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Epsilon proof trouble

  1. Sep 28, 2008 #1
    [tex]Suppose $\lim_n \frac{a_n -1}{a_n +1} = 0$. Prove that $\lim_n a_n = 1$.[/tex]

    I am trying to do the algebra so that -a_n < ?? < a_n , but I am having trouble. Am I going about this correctly?

    I have also tried to solve each separate side of the inequality. I get a_n < (e+1)/(1-e), but this is not quite fitting.

    Can somebody give me a clue please. Thanks

    Edit: There is a hint in the book that says to set,

    [tex]\(x_{n}=\frac{a_n-1}{a_n+1}\)[/tex] and then solve.

    I have done this but I don't know what to do next.
    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2


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    I presume the hint said set [itex]x_n= (a_n-1)/(a_n+ 1)[/itex] and solve for xn. When you did that what did you get? What is the limit of that as x goes to 0?
  4. Sep 28, 2008 #3
    I solved for a_n and I then took the lim of both sides so that

    lim a_n = lim ( -x_n -1 ) / (x_n -1 ), then it was pretty straightforward.

    Assuming I can take the lim of both sides. Can I do that?
  5. Sep 28, 2008 #4


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    Yes, of course. Just use the basic "rules" for limits. If the limits of an and bn exist, and the limit of bn is not 0, then the limit of an/bn exists.
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