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Limit of a function help please

  1. Oct 30, 2012 #1
    Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
    So for ε > 0,exists N>0 so that n>N => |x -a|< ε
    What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε

    I know how to keep this going,but is it correct until now?
     
  2. jcsd
  3. Oct 30, 2012 #2

    Mark44

    Staff: Mentor

    The right side should be |[(-1 )^(n-1)]/n^2| < ε
    |(-1)^(n - 1)| = 1, so all you need to do is find n so that 1/n^2 < ε.
    Put parentheses around your exponent. What you wrote, (-1)^n - 1 would be interpreted as
    $$(-1)^n - 1 $$
     
  4. Oct 30, 2012 #3
    Thank you so much! :)
     
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