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Silly limit of a sequence

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    find the limit and prove it using [itex]\epsilon[/itex] and [itex]N(\epsilon)[/itex] defn of limits f sequences, of this sequence:

    [tex]x_n = \sqrt{n^2+n}-n[/tex]

    i don't know why i'm having so much trouble with this but anyway:

    3. The attempt at a solution

    let's say i divined that the limit is 1/2
    we need to find [itex]N(\epsilon)[/itex] such that [itex]n>N(\epsilon)[/itex] implies

    [tex] \big| \frac{1}{2} - \sqrt{n^2+n}-n \big| < \epsilon [/tex]

    well above implies

    [tex] \big| \frac{1}{2} - \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} \sqrt{n^2+n}-n \big| < \epsilon [/tex]

    which implies

    [tex] \big| \frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n} \big| < \epsilon [/tex]

    which implies

    [tex] \big| \frac{1}{2} - \frac{n}{n\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon [/tex]

    which implies

    [tex] \big| \frac{1}{2} - \frac{1}{\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon [/tex]

    which implies

    [tex] \big| \frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}+1} \big| < 2\epsilon [/tex]

    which implies

    [tex] \big| \frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}+1} \big| < \epsilon' [/tex]

    which implies

    [tex] \big| \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}\big| < \epsilon' [/tex]

    but

    [tex] \big| \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{\sqrt{1+\frac{1}{n}}+1}\big| < \big| 1 - \frac{1}{\sqrt{1+\frac{1}{n}}}\big| [/tex]

    and by triangle inequality ( i think this is a mistake )

    [tex] \big| 1 - \frac{1}{\sqrt{1+\frac{1}{n}}+1} \big| < 1 + \big| \frac{1}{\sqrt{1+\frac{1}{n}}} \big| [/tex]

    back to the epsilon, above implies we need to find n such that blah blah:


    [tex] 1 + \big| \frac{1}{\sqrt{1+\frac{1}{n}}} \big| < \epsilon' [/tex]

    which implies

    [tex] 1+ \frac{1}{n} > \frac{1}{(\epsilon - 1)^2}[/tex]

    which implies:

    [tex] n > \frac{ (\epsilon - 1)^2 }{1- (\epsilon - 1)^2} [/tex]

    so is this valid? intuitively it doesn't look immediately wrong to me? the term on the right is positive and increasing as epsilon gets smaller...
     
  2. jcsd
  3. Oct 15, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi ice109! :smile:

    (nice LaTeX, btw, but you can make it even nicer by typing \left| and \right| for big |s :wink:)

    Sorry, I lost interest less than half-way through. :redface:

    Why not use the same technique at the start, by putting √(n2 + n) = n√(1 + 1/n)? :smile:

    (or maybe compare it with yn = √(n2 + n + 1/4) - n :wink:)
     
  4. Oct 15, 2009 #3
    if you hadn't lost interest you would have noticed i did do the first thing.

    and your hint shows that my expression is less than 1/2 but it doesn't help me show that my sequence converges to 1/2 but that's cause i don't know quite how to use it to help me show that.
     
  5. Oct 15, 2009 #4
    I've lost interest as well, but I think your:

    [tex]
    \big| \frac{1}{2} - \sqrt{n^2+n}-n \big| < \epsilon
    [/tex]

    should be:

    [tex]
    \big| \frac{1}{2} - \left( \sqrt{n^2+n}-n \right) \big| < \epsilon
    [/tex]
     
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