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Finding the limit as n -> inf

  1. Jan 3, 2012 #1
    302xk61.png

    I believe the second is simply 1, as I can ignore a here.

    Not sure about the first, I believe it tends to 0 because of the power of n, and (n/n+1) < 1

    Any help appreciated, thanks!
     
  2. jcsd
  3. Jan 3, 2012 #2
    For the first:
    First of all, you need to fiddle around with the expression so that you get something you recognize. Now

    n/(n+1) = 1 / ( (n+1)/n ) = 1 / ( (1+1/n) ),

    so if you plug this in the original you get

    ( 1 / ( (1+1/n) ) )^n = 1 /( ( (1+1/n) ) )^n

    and now you should recognize the limit of this.
     
  4. Jan 3, 2012 #3
    Looks like 0 to me, but only because 1+1/n > 1 and the power of n

    Which is the same reasoning as I had before.. so I'm still lost!
     
  5. Jan 3, 2012 #4

    SammyS

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    Staff Emeritus
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    In the Calculus & Beyond section, you should recognize:
    [itex]\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\,.[/itex]​

    If not, try increasingly larger numbers for n in your calculator: 1 , 2 , 3 , 5 , 10 , 1000 , 1000000 , ...
     
  6. Jan 3, 2012 #5
    Have you encountered the limit of
    (1 + 1/n)^n
    before? I'm sure it has been mentioned somewhere in your study material. The limit is Euler's number [itex]e\ =\ 2.7182...[/itex].
     
  7. Jan 3, 2012 #6

    Mark44

    Staff: Mentor

    The first limit is an example of the indeterminate form [1].
     
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