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Real Analysis Limit Proof

  1. May 14, 2013 #1
    I am trying to prove that [tex] \lim\left[\sqrt{n^2+n}-n\right]=\frac{1}{2}[/tex]
    Where [itex] n \in \mathbb{N} [/itex] and [itex]\lim[/itex] is the limit of a sequence as [itex]n\to\infty[/itex].

    From the definition of a limit, I know that I need to show that [tex]\exists{N}:n>N\Rightarrow\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon[/tex]

    Through algebraic manipulation I was able to arrive at
    [tex] \left| \sqrt{n^2+n} - n - \frac{1}{2} \right| = \left| \frac{n-\sqrt{n^2+n}}{n+ \sqrt{n^2+n}} \right|[/tex]

    Which now means that I must show that
    [tex]\exists N : n>N \Rightarrow \left| \frac{n-\sqrt{n^2+n}}{n+ \sqrt{n^2+n}} \right| < 2\epsilon[/tex]

    The problem here is I'm not sure how this makes the proof any easier. It seems like now I must start with the assumption that this part is less than [itex]2\cdot\epsilon[/itex] but I don't see how to isolate n in this inequality. Any nudge in the right direction (no spoilers, please!) would be greatly appreciated.
     
    Last edited: May 14, 2013
  2. jcsd
  3. May 14, 2013 #2
    ...Limit of what? You need to define what your limit is doing.
     
  4. May 14, 2013 #3
    Fixed.
     
  5. May 14, 2013 #4

    LCKurtz

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    I would try dividing numerator and denominator of the left side by ##\sqrt{n^2+n}## and see if that helps. Maybe you can get all the ##n##'s under the square root.

    [Edit] Apparently you solved it while I was typing.
     
  6. May 14, 2013 #5
    Nope, still haven't solved this one. Came very close, filled up my hand-held whiteboard with stuff, but it fell apart at the last minute because of one little error at the beginning. I'm still working on it now.
     
  7. May 14, 2013 #6
    I think I have [itex] \left| \sqrt{n^2+n} - n - \frac{1}{2} \right| < 2\epsilon \Rightarrow \left( \sqrt{n+1} - \sqrt{n} \right)^2 < 2\epsilon [/itex] but now I'm not sure where to go from here, or if this is even a useful place to be at. I get a nagging feeling that I am missing something very simple and elegant at the beginning of this problem. I have spent HOURS on this today...starting to get pretty demoralized. Not sure how hard this is "supposed" to be.
     
  8. May 14, 2013 #7

    Dick

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    I would divide numerator and denominator by n. The denominator will be greater than 1, so don't worry about it. Just try and make the numerator small.
     
  9. May 14, 2013 #8
    Okay, I think what you're saying is this: [itex] \left| \frac{n - \sqrt{n^2+n}}{\sqrt{n^2+n}+n} \right| = \left| \frac{1- \sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} \right| [/itex]. The problem for me here is that I don't see how to get an inequality of the form [itex]n > f(\epsilon)[/itex]. Keep in mind I am not using the limit theorems in this problem, this is an epsilon proof, also each algebraic step must be logically reversible, i.e. each implication must be bidirectional.
     
  10. May 14, 2013 #9

    Dick

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    Just concentrate on the numerator. How large does n have to be to make ##\sqrt{1+\frac{1}{n}}-1 \lt \epsilon##?
     
  11. May 14, 2013 #10
    Alright I have
    [tex] \sqrt{1+\frac{1}{n}} - 1 < \epsilon[/tex]
    [tex] \Rightarrow \sqrt{1+\frac{1}{n}} < \epsilon + 1[/tex]
    [tex]\Rightarrow 1 + \frac{1}{n} < \left( \epsilon + 1 \right) ^2 [/tex]
    [tex]\Rightarrow \frac{1}{n} < \left(\epsilon + 1 \right)^2 - 1 [/tex]
    [tex]\Rightarrow n > \frac{1}{\left(\epsilon + 1 \right)^2 - 1} [/tex]

    I don't understand why I can forget about the denominator.

    Edit: Oh! If something is less than epsilon, then something smaller than that is also less than epsilon! Is that the idea?
     
    Last edited: May 14, 2013
  12. May 14, 2013 #11

    Dick

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    The denominator is greater than 2. If the numerator is less than ε, so is the fraction. ##a < \epsilon \Rightarrow \frac{a}{b} < \epsilon## if b>2.
     
  13. May 14, 2013 #12

    Dick

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    Of course!
     
  14. May 14, 2013 #13
    Ok I think I'm with you. I was getting mixed up over the reversibility thing, but what your saying makes perfect sense to me now. Thank you! I will try to knock out the actual proof now.
     
  15. May 14, 2013 #14
    It's solid! Thank you so much!
     
  16. May 24, 2013 #15
    An alternate idea. Let an = [itex]\sqrt{n^2 + n}[/itex] and bn = [itex]\frac{1}{2} +n[/itex] Then bn2 - an2 = (bn - an)(bn + an) = 1/4.

    As limn->∞(bn + an) = ∞ as we must have limn->∞(bn - an) = 0.
     
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