Edit] Ignore the drop the -1 part. Maybe change the -1 to -n?
     
    Last edited: Mar 27, 2014
  4. Mar 27, 2014 #3

    vela

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    In a similar vein, you might want to show that ##0<\frac{n-1}{2(2n-1)}<1## for n>1.
     
  5. Mar 28, 2014 #4

    PeroK

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    The point behind these ideas is that it can be difficult to find N for a given ε unless the expression is quite simple, so you estimate the expression you have with something simpler.

    E.g. suppose you wanted to show that 1/(n^2 + 6) -> 0. One way to do this is to note that:

    0 < 1/(n^2+6) < 1/n

    So, you can take N = 1/ε. Which is much simpler than trying to find a more precise N for the original sequence.

    For your example, you want to find a simple expression that is greater than the one you have but still -> 0. That's the strategy, anyway. (See Vela's idea for how to do this.)
     
  6. Mar 29, 2014 #5
    Thanks for the help! Does this look right?

    Given ε>0 we can determine an N[itex]\in[/itex]N so that |an-L|<ε for n≥N. We have:
    |an-L|=|[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|=|[itex]\frac{2n^{2}}{2(2n^{2}+n-1}[/itex]-[itex]\frac{2n^{2}+n-1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n^{2}+n-1}[/itex]|=|[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<ε.
    Now take [itex]\frac{n-1}{2(2n-1)}[/itex]. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 [itex]\Rightarrow[/itex] 0<[itex]\frac{n-1}{2(2n-1)}[/itex]<1 when n>1. As long as we have n>1, we have that |[itex]\frac{-n+1}{2(2n-1)(n+1)}[/itex]|<[itex]\frac{n-1}{2(2n-1)}[/itex]. We want to bound this by ε>0. Take ε>0. If we choose N to be the smallest integer greater than or equal to max{1,1/ε} then we have that, for n≥N:
    |[itex]\frac{n^{2}}{2n^{2}+n-1}[/itex]-[itex]\frac{1}{2}[/itex]|<[itex]\frac{1}{n}[/itex]<[itex]\frac{1}{N}[/itex]≤ε
    where the last line follows directly if N≥1/ε and is also satisfied for N=1>1/ε since this implies 1/N<ε.
     
  7. Mar 29, 2014 #6

    vela

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    Is that last inequality what you intended?

     
  8. Mar 29, 2014 #7
    What is wrong with that inequality?
     
  9. Mar 30, 2014 #8

    vela

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    How are you going from
    to
     
  10. Mar 31, 2014 #9
    n-1< 4n -2 => 1<3n which is true for n>1 get it?
     
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