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I Understanding limits

  1. Feb 6, 2017 #1
    I'm trying to understand why the ## \lim_{n \rightarrow \infty}
    ( \displaystyle \frac {n+1}{n-1} )
    ## equals the indeterminate form 1?

    I ask because we have started going over sequences and it was used as an example. I understand how to go from here- taking the ln of both sides and using L'Hopital's rule to get the limit of e2.
     
  2. jcsd
  3. Feb 6, 2017 #2

    pwsnafu

    User Avatar
    Science Advisor

    Could you write down what you teacher said in full. I have no idea what is going on.

    The standard way to evaluate the limit is just observe ##\frac{n+1}{n-1} = \frac{n-1+2}{n-1} = 1 + \frac2{n-1}## and then it becomes obvious.

    Edit: alternative ##\frac{n+1}{n-1} = \frac{1+n^{-1}}{1-n^{-1}}## and then apply limit laws. Again the limit is 1.
     
  4. Feb 6, 2017 #3
    We were discussing using L'Hopital's rule for indeterminate forms to evaluate the sequence below. I just don't understand how that limit is
    1. I guess I'm still fuzzy on how to figure out limits?

    20170206_225627.png
     
  5. Feb 6, 2017 #4

    pwsnafu

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    Science Advisor

    Oh, the limit you want to evalute is ##\left(\frac{n+1}{n-1}\right)^{n}## and not ##\frac{n+1}{n-1}##. Yes that's ##1^\infty## indeterminate form.
    This is what happens when students do it naively:

    ##\lim_{n\to\infty}\left(\frac{n+1}{n-1}\right)^{n} = \left(\lim_{n\to\infty}\frac{n+1}{n-1}\right)^{\lim_{n\to\infty} n} = \left(1+\lim_{n\to\infty}\frac2{n-1}\right)^{\lim_{n\to\infty} n} = 1^\infty##.
     
  6. Feb 6, 2017 #5

    Thank you so much for the explanation!!!!
     
  7. Feb 7, 2017 #6

    Mark44

    Staff: Mentor

    It doesn't. The limit here, which apparently isn't the one you really meant to ask about, is 1.
    ##\lim_{n \to \infty}\frac{n + 1}{n - 1} = \lim_{n \to \infty}\frac n n \frac{1 + 1/n}{1 - 1/n} = 1##
    For any finite value of n, n/n is 1, and as n gets larger, the other fraction approaches 1 in value, making the limit equal to 1.
     
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