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Limit of a certain function of n as n goes to infinity

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\lim_{n->\infty} 3(\frac{n}{n+1})^n[/itex]


    3. The attempt at a solution

    Ok, I know that the answer is 3/e, because this limit was solved a year ago when I took calculus 1 by my teacher, and I foolishly copied only the answer, thinking I would never forget and have to go back.

    I can't for the life of me understand how he did that, can someone help me, please?

    Thanks in advance!
     
    Last edited: Jan 27, 2013
  2. jcsd
  3. Jan 27, 2013 #2

    mfb

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    $$\frac{n}{n+1} = 1-\frac{1}{n+1}$$
    Shifting the index by 1 (and ignoring the 3), the limit becomes $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{-1}$$ which is easy to study.
     
  4. Jan 27, 2013 #3

    jbunniii

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    Another way to look at it:
    $$3 \left(\frac{n}{n+1}\right)^n = \frac{3}{\left(\frac{n+1}{n}\right)^n} = \frac{3}{\left(1 + \frac{1}{n}\right)^n}$$
    The expression in the denominator on the right hand side should look familiar.
     
  5. Jan 27, 2013 #4
    Oh yeah, of course! It's a fundamental limit, Thanks to you both, but I have to say that jbunniii's representation made it perfectly clear.
     
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