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Math question here limit please?

  1. Oct 30, 2012 #1
    Prove that the limit when x--> infinite of (2n+1)/(n+1) =2
    So for ε > 0,exists N>0 so that n>N => |x -a|< ε
    What I do is | (2n+1)/(n+1) |< ε, I do the math actions and I have |-1/(n+1)| < ε..... NOW,what I dont get,when I remove the absolute value,do I get 1/(n+1)<ε or NOT?
     
  2. jcsd
  3. Oct 30, 2012 #2

    Zondrina

    User Avatar
    Homework Helper

    I believe you want to use this definition :

    [itex]\forall \epsilon>0, \exists N | n>N \Rightarrow |a_n - L| < \epsilon[/itex]

    So what is [itex]|a_n - L|[/itex]? Plug in your info and start massaging it into a suitable expression.
     
  4. Oct 30, 2012 #3
    I know what to do,I just want to know if the part when I remove the absolute value is correct :)
     
  5. Oct 30, 2012 #4

    Mark44

    Staff: Mentor

    This is incorrect. You should start with
    |(2n + 1)/(n + 1) - 2| < ε

    It looks like the above is what you were working with, but didn't write it correctly.
    Yes.
     
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