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Simple way to get a complex Limit?

  1. Mar 11, 2008 #1
    I'm helping a few people out with school maths and I came across this limit (in connection with the Ratio Test):

    [tex] \lim_{n \to \infty}\left(\frac{n}{n+1}\right)^n [/tex]

    now I can figure out the answer to this myself using l'Hôpital's Rule but the student's I'm teaching wouldn't have come across it yet, so it's either a misprint in their book - or there's a simple way of figuring it out which I can't see. Anyone have any ideas?
     
  2. jcsd
  3. Mar 11, 2008 #2
    could we look at the limit of [tex] (\left(\frac{n}{n+1}\right)^n - \left(\frac{(n-1)}{(n-1)+1}\right)^{n-1} )[/tex] ?

    I mean if [tex]\lim_{n \to \infty} (a_{n} - a_{n-1}) \to \infty [/tex] then doesn't that tell us something?
     
    Last edited: Mar 11, 2008
  4. Mar 11, 2008 #3
    Sorry, I probably shouldn't have mentioned the ratio test at all, I'm just looking to find the limit, which equals [tex] e^{-1} [/tex] which I got by taking the natural log of the term above and applying l'Hôpital's Rule.
     
  5. Mar 11, 2008 #4
    you want to use the squeeze theorem
     
  6. Mar 11, 2008 #5
    OK, any ideas of what two functions to pick?
     
  7. Mar 11, 2008 #6

    Mute

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    Homework Helper

    Surely they have come across the definition of [itex]e[/itex]? If so, it's simple enough:

    [tex]\left(\frac{n}{n+1}\right)^n = \left(\frac{n+1}{n}\right)^{-n} = \left[\left(1 + \frac{1}{n}\right)^n\right]^{-1}[/tex]

    Taking the limit of the last expression and using the definition of [itex]e[/itex] gives [itex]e^{-1}[/itex].

    Is there a reason why they wouldn't be allowed to do it this way?
     
  8. Mar 12, 2008 #7
    mute is correct
     
  9. Mar 12, 2008 #8
    Excellent, yes they would have come across that definition of [itex] e [/itex]. But as for myself on the other hand...let's just say it's been a while.

    Thanks Mute and grmnsplx.
     
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