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Proving a sequence has the given limit.

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that the following sequence has the following limit:

    [tex]\lim_{n\rightarrow \infty} {{{\frac{n}{n^2+3n+1}}} = 0 [/tex]


    2. Relevant equations
    N/A


    3. The attempt at a solution

    First I did the following:

    [tex]{\frac{n}{n^2+3n+1} < {\frac{n}{n^2+3n} = {\frac{n}{n(n+3)} = {\frac{1}{n+3}[/tex]

    Then:
    [tex]{\frac{1}{n+3} < \epsilon[/tex]

    [tex]{{n+3} > \frac{1}{\epsilon}[/tex]

    [tex]{{n} > \frac{1}{\epsilon}-3[/tex]


    Putting it all together:

    Let [tex]\epsilon > 0[/tex] be given

    Let [tex]{{n} > \frac{1}{\epsilon}-3[/tex]

    Then: [tex]{\left|\frac{n}{n^2+3n+1}\right| - 0 = \frac{n}{n^2+3n+1} < \frac{n}{n^2+3n} = \frac{1}{n+3} < {\epsilon }[/tex]

    Therefore [tex]\left|\frac{n}{n^2+3n+1}-0\right| < \epsilon [/tex]
    QED

    Am I going about this the right way?
     
    Last edited: Feb 18, 2010
  2. jcsd
  3. Feb 19, 2010 #2
    Any help would be appreciated. Thanks. :)
     
  4. Feb 19, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since [itex]\epsilon-3< 1/\epsilon[/itex] it would be sufficient to take [itex]n> 1/\epsilon[/itex].

    Other than that, it looks good to me.

     
  5. Feb 19, 2010 #4
    I don't follow that. How did you get [itex]\epsilon-3< 1/\epsilon[/itex] and why can you take [itex]n> 1/\epsilon[/itex] from it?
     
  6. Feb 19, 2010 #5

    Mark44

    Staff: Mentor

    I think HallsOfIvy meant [tex]1/\epsilon - 3 < 1/\epsilon[/tex]
     
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